1.1 It could be considered either way. On the one hand, the deficiency of the enzyme G6PD is hereditary; on the other hand, the disease itself has an environmental trigger (eating broad beans). Both answers are too simple, because the disease actually results from an interaction of a particular genetic constitution with a particular factor in the environment.
1.2 Replication results in two daughter DNA duplexes, each identical in base sequence to the parental molecule (except for possible mutations in the sequence). Each of the daughter molecules contains one of the original intact parental strands.
1.3 The messenger RNA carries the genetic information in DNA (in the form of a specific base sequence) to the ribosome. The ribosome is the physical structure in the cell on which a polypeptide chain is formed. The transfer RNA molecules are adapters that enable each codon in the messenger RNA to specify the presence of a particular amino acid at the corresponding position in the polypeptide chain. There is only one type of ribosome, but there are many different types of transfer RNA (typically, more than one for each amino acid).
1.4 A mixture of heat-killed S cells and living R cells causes pneumonia in mice, but neither heat-killed S cells nor living R cells alone do so.
1.5 The substance was destroyed by DNA-degrading enzymes but not by protein-degrading enzymes.
1.6 The inside of the head is mainly DNA, and the rest of the phage is mainly protein. When a bacterial cell is infected, the head contents are transferred into the cell, but the outer "ghost" of the phage remains attached to the external surface of the cell.
1.7 A small amount of protein was also transferred into the bacterial cell during infection, and some of this was even recovered in the progeny phage. A die-hard advocate of protein as the genetic material could claim that the transmitted protein, though small in amount, was critical in the hereditary process.
1.8 Because the amount of A equals that of T, and the amount of G equals that of C, it can be inferred that the DNA in this bacteriophage is double-stranded; 56 percent of the base pairs are A-T pairs, and 44 percent are G-C pairs.
1.9 Because, in this case, the amount of A does not equal that of T, and the amount of G does not equal that of C, it can be concluded that the DNA molecule is single-stranded, not double-stranded.
1.10 Because A pairs with T, and G pairs with C, the base composition of the other strand must be 24 percent T, 28 percent A, 22 percent C, and 26 percent G.
1.11 Yes, it is possible, because in some viruses the genetic material is RNA.
1.12 Because of A-T and G-C base pairing, the complementary strand has the sequence
Note that the paired strands have opposite 5'-to-3' polarity.
1.13 In this region the complementary strand has the sequence
1.14 The probability that four particular bases have the sequence 5'-GGCC-3' is (1/4)4 = 1/256, so 256 base pairs is the average spacing between consecutive occurrences; similarly, the average spacing for the sequence 5'-GAATTC- 3' is 4096 base pairs.
1.15 The corresponding region of RNA will contain no U (uracil), because U in RNA pairs with A in DNA.
1.16 The top strand is the RNA because it contains U instead of T; the mismatched base pair is the sixth from the
1.17 The initial part of the RNA transcript matches the DNA template starting at the tenth nucleotide. Hence the completed transcript has the sequence
1.18 The complementary sequence is 3'-UAUGCUAU-5'.
1.19 The codon for phenylalanine must be 5'-UUU-3'.
1.20 The codon for leucine must be 5'-UUA-3'.
1.21 Three with in vitro translation:
5'-CGC/UUA/CCA/CAU/GUC/GCG/AAC/UCG-3' 5'-C/GCU/UAC/CAC/AUG/UCG/CGA/ACU/CG-3' 5'-CG/CUU/ACC/ACA/UGU/CGC/GAA/CUC/G-3'
With in vivo translation, there is only one
5'-CGCUUACCAC > AUG/UCG/CGA/ACU/CG-3'
where the symbol > means "start translation with next codon."
1.22 Six. Either DNA strand could be transcribed, and each transcript could be translated in any one of three reading frames.
1.23 The amino acids alternate, because with a triplet code, the codons alternate: 5'-
UCU/CUC/UCU/CUC/UCU/CUC-3'. From this result we can conclude that Ser and Leu are encoded by 5'-UCU-3' and 5'-CUC-3'; we cannot specify which codon corresponds to which amino acid, because in vitro translation begins with either 5'-UCU-3' or 5'-CUC-3'.
1.24 The result means that an mRNA is translated in nonoverlapping groups of three nucleotides: the genetic code is a triplet code.
1.25 There are 64 possible codons and only 20 amino acids; hence some amino acids (most, in fact) are specified by two or more codons. A mutation that changes a codon for a particular amino acid into a synonymous codon for the same amino acid does not change the amino acid sequence.
2.1 The strain or variety must be homozygous for all genes that affect the trait.
2.2 Aa yields A and a gametes, Bb yields B and b gametes, and Aa Bb yields A B, A b, a B, and a b gametes.
2.3 Multiply the number of possible gametes formed for each gene (one for each homozygous gene and two for each heterozygous gene). In the case of AA Bb Cc Dd Ee, there are a total of 1 x2 x2 x2 x2 = 16 possible gametes. With n homozygous genes and m heterozygous genes, the number of possible gametes is 1" x 2m, which equals 2m.
2.4 What Mendel apparently means is that the gametes consist of equal numbers of all possible combinations of the alleles present in the true-breeding parents; in other words, the cross AA BB x aa bb produces F1 progeny of genotype Aa Bb, which yields the gametes A B, A b, a B, and a b in equal numbers. Segregation is illustrated by the 1 : 1 ratio of A : a and B : b gametes, and independent assortment is illustrated by the equal numbers of A B, A b, a B, and a b gametes.
2.5 The round seeds in the F2 generation consist of 1/3 AA and 2/3 Aa. Only the latter produce a-bearing pollen, so the fraction of aa seeds expected is 2/3 x 1/2 = 1/3.
2.6 With dominance, there are two phenotypes, corresponding to the genotypes RR or Rr (dominant) and rr (recessive), in the ratio 3 : 1. With no dominance, there are three phenotypes, corresponding to the genotypes RR, Rr, and rr, in the ratio 1 : 2 : 1.
2.7 The production of a homozygous genotype would require that the pollen and egg contain identical self-sterility alleles, but this is impossible because pollen cannot function on plants that contain the same self-sterility allele.
2.8 1/2; 1/2. The probability argument is that each birth is independent of all the previous ones, so the number of girls and boys already born has no influence on the sexes of future children.
2.9 Because the probability of each child's being a girl is 1/2, and the two children are independent, the probability of two girls in a row is (1/2) x (1/2) = 1/4. The probability of a girl and a boy (not necessarily in that order) equals 2 x (1/4) x (1/4) = 1/2. The reason for the factor of 2 in this case is that the sexes may occur in either of two possible orders: girl-boy or boy-girl. Each of these has a probability of 1/4, so the total is 1/4 + 1/4 = 1/2.
2.10 (a) The parent with the dominant phenotype must carry one copy of the recessive allele and hence must be heterozygous. (b) Because some of the progeny are homozygous recessive, both parents must be heterozygous. (c) The parent with the dominant phenotype could be either homozygous dominant or heterozygous; the occurrence of no homozygous recessive offspring encourages the suspicion that the parent may be homozygous dominant, but because there are only two offspring, heterozygosity cannot be ruled out.
2.11 An Ab gamete can be formed only if the parent is AA Bb, which has probability 1/2, and in this case, 1/2 of the gametes are Ab; overall, the probability of an Ab gamete is (1/2) x (1/2) = 1/4. AB gametes derive from AA BB parents with probability 1 and from AA Bb parents with probability 1/2; overall, the probability of an AB gamete is (1/2) x 1 + (1/2) x (1/2) = 3/4. (Alternatively, the probability of an AB gamete may be calculated as 1 - 1/4 = 3/4,
because Ab and AB are the only possibilities.)
2.12 (a) Two phenotypic classes are expected for the A, a pair of alleles (A- and aa), two for the B, b pair (B- and bb), and three for the R, r pair (RR, Rr, and rr), yielding a total number of phenotypic classes of 2 x 2 x 3 = 12. (b) The probability of an aa bb RR offspring is 1/4 (aa) x 1/4 (bb) x 1/4 (RR) = 1/64. (c) Homozygosity may occur for either allele of each of the three genes, yielding such combinations as AA BB rr, aa bb RR, AA bb rr, and so forth. Because the probability of homozygosity for either allele is 1/2 for each gene, the proportion expected to be homozygous for all three genes is (1/2) x (1/2) x (1/2) = 1/8.
2.13 The probability of a heterozygous genotype for any one of the genes is 1/2, and so for all four together, the probability is (1/2)4= 1/16.
2.14 Because both parents have solid coats but produce some spotted offspring, they must be Ss. With respect to the A, a pair of alleles, the female parent (tan) is aa, and because there are some tan offspring, the genotype of the black male parent must be Aa. Thus the parental genotypes are Ss aa (solid tan female) and Ss Aa (solid black male).
2.15 Because one of the children is deaf (genotype dd, in which d is the recessive allele), both parents must be heterozygous Dd. The progeny genotypes expected from the mating Dd x Dd are 1/4 DD, 2/4 Dd, and 1/4 dd. The son is not deaf and hence cannot be dd. Among the nondeaf offspring, the genotypes DD and Dd are in the proportions 1 : 2, so their relative probabilities are 1/3 DD and 2/3 Dd. Therefore, the probability that the normal son is heterozygous is 2/3.
2.16 (a) Because the trait is rare, it is reasonable to assume that the affected father is heterozygous HD/hd, where hd represents the normal allele. Half of his gametes contain the HD allele, so the probability is 1/2 that the son received the allele and will later develop the disorder. (b) We do not know whether the son is heterozygous HD/hd, but the probability is 1/2 that he is; if he is heterozygous, half of his gametes will contain the HD allele. Therefore, the overall probability that his child has the HD allele is (1/2) x (1/2) = 1/4.
2.17 Both parents must be heterozygous (Aa) because each had an albino (aa) parent. Therefore, the probability of an albino child is 1/4, and the probability of two homozygous recessive children is (1/4) x (1/4) = 1/16. The probability that at least one child is an albino is the probability that exactly one is albino plus the probability that both are albinos. The probability that exactly one is an albino is 2 x (3/4) x (1/4) = 6/16, in which the factor of 2 comes from the fact that there may be two birth orders: normal-albino or albino-normal. Therefore, the overall probability of at least one albino child equals 1/16 + 6/16 = 7/16. Alternatively, the probability of at least one albino child may be calculated as 1 minus the probability that both are nonalbino, or 1 - (3/4)2 = 7/16.
2.18 Compatible transfusions are A donor with A or AB recipient, B donor with B or AB recipient, AB donor with AB recipient, and O donor with any recipient.
2.19 (a) The cross RR BB x rr bb yields F1 progeny of genotype Rr Bb, which have red kernels. (b) Because there is independent segregation (independent assortment), the F2 genotypes are R-B-, R-bb, rr B-, and rr bb. These genotypes are in the proportions 9 : 3 : 3 : 1, and they produce kernels that are red, brown, brown, and white, respectively. Therefore, the F2 plants have red, brown, or white seeds in the proportions 9/16 : 6/16 : 1/16.
2.20 Because the genes segregate independently, they can be considered separately. For the Cr, cr pair of alleles, the genotypes of the zygotes are Cp Cp. Cp cp, and cp cp in the proportions 1/4, 1/2, and 1/4, respectively. However, the Cp Cp zygotes do not survive, so among the survivors the genotypes are Cp cp (creeper) and cp cp (noncreeper), in the proportions 2/3 and 1/3, respectively. For the W, w pair of alleles, the progeny genotypes are W- (white) and ww (yellow), in the proportions 3/4 and 1/4, respectively. Altogether, the phenotypes and proportions of the surviving offspring can be obtained by multiplying (2/3 creeper + 1/3 noncreeper) x (3/4 white + 1/4 yellow), which yields 6/12 creeper, white + 2/12 creeper, yellow + 3/12 noncreeper, white + 1/12 noncreeper, yellow. Note that the proportions sum to 12/12 = 1, which serves as a check.
2.21 Colored offspring must have the genotype C- ii; otherwise, color could not be produced (as in cc) or would be suppressed (as in I-). Among the F2 progeny, 3/4 of the offspring have the C- genotype, and 1/4 of the offspring have the ii genotype. Because the genes undergo independent assortment, the overall expected proportion of colored offspring is (3/4) x (1/4) = 3/16.
2.22 The 9 : 3 : 3 : 1 ratio is that of the genotypes A- B-, A- bb, aa B-, and aa bb. The modified 9 : 7 ratio implies that all of the last three genotypes have the same phenotype. The F1 testcross is between the genotypes Aa Bb x aa bb, and the progeny are expected in the proportions 1/4 Aa Bb, 1/4 Aa bb, 1/4 aa Bb, and 1/4 aa bb. Again, the last three genotypes have the same phenotype, so the ratio of phenotypes among progeny of the testcross is 1 : 3.
2.23 For the child to be affected, both III-1 and III-2 must be heterozygous Aa. For III-1 to be Aa, individual II-2 must be Aa and must transmit the recessive allele to III-1. The probability that II-2 is Aa equals 2/3, and then the probability of transmitting the a allele to III-1 is 1/2. Altogether, the probability that III-1 is a carrier equals (2/3) x (1/2) = 1/3. This is also the probability that III-2 is a carrier. Given that both III-1 and III-2 have genotype Aa, the probability of an aa offspring is 1/4. Overall, the probability that both III-1 and III-2 are carriers and that they have an aa child is (1/3) x (1/3) x (1/4) = 1/36. The numbers are multiplied because each of the events is independent. (The reason for the 2/3 is as follows: Because II-3 is affected, the genotypes of I-1 and I-2 must be Aa. Among nonaffected individuals from this mating (individuals II-2 and II-4), the ratios of AA and Aa are 1/4 : 2/4, so the probability of Aa is 2/3.)
2.24 For any individual offspring, the probabilities of black B- and white bb are 3/4 and 1/4, respectively, (a) The order white-black-white occurs with probability (1/4) x (3/4) x (1/4) = 3/64. The order black-white-black occurs
with probability (3/4) x (1/4) x (3/4) = 9/64. Therefore, the probability of either the order white-black-white or the order black-white-black equals 3/64 + 9/64 = 3/16. The probabilities are summed because the events are mutually exclusive. (b) The probability of exactly two white and one black equals 3 x (1/4)2 x (3/4)1 = 9/64. The factor of 3 is the number of birth orders of two white and one black (there are three, because the black offspring must be the first, second, or third), and the rest of the expression is the probability that any one of the birth orders occurs (in this case, 3/64).
2.25 The probability of all boys is (1/2)4 = 1/16. The probability of all girls is also 1/16, so the probability of either all boys or all girls equals 1/16 + 1/16 = 1/8. The probability of equal numbers of the two sexes is 6 x (1/2)4 = 3/8. The factor of 6 is the number of possible birth orders of two boys and two girls (BBGG, BGBG, BGGB, GGBB, GBGB, and GBBG).
2.26 Black and splashed white are the homozygous genotypes and slate blue the heterozygote. The probabilities of each of these phenotypes from the mating between two heterozygotes are 1/4 black, 1/2 slate blue, and 1/4 splashed white. The probability of occurrence of the particular birth order black, slate blue, and splashed white is (1/4) x (1/2) x (1/4) = 1/32, but altogether there are six different orders in which exactly one of each type could be produced. (The black offspring could be first, second, or third, and in each case, the remaining phenotypes could occur in either of two orders.) Consequently, the overall probability of one of each phenotype, in any order, is 6 x (1/32) = 3/16.
2.27 The probability that one or more offspring are aa equals 1 minus the probability that all are A-. In the mating Aa x Aa, the probability that all of n offspring will be Aa equals (3/4)n, and the question asks for the value of n such that 1 - (3/4)n > 0.95. Solving this inequality yields n > log (1 - 0.95)/log(3/4) = 10.4. Therefore, 11 is the smallest number of offspring for which there is a greater than 95 percent chance that at least one will be aa. As a check, note that (3/4)10 = 0.056 and (3/4)11 = 0.042, so with 10 offspring, the chance of one or more aa equals 0.944, and with 11 offspring, the chance of one or more aa equals 0.958.
2.28 The ratio of probabilities that the sire is AA : Aa is 1/3 : 2/3. The ratio of the probabilities of producing n pups, all A-, for AA : Aa sires is 1 : (1/2)n. Hence, for n A- pups, the ratio of probabilities of AA : Aa sires is 1/3 : (2/3) (1/2)n. Therefore, the probability that the sire is AA, given that he had a litter of n A- pups, equals (1/3)/[(1/3) + (2/3) (1/2)n]. This is the degree of confidence you can have that the sire is AA. For n = 1 to 15, the answer is shown in the accompanying table. It is interesting that 6 progeny are required for 95 percent confidence and 8 for 99 percent confidence.
1 0.5000 6
0.9697 11 0.9990
2 0.6667 7
0.9846 12 0.9995
3 0.8000 8
0.9922 13 0.9998
4 0.8889 9
0.9961 14 0.9999
5 0.9412 10 0.9981 15 0.9999
2.29 Make a Punnett square with the ratio of D : d along each margin as 3/4 : 1/4. (a) The progeny genotypes DD, Dd, and dd therefore occur in the proportions 9/16 : 6/16 : 1/16. (b) If D is dominant, the ratio of dominant: recessive phenotypes is 15 : 1. (c) Among the D- genotypes, the ratio DD : Dd is 9 : 6. (d) If meiotic drive happens in only one sex, then the Punnett square has 3/4 : 1/4 along one margin and 1/2 : 1/2 along the other. The progeny genotypes DD, Dd, and dd are in the ratio 3/8 : 1/2 : 1/8. The ratio of dominant: recessive phenotypes is 7 : 1. Among D- genotypes, the ratio DD : Dd is 3 : 4.
3.1 Chromosome replication takes place in the S period, which is in the interphase stage of mitosis and the interphase I stage of meiosis. (Note that chromosome replication does not take place in interphase II of meiosis.) Chromosomes condense and first become visible in the light microscope in prophase of mitosis and prophase I of meiosis.
3.2 Each chromosome present in telophase of mitosis is identical to one of a pair of chromatids present in the preceding metaphase that were held together at the centromere. In this example, there are 23 pairs of chromosomes present after telophase, which implies 23 x 2 = 46 chromosomes altogether. Therefore, at the preceding metaphase, there were 46 x 2 = 92 chromatids.
3.3 In leptotene, the chromosomes first become visible as thin threads. In zygotene, they become paired threads because of synapsis. The chromosomes continue to condense to become thick threads in pachytene. In diplotene, the sister chromatids become evident, and each chromosome is seen to be a paired thread consisting of the sister chromatids. (Early signs of the bipartite nature of each chromosome are evident in pachytene.) In diakinesis, the homologous chromosomes repulse each other and move apart; they would fall apart entirely were they not held together by the chiasmata.
3.4 The number of chromosomes. Each centromere defines a single chromosome; two chromatids are considered part of a single chromosome as long as they share a single centromere. As soon as the centromere splits in anaphase, each chromatid is considered a chromosome in its own right. Meiosis I reduces the chromosome number from diploid to haploid because the daughter nuclei contain the haploid number of chromosomes (each chromosome consisting of a single centromere connecting two chromatids.) In meiosis II, the centromeres split, so the numbers of chromosomes are kept equal.
3.5 (a) 20 chromosomes and 40 chromatids; (b) 20 chromosomes and 40 chromatids; (c) 10 chromosomes and 20 chromatids.
3.6 (a) For each centromere, there are two possibilities (for example, A or a); so altogether there are 27 = 128 possible gametes. (b) The probability of a gamete's receiving a particular centromere designated by a capital letter is 1/2 for each centromere, and each centromere segregates independently of the others; hence the probability for all seven simultaneously is (1/2)7 = 1/128.
3.7 The wheat parent produces gametes with 14 chromosomes and the rye parent gametes with 7, so the hybrid plants have 21 chromosomes.
3.8 The crisscross occurs because the X chromosome in a male is transmitted only to his daughters. The expression is misleading because any X chromosome in a female can be transmitted to a son or a daughter.
3.9 Because there is an affected son, the phenotypically normal mother must be heterozygous. This implies that the daughter has a probability of 1/2 of being heterozygous.
3.10 The Bar mutation is an X-linked dominant. Because the sexes are affected unequally, some association with the sex chromosomes is suggested. Mating (a) provides an important clue. Because a male receives his X chromosome from his mother, the wildtype phenotype of the sons suggests that the Bar gene is on the X chromosome. The fact that all daughters are affected is also consistent with X-linkage, provided that the Bar mutation is dominant. Mating (b) confirms the hypothesis, because the females from mating (a) would have the genotype Bar/+ and so would produce the indicated progeny.
3.11 (a) Cross is v/Y x +/+, in which Y represents the Y chromosome; progeny are 1/2 +/Y males (wildtype) and 1/2 v/+ females (phenotypically wildtype, but heterozygous. (b) Mating is v/v x +/Y; progeny are 1/2 v/Y males (vermilion) and 1/2 v/+ females (phenotypically wildtype, but heterozygous). (c) Mating is v/+ x +/Y; progeny are 1/4 v/+ females (phenotypically wildtype), 1/4 ++/+ females (wildtype), 1/4 v/Y males (vermilion), and 1/4 +/Y males (wildtype). (d) Mating is v/+ x v/ Y; progeny are 1/4 v/v females (vermilion), 1/4 v/+ females (wildtype), 1/4 v/Y males (vermilion), and 1/4 +/Y males (wildtype).
3.12 All the females will be v/v+; bw/bw and will have brown eyes; all the males will be v/Y; bw/bw and will have white eyes.
3.13 Because the sex-chromosome situation is the reverse of that in mammals, female chickens are ZW and males ZZ, and females receive their Z chromosome from their father. Therefore, the answer is to mate a gold male (ss) with a silver female (S). The male progeny will all be Ss (silver plumage), the female progeny will be s (gold plumage), and these are easily distinguished.
3.14 (a) The mother is a heterozygous carrier of the mutation, so the probability that a daughter is a carrier is 1/2. The probability that both daughters are carriers is (1/2)2 = 1/4. (b) If the daughter is not heterozygous, the probability of an affected son is 0, and if the daughter is heterozygous, the probability of an affected son is 1/2; the overall probability is therefore 1/2 (that is, the chance that the daughter is a carrier) x 1/2 (that is, the probability of an affected son if the daughter is a carrier) = 1/4.
3.15 Let A and a represent the normal and mutant X-linked alleles. Y represents the Y chromosome. The genotypes are as follows: I-1 is Aa (because there is an affected son), I-2 is AY, II-1 is AY II-2 is aY, II-3 is Aa (because there is an affected son), II-4 is AY, and III-1 is aY.
3.16 20, because in females there are 5 homozygous and 10 heterozygous genotypes, and in males there are 5 hemizygous genotypes.
3.17 The accompanying Punnett square shows the outcome of the cross. The X and Y chromosomes from the male are denoted in red, the attached-X chromosomes (yoked Xs) and Y from the female in black. The red X chromosome also contains the w mutation. The zygotes in the upper left and lower right corners (shaded) fail to survive because they have three X chromosomes or none. The surviving progeny consist of white-eyed males (XY) and red-eyed females (XXY) in equal proportions. Attached-X inheritance differs from the typical situation in that the sons, rather than the daughters, receive their fathers' X chromosome.
3.18 The genotype of the female is Cc, where c denotes the allele for color blindness. In meiosis, the c-bearing chromatids undergo nondisjunction and produce an XX-bearing egg of genotype cc. Fertilization by a normal Y- bearing sperm results in an XXY zygote with genotype cc, which results in color blindness.
3.19 The female produces 1/2 X-bearing eggs and 1/2 O-bearing eggs ("O" means no X chromosome), and the male produces 1/2 X-bearing sperm and 1/2 Y-bearing sperm. Random combinations result in 1/4 XX, 1/4 XO, 1/4 XY, and 1/4 YO, and the last class dies because no X chromosome is present. Among the survivors, the expected progeny are 1/3 XX females, 1/3 XO females, and 1/3 XY males.
3.20 (a) II-2 must be heterozygous, because she has an affected son, which means that half of her sons will be affected. The probability of a nonaffected child is therefore 3/4, and the probability of two nonaffected children is (3/4)2 = 9/16. (b) Because the mother of II-4 is heterozygous, the probability that II-4 is heterozygous is 1/2. If she is heterozygous, half of her sons will be affected. Therefore, the overall probability of an affected child is 1/2 x 1/4 = 1/8.
3.21 (a) This question is like asking for the probability of the sex distribution GGGBBB (in that order), which equals (1/2)6 = 1/64. (b) This question is like asking for the probability of either sex distribution GGGBBB or BBBGGG, which equals 2(1/2)6 = 1/32.
3.22 With 1 : 1 segregation, the expected numbers are 125 in each class, so the X2 value equals (140 - 125)2/125 + (110 - 125)2/125 = 3.6, and there is 1 degree of freedom because there are two classes of data. The associated P value is approximately 6 percent, which means that a fit at least as bad would be expected 6 percent of the time even with Mendelian segregation. Therefore, on the basis of these data, there is no justification for rejecting the hypothesis of 1 : 1 segregation. (On the other hand, the observed P value is close to 5 percent, which is the conventional level for
rejecting the hypothesis, so the result should not inspire great confidence.)
3.23 The total number of progeny equals 800, and on the assumption of a 9 : 3 : 3 : 1 ratio, the expected numbers are 450, 150, 150, and 50, respectively.
The X2 equals (462 - 450)2/450 + (167 - 150)2/150 + (127 - 150)2 150 + (44 - 50)2/50 = 6.49 and there are 3 degrees of freedom (because there are four classes of data). The P value is approximately 0.09, which is well above the conventional rejection level of 0.05. Consequently, the observed numbers provide no basis for rejecting the hypothesis of a 9 : 3 : 3 : 1 ratio.
4.1 Because the progeny are wildtype, the mutations complement each other, and both m1 and m2 must be heterozygous. This indicates that the genotype of the progeny is m1 +/+ m2, which implies that m1 and m2 are mutations in different genes.
4.2 (a) A b, a B, A B, and a b. (b) A b and a B.
4.3 The nonrecombinant gametes are A B and a b, and the recombinant gametes are A b and a B. The nonrecombinant gametes are always more frequent than the recombinant gametes.
4.4 The gametes from the A B/a b parent are A B and a b; those from the other parent are A b and a B. The offspring genotypes are expected to be A B/A b, A B/a B, a b/A b, and a b/a B, in equal numbers. The phenotypes are A- B-, Abb, and aa B-, which are expected in the ratio 2 : 1 : 1.
4.5 The cis configuration is a b/A B.
4.6 There is no crossing-over in male Drosophila, so the gametes will be A B and a b in equal proportions. In the female, each of the nonrecombinant gametes A B and a b is expected with a frequency of (1 - 0.05)/2 = 0.475, and each of the recombinant gametes A b and a B is expected with a frequency of 0.05/2 = 0.025.
4.7 (a) 17; (b) 1; (c) in diploids there is one linkage group per pair of homologous chromosomes, so the number of linkage groups is 42/2 =21.
4.8 Recombination at a rate of 6.2 percent implies that double crossovers (and other multiple crossovers) occur at a negligible frequency; in such a case, the distance in map units equals the frequency of recombination, so the map distance between the genes is 6.2 map units.
4.9 Each meiotic cell produces four products: namely, A B and a b, which contain the chromatids that do not participate in the crossover, and A b and a B, which contain the chromatids that do take part in the crossover. Therefore, the frequency of recombinant gametes is 50 percent. Because one crossover occurs in the region between the genes in every cell, the occurrence of multiple crossing-over does not affect the recombination frequency as long as the chromatids participating in each crossover are chosen at random.
4.10 The aa bb genotype requires an a b gamete from each parent. From the A b/a B parent, the probability of an a b gamete is 0.16 / 2 = 0.08, and from the A B/a b parent, the probability of an a b gamete is (1 - 0.16)/2 = 0.42; therefore, the probability of an aa bb progeny is 0.08 x 0.42 = 0.034, or 3.4 percent.
4.11 All the female gametes are bz m. The male gametes are bz+M with frequency (1 - 0.06)/2 = 47 percent, bz m with frequency 47 percent, bz+m with frequency 0.06/2 = 3 percent, and bz M with frequency 3 percent. The progeny are therefore black males (bz+ M/bz m) with frequency 47 percent, bronze females (bz m/bz m) with frequency 47 percent, black females (bz+ m/bz m) with frequency 3 percent, and bronze males (bz M/bz m) with frequency 3 percent).
4.12 The most frequent gametes from the double heterozygous parent are the nonrecombinant gametes, in this case Gl ra and gl Ra. The genotype of the parent was therefore Gl ra/gl Ra. The recombinant gametes are Gl Ra and gl ra, and the frequency of recombination is calculated as the ratio (6 + 3)/(88 + 103 + 6 + 3) = 4.5 percent.
4.13 There is no recombination in the male, so the male gametes are ++ and b cn, each with a probability of 0.5. A map distance of 8 units means that the recombination frequency between the genes is 0.08. The female gametes are ++, b cn [the nonrecombinants, each of which occurs at a frequency of (1 - 0.08)/2 = 0.46] and + cn and b + [the
recombinants, each of which occurs at a frequency of 0.08/2 = 0.04]. When the male and female gametes are combined at random, their frequencies are multiplied, and the progeny and their frequencies are as follows:
++/++ 0.23 ++/b cn 0.23
b cn/++ 0.23 b cn/b cn 0.23
b+/++ 0.02 b+/b cn 0.02
+cn/++ 0.02 +cn/b cn 0.02
Note that b cn/+ + is the same as + + b cn, so the overall frequency of this genotype is 0.46. With regard to phenotypes, the progeny are wildtype (73 percent), black cinnabar (23 percent), black (2 percent), and cinnabar (2 percent).
4.14 The most frequent classes of gametes are the nonrecombinants, and so the parental genotype was A B c/a b C. The least frequent gametes are the double recombinants, and the allele that distinguishes them from the nonrecombinants (in this case, C and c) is in the middle. Therefore, the gene order is A-C-B.
4.15 Comparison of the first three numbers implies that r lies between c andp, and the last two numbers imply that s is closer to r than to c. The genetic map is therefore c-10-r-3-p-5-s. From this map, you would expect the recombination frequency between c and p to be 13 percent
and that between c and s to be 18 percent. The observed values are a little smaller because of double crossovers.
4.16 (a) The nonrecombinant gametes (the most frequent classes) are y v+ sn and y+ v sn+, and the double recombinants (the least-frequent classes) indicate that sn is in the middle. Therefore, the correct order of the genes is y-sn-v, and the parental genotype was y sn v+/y+sn+ v. The total progeny is 1000. The recombination frequency in the y-sn interval is (108 + 5 + 95 + 3)/1000 = 21.1 percent, and the recombination frequency in the sn-v interval is (53 + 5 + 63 + 3)/1000 = 12.4 percent. The genetic map is y-21.1-sn-12.4-v. (b) On the assumption of independence, the expected frequency of double crossovers is 0.211 x 0.124 = 0.0262, whereas the observed frequency is (5 + 3)/1000 = 0.008. The coincidence is the ratio 0.008/0.0262 = 0.31, so the interference is 1 - 0.31 = 0.69, or 69 percent.
4.17 The nonrecombinant gametes are c wx Sh and C Wx sh, and the double recombinants indicate that sh is in the middle. The parental genotype was therefore c Sh wx/C sh Wx. The frequency of recombination in the c-sh region is (84 + 20 + 99 + 15)/6708 = 3.25 percent, and that in the sh-wx region is (974 + 20 + 951 + 15)/6708 = 29.2 percent. The genetic map is therefore c-3.25-sh-29.2-wx.
4.18 Let x be the observed frequency of double crossovers. Then single crossovers in the a-b interval occur with frequency 0.15 - x, and single crossovers in the b-c interval occur with frequency 0.20 - x. The observed recombination frequency between a and c equals the sum of the single-crossover frequencies (because double crossovers are undetected), which implies that 0.15 - x + 0.20 - x = 0.31, or x = 0.02. The expected frequency of double crossovers equals 0.15 x 0.20 = 0.03, and the coincidence is therefore 0.02/0.03 = 0.67. The interference equals 1 - 0.67 = 33 percent.
4.19 b and st are unlinked, as are hk and st. The evidence is the 1 : 1 : 1 : 1 segregation observed. For b and st, the comparisons are 243 +10 (black, scarlet) versus 241 + 15 (black, nonscarlet) versus 226 +12 (nonblack, scarlet) versus 235 + 18 (nonblack, nonscarlet). For hk and st, the comparisons are 226 + 10 (hook, scarlet) versus 235 + 15 (hook, nonscarlet) versus 243 + 12 (nonhook, scarlet) versus 241 + 18 (nonhook, nonscarlet). On the other hand, b and hk are linked, and the frequency of recombination between them is (15 + 10 + 12 + 18)/1000 = 5.5 percent.
4.20 The frequency of dpy-21 unc-34 recombinants is expected to be 0.24/2 = 0.12 among both eggs and sperm, so the expected frequency of dpy-21 unc-34 / dpy-21 unc-34 zygotes is 0.122 = 1.44 percent.
4.21 The genes must be linked, because with independent assortment, the frequency of dark eyes (R- P-) would be 9/16 in both experiments. The first experiment is a mating of R P/rp x r p r p. The dark-eyed progeny represent half the nonrecombinants, so the recombination frequency r can be estimated as (1 - r)/2 = 628/(628 + 889), or r = 0.172. The second experiment is a mating of R p/r P x r p/r p. In this case, the dark-eyed progeny represent half the recombinants, so r can be estimated as r/2 = 86/(86 + 771), or r = 0.201. It is not uncommon for repeated experiments or crosses carried out in different ways to yield somewhat different estimates of the recombination frequency—in this problem, 0.17 and 0.20. The most reliable value is obtained by averaging the experimental values—in this case the average is (0.171 + 0.201)/2 = 0.186.
4.22 Consider row 1: there are - entries for 3 and 7, and so 1, 3, and 7 form one complementation group. Row 2: all +, and so 2 is the only representative of a second complementation group. Row 3: mutation already classified. Row 4: there is a - for 6 only and no other - in column 4, and so 4 and 6 make up a third complementation group. Row 5: - only for 9 and no other - in column 5, and so 5 and 9 form a fourth complementation group. Rows 6 and 7: mutations already classified. Row 8: all +, and so 8 is the only member of a fifth complementation group. In summary, there are five complementation groups as follows: group 1 (mutations 1, 3, and 7); group 2 (mutation 2); group 3 (mutations 4 and 6); group 4 (mutations 5 and 9); group 5 (mutation 8).
4.23 Consider each gene in relation to first- and second-division segregation. Gene a gives 1766 asci with first- division segregation and 234 asci with second-division segregation; the frequency of second-division segregation is 234/(1766 + 234) = 0.117, which implies that the distance between a and the centromere is 0.117/2 = 5.85 percent. Gene b gives 1780 first-division and 220 second-division segregations, for a frequency of second-division segregation of 220/(1780 + 220) = 0.110. The distance between b and the centromere is therefore 0.110/2 = 5.50 map units. If we consider a and b together, there are 1986 PD asci, 14 TT asci, and no NPD. Because NPD << PD, genes a and b are linked. The frequency of recombination between a and b is [(1/2) x 14]/2000 = 0.35 percent. Comparing this distance with the gene-centromere distances calculated earlier results in the map a-0.35-b-5.50— centromere.
4.24 (a) On the assumption of independence, (0.30 x 0.10)/2 = 0.015. (Division by 2 is necessary because sh wx gl gametes represent only one of the two classes of double recombinants.) (b) If the interference is 60 percent, then the
coincidence equals 1 - 0.60 = 0.40. The observed frequency of doubles is therefore 40 percent as great as with independence, so the expected frequency of sh wx gl gametes is 0.015 x 0.40 = 0.006.
4.25 Considered individually, gene c yields 90 first-division segregation asci and 10 (1 + 9) second-division segregation asci, so the frequency of recombination between c and the centromere is (10/100) x 1/2 = 5 percent. Gene v yields 79 first-division and 21 (20 + 1) second-division asci, so the frequency of recombination between v and the centromere is (21/100) x 1/2 = 10.5 percent. If the genes are taken together, the asci include 34 PD, 36 NPD, and 30 (20 + 1 + 9) TT. Because NPD = PD, the genes are unlinked.
In summary, c and v are in different chromosomes, c being 5 map units from the centromere and v being 10.5 map units from the centromere.
4.26 Analysis of cross 1. Ascospore arrangements 1, 2, 4, and 5 result when there is a crossover between the gene and the centromere. Types 3 and 6 result when there is no crossing-over between the gene and the centromere. The map distance between the gene and its centromere is (1/2) (Frequency of second-division segregation) x 100 = (1/2) [(6 + 6 + 6 + 6)/120] x 100 = 10 centimorgans. Analysis of cross 2. There is no crossing-over between the gene and its centromere. Analysis of cross 3. The proportion of second-division segregation (asci types 1, 2, 4, and 5) equals 80/120 = 67 percent. This distribution of asci results when crossing-over is frequent enough to randomize the position of alleles with respect to their centromeres. The gene is said to be "unlinked" with its centromere.
4.27 For each pair of genes, classify the tetrads as PD, NPD, or TT, and tabulate the results as follows:
leu2-trp1 PD 230 + 235 = 465
NPD 215 + 220 = 435 TT 54 + 46 = 100 leu2-met14 PD 235 + 220 = 455
NPD 230 + 215 = 445 TT 54 + 46 = 100 trp1-met14 PD 235 + 215 + 46 = 496
NPD 230 + 220 + 54 = 504 TT 0
For leu2-trp1, PD = NPD; therefore, leu2 and trp1 are unlinked. Similarly, for leu2-met14, PD = NPD; these genes are also unlinked. Likewise, trp1 and met14 are unlinked because PD = NPD. However, for the trp1-met14 gene pair, there are no tetratypes, which means that each gene is closely linked to its own centromere. Because they do not recombine with their centromeres, the segregation of trp1 and met14 can be used as genetic markers of centromere segregation; their presence marks the sister spores created by segregation in the first meiotic division. Using the trp1 and met14 markers in this fashion allows the remaining leu2 gene to be mapped with respect to its centromere; all tetratype tetrads will result from crossing-over between leu2 and its centromere because neither of the other two genes recombines with its centromere. The map distance between leu2 and its centromere is therefore (1/2) (100/1000) x 100 = 5 centimorgans.
4.28 For the rad6-trp5 gene pair, PD = 188 + 206 = 140 + 154 + 154 + 109 = 797, NPD = 0, and TT = 105 + 92 + 3 + 2 + 1 = 203. Because PD >> NPD, the genes are linked. In this case, TT > 0 and NPD = 0, so the map distance is calculated as (1/2)(203/1000) x 100 = 10.15 centimorgans. For the trp5-leu1 gene pair, PD = 188 + 206 + 105 + 92 + 109 = 700, NPD = 0, and TT = 140 + 154 + 3 + 2 + 1 = 300; here again, PD >> NPD, TT > 0, and NPD = 0, so the map distance is calculated as (1/2)(300/1000) x 100 = 15 centimorgans. For the rad6-leu1 gene pair, PD = 188 + 206 + 109 = 503, NPD = 1, and TT = 105 + 92 + 140 + 154 + 3 + 2 = 496. In this case, PD >> NPD, but TT > 0 and NPD > 0. The map distance is calculated as (1/2)[(496 + 6 x 1)/1000] x 100 = 25.1 centimorgans. For the leu1 - me14 gene pair, PD = 188 + 105 + 140 + 2 = 435, NPD = 206 + 92 +154 + 3 + 1 = 456, and TT = 109. In this case PD = NPD, so leu1 is not linked to met14. However, we know from the previous problem that met14 is tightly linked with its centromere, and this information can be used to calculate the map distance between leu1 and its centromere. Because TT = 109 and also TT < 2/3, the map distance between leu1 and its centromere is (1/2)(109/1000) x 100 = 5.45 centimorgans. For the rad6-me14 gene pair, PD = 188 + 1 = 189, NPD = 206, and TT = 105 + 92 + 140 + 154 + 109 + 3 + 2 = 605. Here again, PD = NPD, which indicates that there is no linkage between the genes. Because
met14 is tightly linked to its centromere, and TT = 605 but TT < 2/3, the map distance between rad6 and its centromere is (1/2)(605/1000) x 100 = 30.25 centimorgans. For the trp5 - met14 gene pair, PD = 188 + 105 = 293, NPD = 206 + 92 = 298, and TT = 140 + 154 + 109 + 3 + 2 + 1 = 409. Once again, PD = NPD, indicating no linkage between the genes. As before, we use met14 as a centromere marker to calculate the distance between trp5 and its centromere as (1/2) (409/1000) x 100 = 20.5 centimorgans. The overall conclusion is that rad6, trp5, and leu1 are all linked to each other; met14 is tightly linked to its centromere but not to the other three genes. The genetic map based on these data is shown below:
5.1 The complementary strand is 5'-T-C-C-G-A-G-3'.
5.2 A nuclease is an enzyme capable of breaking a phosphodiester bond. An endonuclease can break any phosphodiester bond, whereas an exonuclease can only remove a terminal nucleotide.
5.3 Movement along the template strand is from the 3' end to the 5' end, because consecutive nucleotides are added to the 3' end of the growing chain. In double-stranded DNA, one strand is replicated continuously, and the other strand is replicated in short segments that are later joined.
5.4 DNA polymerase joins a 5'-triphosphate with a 3'-OH group, and the outermost two phosphates are released; the resulting phosphodiester bond contains one phosphate. DNA ligase joins a single 5'-P group with a 3'-OH group.
5.5 The enzyme has polymerizing activity, it is a 3' '5' exonuclease (the proofreading function), and it is a 5' 3' exonuclease.
5.6 Neither RNA polymerases nor primases require a primer to get synthesis started.
5.7 Smaller molecules move faster because they can penetrate the pores of the gel more easily.
5.8 (a) In rolling-circle replication, one parental strand remains in the circular part and the other is at the terminus of the branch. Therefore, only half of the parental radioactivity will appear in progeny. (b) If we are assuming no recombination between progeny DNA molecules, one progeny phage will be radioactive. If a single recombination occurs, two particles will be radioactive. If recombination is frequent and occurs at random, the radioactivity will be distributed among virtually all the progeny.
5.9 Rolling-circle replication must be initiated by a single-stranded break; 0 replication does not need such a break.
5.10 (a) Because 18 percent is adenine, 18 percent is also thymine; so [T] = 18 percent. (b) Because [A] + [T] = 36 percent, [G] + [C] equals 64 percent, or [G] = [C] = 32 percent each. Overall, the base composition is [A] = 18 percent, [T] = 18 percent, [G] = 32 percent, and [C] = 32 percent; the [G] + [C] content is 64 percent.
5.11 (a) The distance between nucleotide pairs is 3.4 A, or 0.34 nm. The length of the molecule is 34 |i, or 34 x 103 nm. Therefore, the number of nucleotide pairs equals (34 x 103)/0.34 = 105. (b) There are ten nucleotide pairs per turn of the helix, so the total number of turns equals 105/10 = 104.
5.12 (a) First, note that the convention for writing polynucleotide sequences is to put the 5' end at the left. Thus the sequence complementary to AGTC (which is 5'-AGTC-3') is 3'-TCAG-5', written in conventional format as GACT (that is, 5'-GACT-3'). Hence the frequency of CT is the same as that of its complement AG, which is given as 0.15. Similarly, AC = 0.03, TC = 0.08, and AA = 0.10. (b) If DNA had a parallel structure, then the 5' ends of the strands would be together, as would the 3' ends, so the sequence complementary to 5'-AGTC-3' would be 5'-TCAG-3'. Thus AG = 0.15 implies that TC = 0.15, and similarly, the other observed nearest neighbors would imply that CA = 0.03, CT = 0.08, and AA = 0.10.
5.13 Remember that the chemical group at the growing end of the leading strand is always a 3'-OH group, because DNA polymerases can add nucleotides only to such a group. Because DNA is antiparallel, the opposite end of the leading strand is a 5'-P group. Therefore, (a) 3'-OH; (b) 3'-OH; (c) 5'-P.
5.14 Rolling-circle replication begins with a single-strand break that produces a 3'-OH group and a 5'-P group. The polymerase extends the 3'-OH terminus and displaces the 5'-P end. Because the DNA strands are antiparallel, the strand complementary to the displaced strand must be terminated by a 3'-OH group.
5.15 (a) The first base in the DNA that is copied is a T, so the first base in the RNA is an A. RNA grows by addition to the 3'-OH group, so the 5'-P end remains free. Therefore, the sequence of the eight-nucleotide primer is
(b) The C at the 3' end has a free -OH; the A at the 5' end has a free triphosphate. (c) Right to left, because this would require discontinuous replication of the strand shown.
5.16 (a) The cloned SalI fragment is 20 kb in length, and there are no SalI sites within it. (b) There is a single EcoRI site 7 kb from one end. (c) There are two HindIII sites within the fragment. (d) The HindIII sites must flank the EcoRI site, dividing the 7-kb EcoRI fragment into 3-kb and 4-kb subfragments and the 13-kb EcoRI fragment into 5kb and 8-kb subfragments. The 3-kb and 8-kb fragments must be adjacent, because digestion with SalI and HindIII alone produces an 11 -kb fragment. Except for the left-to-right orientation, the inferred restriction map must be as follows:
5.17 The bands in the gel result from incomplete chains whose synthesis was terminated by incorporation of the dideoxynucleotide indicated at the top. The smaller fragments are at the bottom of the gel, the larger ones at the top. Because DNA strands elongate by addition to their 3' ends, the strand synthesized in the sequencing reactions has the sequence, from bottom to top, of
The template strand is complementary in sequence, and antiparallel, so its sequence is
5.18 For each site, the probability equals the product of the probability of each of the nucleotide pairs in turn, 1/4 for any of the specified pairs and 1/2 for R-Y. The average spacing is the reciprocal of the probability. Hence for TaqI, the average distance between sites is 44 = 256 base pairs; for BamHI, it is 46 = 4096 base pairs; and for HaeII, it is 44 22= 1024 base pairs.
6.1 The 1.3-kb insertion is probably a transposable element. (It is a transposable element, called mariner.)
6.2 Transposition in somatic cells could cause mutations that kill the host or decrease the ability of the host to reproduce, which reduces the chance of transmission of the transposable element to the next generation. Restriction of activity to the germ line lessens these potentially unfavorable effects.
6.3 Insertion is initiated by a staggered cut in the host DNA sequence, and the overhanging single-stranded ends are later filled in by a DNA polymerase. This process creates a direct repeat.
6.4 E. coli: 4,700,000 bp x 3.4 A/bp x 10-7 mm/A = 1.6 mm. Human beings: 3 x 109 bp x 3.4 A/bp x 10-7 mm/A = 1020 mm; in other words, there is approximately 1 meter of DNA in a human gamete.
6.5 DNA structure is a long, thin thread, and its total length in most cells greatly exceeds the diameter of a nucleus, even allowing for a large number of fragments per cell. This forces you to conclude that each DNA molecule must be folded back on itself repeatedly.
6.6 Matching of chromomeres, which are locally folded regions of chromatin. Polytene chromosomes do not divide because the cells in which they exist do not divide.
6.7 No, because the E. coli chromosome is circular. There are no termini.
6.8 The repeated sequences are located at the ends of the element. They may be in direct or inverted orientation, depending on the particular transposable element.
6.9 (a) There are ten base pairs for every turn of the double helix. With four turns of unwinding, 4 x 10 = 40 base pairs are broken, (b) Each twist compensates for the underwinding of one full turn of the helix; hence, there will be four twists.
6.10 The supercoiled form is in equilibrium with an underwound form having many unpaired bases. At the instant that a segment is single-stranded, S1 can attack it. Because a nicked circle lacks the strain of supercoiling, there is no tendency to have single-stranded regions, and so a nicked circle is resistant to S1 cleavage.
6.11 The rate of movement in the gel is determined by the overall charge and by the ability of the molecule to penetrate the pores of the gel. Supercoiled and relaxed circles migrate at different rates because they have different conformations.
6.12 (a) Renaturation is concentration-dependent because it requires complementary molecules to collide by chance before the formation of base pairs can occur. (b) Here, the lengths of the GC tracts are important. Molecule 2 has a long GC segment, which will be the last region to separate. Hence molecule 1 has the lower temperature for strand separation.
6.13 The hybrid molecules that can be produced from single-stranded DNA fragments from species A and B are AA, AB, BA, and BB, and these are expected in equal amounts. The molecules with a hybrid (14N/15N) density are AB and BA, and these account for 5 percent of the total. The same sequences can also renature as AA or BB which together must make up another 5 percent, so a total of 10 percent of the base sequences are common to the two species.
6.14 In a direct repeat, the sequence is repeated in the same 5'-to-3' orientation. In an inverted repeat, the sequence is repeated in the reverse orientation and on the opposite strand, which is necessary to preserve the correct 5'-to-3' polarity in view of the antiparallel nature of DNA strands. The accompanying figure shows a direct and inverted repeat of the example sequence. The dashed lines represent unspecified DNA sequences between the repeats.
6.15 All such mutations are probably lethal. The fact that the amino acid sequence of histones is virtually the same in all organisms suggests that functional histones are extremely intolerant of amino acid changes.
6.16 The relatively large number of bands and the diverse locations in the genome among different flies suggests very strongly that the DNA sequence is a transposable element of some kind.
6.17 In the first experiment, the nucleosomes form at random positions in different molecules. In the second experiment, the protein binds to a unique sequence in all the molecules, and the nucleosomes form by sequential addition of histone octamers on both sides of the protein-binding site. Because nuclease attacks only in the linker regions, the cuts are made in small regions that are highly localized.
7.1 (a) The Klinefelter karyotype is 47,XXY; hence one Barr body. (b) The Turner karyotype is 45,X; hence no Barr bodies. (c) The Down syndrome karyotype is 47,+21; people with this condition have one or zero Barr bodies, depending on whether they are female (XX) or male (XY). (d) Males with the karyotype 47,XYY have no Barr bodies. (e) Females with the karyotype 47,XXX have two Barr bodies.
7.2 An autopolyploid series is formed by the combination of identical sets of chromosomes. Therefore, the chromosome numbers must be even multiples of the monoploid number, or 10 (diploid), 20 (tetraploid), 30, 40, and 50.
7.3 Species S is an allotetraploid of A and B formed by hybridization of A and B, after which the chromosomes in the hybrid became duplicated. The univalent chromosomes in the S x A cross are the 12 chromosomes from B, and the univalent chromosomes in the S x B cross are the 14 chromosomes from A.
7.4 The chromosomes underwent replication with no cell division (endoreduplication), resulting in an autotetraploid.
7.5 The only 45-chromosome karyotype found at appreciable frequencies in spontaneous abortions is 45,X, and so 45,X is the probable karyotype. Had the fetus survived, it would have been a 45,X female with Turner syndrome.
7.6 Because the X chromosome in the 45,X daughter contains the color-blindness allele, the 45,X daughter must have received the X chromosome from her father through a normal X-bearing sperm. The nondisjunction must therefore have occurred in the mother, resulting in an egg cell lacking an X chromosome.
7.7 The mother has a Robertsonian translocation that includes chromosome 21. The child with Down syndrome has 46 chromosomes, including two copies of the normal chromosome 21 plus an additional copy attached to another chromosome (the Robertsonian translocation). This differs from the usual situation, in which Down syndrome children have trisomy 21 (karyotype 47,+21).
7.8 The inversion has the sequence A B E D C F G, the deletion A B F G. The possible translocated chromosomes are (a) A B C D E T U V and MN 0 P Q R S F G or (b)A B C D E S R Q P O NM and V U TF G. One of these possibilities includes two monocentric chromosomes; the other includes a dicentric and an acentric. Only the translocation with two monocentrics is genetically stable.
7.9 The order is a e d f c b or the other way around.
7.10 Genes b, a, c, e, d, andf are located in bands 1, 2, 3, 4, 5, and 6, respectively. The reasoning is as follows. Because deletion 1 uncovers any gene in band 1 but the other deletions do not, the pattern -++++ observed for gene b puts b in band 1. Deletions 1-3 uncover band 2 but deletions 4-5 do not, so the pattern —++ observed for gene a localizes gene a to band 2. Genes in band 3 are uncovered by deletion 1-4 but not deletion 5, so the pattern —+ implies that gene c is in band 3. Band 4 is uncovered by deletions 3-5 but not deletions 1-2, which means that gene e, with pattern ++—, is in band 4. Genes in band 5 are uncovered by deletions 4-5 but not deletions 1-3, so the pattern +++-- observed for gene d places it in band 5. Genes in band 6 are uncovered by deletion 5 but not deletions 1-4, so the pattern ++++- puts gene f in band 6.
7.11 The most probable explanation is an inversion. In the original strain the inversion is homozygous, so no problems arise in meiosis. The F1 is heterozygous, and crossing-over within the inversion produces the dicentrics and acentrics. Because the crossover products are not recovered, the frequency of recombination is greatly reduced in the chromosome pair in which the inversion is heterozygous, so the inversion is probably in chromosome 6.
7.12 Starting with chromosome (c), compare the chromosomes pairwise to find those that differ by a single inversion. In this manner, you can deduce that inversion of the i-d-c region in (c) gave rise to (d), inversion of the h- g-c-d region in (d) gave rise to (a), and inversion of the c-d-e-f region in (a) gave rise to (b). Because you were told
that (c) is the ancestral sequence, the evolutionary ancestry is (c) ' (d) ' (a) ' (b).
7.13 The group of four synapsed chromosomes implies that a reciprocal translocation has taken place. The group of four includes both parts of the reciprocal translocation and their nontranslocated homologs.
7.14 Translocation heterozygotes are semisterile because adjacent segregation from the four synapsed chromosomes in meiosis produces aneuploid gametes, which have large parts of chromosomes missing or present in excess. The
aneuploid gametes fail to function in plants, and in animals they result in zygote lethality. The translocation homozygote is fully fertile because the translocated chromosomes undergo synapsis in pairs, and segregation is completely regular. In the cross of translocation homozygote x normal homozygote, all progeny are expected to be semisterile.
7.15 In this male, there was a reciprocal translocation between the Curly-bearing chromosome and the Y chromosome. Because there is no crossing-over in the male, all Y-bearing sperm that give rise to viable progeny contain Cy, and all X-bearing sperm that give rise to viable progeny contain y+.
7.16 In this male, the tip of the X chromosome containing y+ became attached to the Y chromosome, giving the genotype y X/y+ Y. The gametes are y X and y+ Y, so all of the offspring are either yellow females or wildtype males.
7.17 The A B c D E chromosome results from two-strand double crossing-over in the B-C and C-D regions. The double recombinant chromosome still has the inversion.
7.18 The unusual yeast strain is not completely haploid but is disomic for chromosome 2. Segregation of markers on this chromosome, like his7, is aberrant, but segregation of markers on the other chromosomes is completely normal.
7.19 The parental classes are wildtype and brachytic, fine-stripe and the recombinant classes are brachytic, fine- stripe+ and brachytic+, fine-stripe. The recombination frequency between brachytic and fine-stripe is therefore equal to (17 + 1 + 6 + 8)/682 = 0.047.
Because semisterility is associated with the presence of the translocation, determining the recombination frequency between each of the mutations and the translocation breakpoint follows the same logic as mapping any other kind of genetic marker. For the parental classes, the mutations and the translocation are on different chromosomes. Members of the recombinant classes are those plants that show both the mutation in question and semisterility or neither semisterility nor the mutation in question. The recombination frequency between brachytic and the translocation breakpoint is therefore (17 + 25 + 19 + 8)/682 = 0.102; the recombination frequency between fine
-stripe and the translocation breakpoint is (19 + 6 + 1 + 25)/682 = 0.075.
8.1 Minimal medium containing galactose and biotin; the galactose would require the presence of gal+, and the biotin would allow both bio+ and bio' to grow.
8.2 The E. coli genome contains approximately 4700 kilobase pairs, which implies approximately 47 kb per minute. Because the l genome is 50 kb, the genetic length of the prophage is approximately 1 minute (more precisely, 47/50 = 0.94 minutes).
8.3 Infect an E. coli l lysogen with P1. The l lysogen contains l prophage, so some of the resulting P1 phage will include the part of the chromosome that contains the prophage.
8.4 After circularization and integration at the bacterial att site, the map is . . . att D EF A B C att . . ..
8.5 The specialized-transducing particles originate from aberrant prophage excision.
8.6 Plate on minimal medium that lacks leucine (selects for Leu+) but contains streptomycin (selects for Str-r). The leu+ allele is the selected marker, and str-r the counterselected marker.
8.7 One plaque per phage. One plaque, because the plating was done before lysis, so the progeny phage are confined to one tiny area.
8.8 The T2 plaques will be turbid because T2 fails to lyse the resistant bacteria. Plaques made by the T2h mutant will be clear because the mutant can lyse both the normal and the resistant cells.
8.9 The mean number of phage per cell equals 1. If you know the Poisson distribution, then it is apparent that the proportion of uninfected cells is e"1 = 0.37. If you do not know the Poisson distribution, then the answer can be calculated as follows. The probability that a bacterial cell escapes infection by a particular phage is 1 - 10-6 = 0.999999, and the probability that it escapes infection by all one million phages is P0 = (0.999999)1000000. Therefore, In P0 = 1000000 x ln(0.999999) = -1, so P0 = e"1 = 0.37.
8.10 Apparently, h and tet are closely linked, so recombinants containing the h+ allele of the Hfr tend also to contain the tet-s allele of the Hfr, and these recombinants are eliminated by the counterselection for tet-r.
8.11 All receive the a+ allele, but whether or not a particular b+ str-r recombinant contains a+ depends on the positions of the genetic exchanges.
8.12 Depending on the size of the F', it could be F' g, F' g h, F' g h i, and so forth, or F' f, F' f e, F' f e d, and so forth.
8.13 The first selection is for Met+ and so lac+ must have been transferred. The probability that a marker that is transferred is incorporated into the recombinants is about 50 percent, so about 50 percent of the recombinants would be expected to be lac+. The second selection is for Lac+, and met+ is a late marker. The great majority of mating pairs will spontaneously break apart before transfer of met+, so the frequency of met+ recombinants in this experiment will be close to zero.
8.14 The order 500 his+, 250 leu+, 50 trp+ implies that the genes are transferred in the order his leu trp. The met mutation is the counterselected marker that prevents growth of the Hfr parent. The medium containing histidine selects for leu+ trp+, and the number is small because both genes must be incorporated by recombination.
8.15 The three possible orders are (1) pur-pro-his, (2)pur-his-pro, and (3) pro-pur-his. The predictions of the three orders follow. (1) Virtually all pur+ his transductants should be pro', but this is not true. (2) Virtually all pur+pro" transductants should be his', but this is not true. (3) Somepur+ pro' transductants will be his', and somepur+ his' transductants will be pro' (depending on where the exchanges occur). Order (3) is the only one that is not contradicted by the data.
8.16 (a) Use m = 2 - 2d1'3, in which m is map distance in minutes and d is cotransduction frequency. With the values of d given, the map distances are 0.74, 0.41, and 0.18 minutes, respectively. (b) Use d = [1 - (m/2)]3. With the values of m given, cotransduction frequencies are 0.42, 0.12, and 0, respectively. For markers greater than 2 minutes apart, the frequency of cotransduction equals zero. (c) Map distance a-b equals 0.66 minutes, b-c equals 1.07 minutes, and these are additive, giving 1.73 minutes for the distance a-c. Predicted cotransduction frequency for this map distance is 0.002.
8.17 It is probable that the amp-r gene in the natural isolate was present in a transposable element. Transposition into l occurs in a small proportion of cases, and when these infect the laboratory strain, the transposable element can transpose into the chromosome before the infecting l is lost.
8.18 The tet-r gene in the phage was included in a transposable element. Transposition occurred in the lysogen to another location in the E. coli chromosome, and when the prophage was lost, the antibiotic resistance remained.
8.19 Both the order of times of entry and the level of the plateaus imply that the order of gene transfer is a b c d. The times of entry are obtained by plotting the number of recombinants of each type versus time and extrapolating back to the time axis. The values are a, 10 minutes; b, 15 minutes; c, 20 minutes; d, 30 minutes. The low plateau value for the d gene probably results from close linkage with str-s, because in this case, the d allele in the Hfr strain would tend to be inherited along with the str-r allele.
8.20 The deletion map is shown in the accompanying illustration. A completely equivalent map can be drawn with the order of the genes reversed.
8.21 The genetic map shows the genetic intervals defined by the deletion endpoints and the locations of the mutations with respect to these intervals.
8.22 The mutations v and z are in the rIIB cistron, whereas the mutations t, u, w, x, and y are in the rIIA cistron. As indicated in the genetic map, the boundary between the cistrons lies between the left end of the c deletion and the right end of the e deletion.
The matrix for complementation among the deletions must be as follows:
.'j b : d c f
8.23 The l prophage itself is being transduced as part of the chromosome. It maps next to gal. Hence when phage 363 is grown on strain D, 6 percent of the Gal+ transductants are lysogenic, whereas none of the Thr+ or Lac+ transductants is lysogenic. The reciprocal cross gives the same basic result, only here l is being removed from the chromosome and being replaced by phage-free DNA. Hence about 8 percent of the Gal+ transductants are nonlysogenic. As expected from the close linkage of the phage l attachment site and gal, no linkage is seen for the other two markers.
9.1 The DNA content per average band equals about 1.1 x 108/5000 = 22 kb, that per average lettered subdivision equals approximately 1.1 x 108/600 = 183 kb, and that per average numbered section equals about 1.1 x 108/100 =
1.1 Mb. A YAC with a 200-kb insert contains the DNA equivalent to about 9 salivary bands, 1.1 lettered subdivisions, and 0.2 numbered section. An 80-kb PI clone contains the DNA equivalent of about 4 salivary bands, 0.48 lettered subdivision, and 0.07 numbered section.
9.2 The ends may be blunt or they may be cohesive (sticky) with either a 3' overhang or a 5' overhang.
9.3 All restriction fragments have the same ends, because the restriction sites are all the same and they are cleaved in the same places. Opposite ends of the same fragment must also be identical, because restriction sites are palindromes.
9.4 No. The complement of 5'-GGCC-3' is 3'-CCGG-5'.
9.5 The insertion disrupts the gene, and sensitivity to the antibiotic shows that the vector has an inserted DNA sequence. The second antibiotic-resistance gene is needed to select transformants.
9.6 (a) The probability of a TaqI site is 1/6 x 1/3 x 1/3 x 1/6 = 1/324, so TaqI cleavage is expected every 324 base pairs; the probability of a .MaeIII site is 1/3 x 1/6 x 1 x 1/6 x 1/3 = 1/324, so .MaeIII cleavage is expected every 324 base pairs. (b) The answers are the same as in part (a), because for both TaqI and MaeIII sites, the number of A + T = the number of G + C.
9.7 Comparison of the genomic and cDNA sequences tells you where the introns are in the genomic sequence. The genomic sequence contains the introns; the cDNA does not.
9.8 Bacterial cells do not normally recognize eukaryotic promoters, and if transcription does occur, the transcript usually contains introns that the bacterial cell cannot remove. If these problems are overcome, the protein may still not be produced because the mRNA lacks a bacterial ribosome-binding site, because the protein might require post- translational processing, or because the protein might be unstable in bacterial cells and subject to degradation.
9.9 The finding is that the 3.6-kb and 5.3-kb fragments in the free phage are joined to form an 8.9-kb fragment in the intracellular form. This suggests that the the 3.6-kb and 5.3-kb fragments are terminal fragments of a linear molecule and that the intracellular form is circular.
9.10 A base substitution may destroy a restriction site and prevent two potential fragments from being separated, or a deletion may eliminate a restriction site.
9.11 The gene probably contains an EcoRI site that disrupts the gene in the process of cloning; the gene does not contain a HindIII site.
9.12 (a) The tet-r gene is not cleaved with BglI, so addition of tetracycline to the medium requires that the colonies be tetracycline-resistant (Tet-r) and hence contain the plasmid. (b) Cells with the phenotype Tet-r Kan-r or Tet-r Kan-s will form colonies. (c) Colonies with the phenotype Tet-r Kan-s contain inserts within the cleaved kan gene.
9.13 An insertion of two bases is generated within the codon, resulting in a frameshift. Therefore, all colonies should be Lac-.
9.14 You must use cDNA under a suitable promoter that functions in the prokaryotic system. Eukaryotic genomic DNA includes regulatory elements that will not work in bacteria and introns that cannot be spliced out of RNA in prokaryotic cells.
9.15 The restriction map is shown in the accompanying figure.
JVT0.7 EcoR r I 0.22 BautMl 0.56 Sail
I — 1- — 1 ^
artip-r 0.03 tei-t \
Rest Of plasmid between Hi T antf Silt I = £.-15 kb
10.1 The start codon is AUG, which codes for methionine; the stop codons are UAA, UAG, and UGA. The probability of a start codon is 1/64, and the probability of a stop codon is 3/64. The average distance between stop codons is 64/3 = 21.3 codons.
10.2 The mRNA sequence is
and translation begins with the first AUG, which codes for methionine. The amino acid sequence coded by this region is Met-Pro-Leu-Ile-Ser-Ala-Ser-Tyr.
10.3 With a G at the 5' end, the first codon is GUU, which codes for valine; and with a G at the 3' end, the last codon
is UUG, which codes for leucine.
10.4 Codons are read from the 5' end of the mRNA, and the first amino acid in a polypeptide chain is at the amino terminus. If the G were at the 5' end, the first codon would be GAA, coding for glutamic acid at the amino terminus of the polypeptide. If the G were at the 3' end, the last codon would be AAG, coding for arginine at the carboxyl terminus. Thus the G was added at the 5' end.
10.5 GUG codes for valine, and UGU codes for cysteine. Therefore, an alternating polypeptide containing only valine and cysteine would be made.
10.6 The artificial mRNA can be read in any one of three reading frames, depending on where translation starts. The reading frames are
GUC GUC GUC . . ., UCG UCG UCG . . ., and CGU CGU CGU . . .
The first codes for polyvaline, the second for polyserine, and the third for polyarginine.
10.7 The possible codons, their frequencies, and the amino acids coded are as follows:
AAA (3/4)3 = 0.421 Lys
AAC (3/4)2(1/4) = 0.141 Asn
ACA (3/4)2(1/4) = 0.141 Thr
CAA (3/4)2(1/4) = 0.141 Gln
CCA (3/4)(1/4)2 = 0.141 Pro
CAC (3/4)(1/4)2 = 0.047 His
ACC (3/4)(1/4)2 = 0.047 Thr
CCC (1/4)3 = 0.015 Pro
Therefore, the amino acids in the random polymer and their frequencies are lysine (42.1 percent), asparagine (14.1 percent), threonine (18.8 percent), glutamine (14.1 percent), histidine (4.7 percent), and proline (6.2 percent).
10.8 Methionine has 1 codon, histidine 2, and threonine 4, so the total number is 1 x 2 x 4 = 8. The sequence AUGCAYACN encompasses them all. Because arginine has six codons, the possible number of sequences coding for Met-Arg-Thr is 1 x 6 x 4 = 24. The sequences are AUGCGNACN (16 possibilities) and AUGAGRACN (8 possibilities).
10.9 In an overlapping code, a single base change affects more than one codon, so changes in more than one amino acid should often occur (but not always, because single amino acid changes can result from redundancy in the code). For this reason, an overlapping code was eliminated from consideration as a likely possibility.
10.10 Because the pairing of codon and anticodon is antiparallel, it is convenient to write the anticodon as 3'-UAI-5'. The first two codon positions are therefore 5'-AU-3', and the third is either A, U, or C (because each can pair with I). The possible codons are 5'-AUA-3', 5'-AUU-3', and 5'-AUC-3', all of which code for isoleucine.
10.11 In theory, the codon 5'-UGG-3' (tryptophan) could pair with either 3'-ACC-5' or 3'-ACU-5'. However, 3'- ACU-5' can also pair with 5'-UGA-3', which is a chain-termination codon. If this anticodon were used for tryptophan, an amino acid would be inserted instead of terminating the chain. Therefore, the other anticodon (3'- ACC-5') is the one used.
10.12 (a) The deletion must have fused the amino-coding terminus of the B gene with the carboxyl-coding terminus of the A gene. The nontranscribed strand must therefore be oriented 5'-B-A-3'. (b) The number of bases deleted must be a multiple of 3; otherwise, the carboxyl terminus would not have the correct reading frame.
10.13 (a) The nontranscribed strand is given, so the wildtype reading frame of the mRNA is
AUG CAU CCG GGC UCA UUA GUC U...
which codes for Met-His-Pro-Gly-Ser-Leu-Val- . . . (b) The mutant X mRNA reading frame is
AUG GCA UCC GGG CUC AUU AGU CU...
which codes for Met-Ala-Ser-Gly-Leu-Ile-Ser-Leu- . . .. (c) The mutant Y mRNA reading frame is
AUG CAU CCG GGC UCU UAG UCU...
which codes for Met-His-Pro-Gly-Ser (the UAG is a termination codon). (d) The double-mutant mRNA reading frame is
AUG GCA UCC GGG CUC UUA GUC U...
which codes for Met-Ala-Ser-Gly-Leu-Leu-Val- . . .. Note that the double mutant differs from wildtype only in the region between the insertion and deletion mutations, in which the reading frame is shifted.
10.14 Mutation X can be explained by a single base change in the start codon for the first Met, in which case translation would begin with the second AUG codon. Mutation Y can be explained by a single base change that converts a codon for Tyr (UAU or UAC) into a chain-termination codon (UAA or UAG). The tripeptide sequence is Met-Leu-His.
10.15 The probability of a correct amino acid at a particular site is 0.999, so the probability of all 300 amino acids being correctly translated is (0.999)300 = 0.74, or approximately 3/4.
10.16 Each loop represents an intron, so the total number is seven introns.
10.17 If transcribed from left to right, the transcribed strand is the bottom one, and the mRNA sequence is
AGA CUU CAG GCU CAA CGU GGU
which codes for Arg-Leu-Gln-Ala-Gln-Arg-Gly. To invert the molecule, the strands must be interchanged in order to preserve the correct polarity, and so the sequence of the inverted molecule is
This codes for an mRNA with sequence
AGA ACG UUG AGC CUG AAG GGU
which codes for Arg-Thr-Leu-Ser-Leu-Lys-Gly.
10.18 The DNA nontemplate strand is shown as sequence (1) in the table below. The corresponding mRMA for this region is shown as sequence (2). Translation in all three possible reading frames yields (in the single-letter amino acid codes) the sequences shown in (3).
(3) * R M L D L Q A I D A S S R
N V C L T S K Q S M P A Q T Y A * P P S N R C Q L K
The sequence is from the middle of an exon, so you do not know the reading frame. However, two of the possible reading frames contain stop codons (*). Therefore, the only sequence without stop codons must be correct, because you know that the sequence comes from the middle of an exon of an active gene. Thus the correct amino acid sequence of the polypeptide is N V C L T S K Q S M P A Q.
10.19 Unlike other 6-fold degenerate amino acids (Arg, Leu), is not possible to change between Ser codons with a single mutation. In particular, to get from the 4-fold degenerate class (UCN, where N is any of four bases) to the 2fold degenerate class (AGY, where Y is any pyrimidine) requires a change in both the first and the second codon positions. It is highly unlikely that two point mutations would occur simultaneously in the same codon, so an organism would have temporarily had an amino acid other than Ser when changing from one codon class to the other. It is believed that the two classes of Ser codons may have evolved independently.
10.20 Mutant 1 indicates the sequence AUGAARUAG, where R means any purine. Mutant 2 indicates the sequence AUGAU[U or C or A]GUNUAA, where N means any nucleotide. Because mutation 2 is a deletion in the second codon, the second codon in the wildtype gene must be either ?AU, A?U, or AU?, where the question mark indicates the deleted nucleotide; also, the third codon in the wildtype gene must be [U or C or A]GU, the fourth codon must be NUA, and the fifth must be A??. Similarly, because mutation 1 is an insertion in the second codon, the second codon in the wildtype gene must consist of three of the four bases AARU; also, the third codon in the wildtype gene must be AG?. The only consistent possibility for codon 2 is AAU, which codes for Asn. The third codon must be AGU, which codes for Ser. That leaves, for codon 4, the possibilities Val and Lys, and of these, only Val is coded by NUA, where N = G. Finally, the fifth codon must be AAR (Lys).
Hence the sequence of the nontranscribed strand of the wildtype gene, as nearly as can be determined from the data given, and separated into codons, is
Mutation 1 has a single nucleotide addition in codon 2 that yields a codon for Lys, so it must be an addition of a purine (R) at position 6 or of an A at either position 4 or position 5. Mutation 2 has a single nucleotide deletion that yields a codon for Ile, which means that the A at either position 3 or position 4 was deleted.
11.1 The enzymes are expressed constitutively because glucose is metabolized in virtually all cells. However, the levels of enzyme are regulated to prevent runaway synthesis or inadequate synthesis.
11.2 The hemoglobin mRNA has a very long lifetime before being degraded.
11.3 (a) Transcription, (b) RNA processing, (c) Translation.
11.4 The repressor gene need not be near because the repressor is diffusible. The trp repressor is one example in which the repressor gene is located quite far from the structural genes.
11.5 The mutant gene should bind more of the activator protein or bind it more efficiently. Therefore, the mutant gene should be induced with lower levels of the activator protein or expressed at higher levels than the wildtype gene, or both.
11.6 (a) ^-galactosidase (lacZ product), lactose permease (lacY product), and the transacetylase (lacA product), (b) The promoter region becomes inaccessible to RNA polymerase. (c) Repressor synthesis is constitutive.
11.7 At 37°C, both lacZ- and lacY- are phenotyipcally Lac-; at 30°C, the lacY- strain is Lac+, and the lacZ- strain is Lac-.
11.8 Inducers combine with repressors and prevent them from binding with the operator regions of the respective operons.
11.9 The cAMP concentration is low in the presence of glucose. Both types of mutations are phenotypically Lac-. The cAMP-CRP binding does not interfere with repressor binding because the DNA binding sites are distinct.
11.10 No; there is often a polar effect in which the relative amount of each protein synthesized decreases toward the 3' end of the mRNA.
11.11 With a repressor, the inducer is usually an early (often the first) substrate in the pathway, and the repressor is inactivated by combining with the inducer. With an aporepressor, the effector molecule is usually the product of the pathway, and the aporepressor is activated by the binding.
11.12 The attenuator is not a binding site for any protein, and RNA synthesis does not begin at an attenuator. An attenuator is strictly a potential termination site for transcription.
11.13 Steroid hormones pass through the cell membrane and combine with specific receptor proteins; thereupon they are transported to the nucleus and act as transcriptional activator proteins that stimulate transcription of their target genes.
11.14 The a cell would be unable to switch to the a mating type.
11.15 The a1 protein functions only in combination with the a2 protein in diploids; hence the MATa haploid cell would appear normal. However, the diploidMATaVMATa has the phenotype of an a haploid. This is because the genotype lacks an active a1-a2 complex to repress the haploid-specific genes as well as repressing a1, and the a1 gene product activates the a-specific genes.
11.16 The mutant diploid MA Ta/MA Ta has the phenotype of an a haploid. As in Problem 11.15, the genotype lacks an active a1-a2 complex to repress the haploid-specific genes as
well as a1, and the a1 gene product activates the a-specific genes.
11.17 In the absence of lactose or glucose, two proteins are bound: the lac repressor and CRP-cAMP. In the presence of glucose, only the repressor is bound.
11.18 (a) The strain is constitutive and has either the genotype lacl' or lacOc. (b) The second mutant must be lacZ' because no ^-galactosidase is made; and because permease synthesis is induced by lactose, the mutant must have a normal repressor-operator system. Therefore, the genotype is lacl+ lacO+ lacZ' lacY+. (c) Because the partial diploid is regulated, the operator in the operon with a functional lacZ gene must be wildtype. Hence the mutant in part (a) must be lacI'.
11.19 The lac genes are transferred by the Hfr cell and enter the recipient. No repressor is present in the recipient initially, so lac mRNA is made. However, the lacI gene is also transferred to the recipient, and soon afterward the repressor is made and lac transcription stops.
11.20 Amino acids with codons one step away from the nonsense codon, because their tRNA genes are the most likely to mutate to nonsense suppressors. For UGA, these are UGC (Cys), UGU (Cys), UGG (Trp), UUA (Leu), UCA (Ser), AGA (Arg), CGA (Arg), and GGA (Gly).
11.21 The mutant clearly lacks a general system that regulates sugar metabolism, and the most likely possibility is the cAMP-dependent regulatory system. Several mutations can prevent activity of this system. For example, the mutant either could make a defective CRP protein (which either fails to bind to the promoter or is unresponsive to cAMP) or could have an inactive adenyl cyclase gene and be unable to synthesize cAMP.
11.22 (a) No enzyme synthesis; the mutation is cis-dominant because lack of RNA polymerase binding to the promoter prevents transcription of the adjacent genes. (b) Enzymes synthesized; this one is cis-dominant because failure to bind the repressor cannot prevent constitutive transcription of the adjacent genes. (c) Enzymes synthesized; this mutation is recessive because the defective repressor can be complemented by a wildtype repressor in the same cell. (d) Enzymes synthesized; the mutation is recessive because the mutant repressor can be complemented by a wildtype repressor in the same cell. (e) No enzyme synthesis; this mutation is dominant because the repressor is always activated and all his transcription is prevented.
11.23 The wildtype mRNA reads
AAA CAC CAC CAU CAU CAC CAU CAU CCU GAC
which codes for Lys-His-His-His-His-His-His-His-Pro-Asp. The mutant mRNA reads AAA ACA CCA CCA UCA UCA CCA UCA UCC UGA C, which codes for Lys-Thr-Pro-Pro-Ser-Ser-Pro-Ser-Ser (the UGA is a normal stop codon). The expected phenotype of the mutant is His- because of lack of attenuation by low levels of histidine. That is, translation of the attenuator occurs (preventing efficient transcription of the operon) even when the level of histidine within the cell is very low. However, low levels of proline or serine will prevent translation of the attenuator and allow transcription of the histidine operon.
12.1 Autonomous developmental fates are controlled by genetically programmed changes within cells. Fates determined by positional information are controlled by interactions with morphogens or with other cells. Cell transplantation and cell ablation (destruction) are two kinds of surgical manipulation that help investigators distinguish between them.
12.2 Lineage mutations affect the developmental fates of one or more cells within a group of cells related by descent that normally undergo a defined pattern of development. Lineage mutations are detected by the occurrence of abnormal types or numbers of cells at characteristic positions within the embryo.
12.3 The Drosophila blastoderm is the flattened hollow ball of cells formed by cellularization of the syncytium formed by the early nuclear divisions. The fate map is a diagram of the blastoderm showing the locations of cells and their developmental fates. The existence of the fate map does not mean that development is autonomous, because many developmental decisions are made according to positional information in the blastoderm. In fact, most Drosophila developmental decisions are nonautonomous.
12.4 A female that is homozygous for a maternal-effect lethal produces defective eggs resulting in inviable embryos.
However, the homozygous genotype itself is viable.
12.5 Cells normally destined to die adopt some other developmental fate, with the result that there are supernumerary copies of one or more cell types.
12.6 In a homeotic mutation, certain segments develop in a manner characteristic of that of other segments, and therefore the lineage effect is one of transformation.
12.7 A sister-cell segregation defect would lead to both daughter cells differentiating as B or to both differentiating as C. A parent-offspring segregation defect would cause both daughter cells to differentiate as A.
12.8 Heterochronic mutations affect the timing of developmental events relative to each other; hence a mutation that uniformly slows or accelerates development is not a heterochronic mutation.
12.9 Because the Drosophila oocyte contains all the transcripts needed to support the earliest stages of development; the mouse oocyte does not.
12.10 In a loss-of-function mutation, the genetic information in a gene is not expressed in some or all cells. In a gain-of-function mutation, the genetic information is expressed at inappropriate times or in inappropriate cells. Both mutations can occur in the same gene: Some alleles may be loss-of-function and others gain-of-function. However, a particular allele must be one or the other (or neither).
12.11 Because the gene is necessary, a loss-of-function mutation will prevent the occurrence of the proper developmental fate. The allele will be recessive, however, because a normal allele in the homologous chromosome will allow the developmental pathway to occur.
12.12 Because the gene is sufficient, a gain-of-function mutation will induce the developmental fate to occur. The allele will be dominant because a normal allele in the homologous chromosome will not prevent the developmental pathway from occurring.
12.13 Developmental abnormalities in gap mutants occur across contiguous regions, resulting in a gap in the pattern of normal development. Developmental abnormalities in pairrule mutants are in even- or odd-numbered segments or parasegments, so the abnormalities appear in horizontal stripes along the embryo.
12.14 The proteins required for cleavage and blastula formation are translated from mRNAs present in the mature oocyte, but transcription of zygotic genes is required for gastrulation.
12.15 The material required for rescue is either mRNA transcribed from the maternal o+ gene or a protein product of the gene that is localized in the nucleus.
12.16 The transplanted nuclei respond to the cytoplasm of the oocyte and behave like the oocyte nucleus at each stage of development.
12.17 XDH activity is needed only at an early stage in development for the eyes to have wildtype pigmentation. The data are explained by a maternal effect in which the mal+/mal; ry/ry females in the cross transmit enought mal+ gene product to allow brief activity of XDH in the ry+/ry embryos. In theory, the malU+ product could be a transcription factor that allows ry+ transcription, but in fact the mal+ gene product participates in the synthesis of a cofactor that combines with the ry+ gene product to make an active XDH enzyme.
12.18 First, examine embryos of strains (c) and (d) to determine the wildtype expression patterns of kni and ftz. Next, cross strains (a) and (d) and strains (b) and (c). Intercross the F1 heterozygotes from the cross (a) x (d), and examine the embryos to detect changes in ftz expression. Intercross the F1 heterozygotes from the cross (b) x (c), and examine the embryos to detect changes in kni expression. Some of the F2 embryos from the (a) x (d) cross will show abnormal expression offtz, indicating that wildtype kni is required for proper ftz expression. The F2 embryos of (b) x (c) will show normal expression of kni, indicating that ftz is not required for proper kni expression.
13.1 Such people are somatic mosaics. This condition can be explained by somatic mutations in the pigmented cells of the iris of the eye or in their precursor cells.
13.2 Causes of a nonreverting mutation include (1) deletion of the gene, (2) multiple nucleotide substitutions in the gene, and (3) complex DNA rearrangement with more than one breakpoint in the gene. Deletions do not revert because the genetic information in the gene is missing, and the others do not revert because the nucleotide substitutions or DNA rearrangements are unlikely to be reversed precisely in one step.
13.3 Transitions are AT ' GC, GC ' AT, TA ' CG and CG ' TA (total four). Transversions are AT ' TA, AT ' CG, GC ' CG, GC ' TA, TA ' AT, TA ' GC, CG ' GC, and CG ' AT (total eight).
13.4 All nucleotide substitutions in the second codon position result in either an amino acid replacement or a terminator codon.
13.5 Both UUR and CUR code for leucine, and both CGR and AGR code for arginine (R stands for any purine, either A or G). Hence, in each of these codons, one of the possible substitutions in the first codon position does not result in an amino acid replacement. Even when an amino acid replacement does occur, the mutant protein may still be functional, because more than one amino acid at a given position in a protein molecule may be compatible with normal or nearly normal function.
13.6 Any single nucleotide substitution creates a new codon with a random nucleotide sequence, so the probability that it is a chain-termination codon (UAA, UAG, or UGA) is 3/64, or approximately 5 percent.
13.7 A frameshift mutagen will also revert frameshift mutations.
13.8 Exactly three. Otherwise, there would be a frameshift, and all downstream amino acids would be different.
13.9 No. Dominance and recessiveness are characteristics of the phenotype. They depend on the function of the gene and on which attributes of the phenotype are examined.
13.10 Photoreactivation and excision repair.
13.11 (a) Development arrests at A at both the high and low temperatures. (b) Development arrests at B at both the high and low temperatures. (c) Development arrests at A at the high temperature and arrests at B at the low temperature. (d) Development arrests at B at the high temperature and arrests at A at the low temperature.
13.12 The mutation frequency in the Lac- culture is at least 0.90 x 10-8 = 9 x 10-9. Because the mutation rate to a nonsense suppressor is independent of the presence of a suppressible mutation (in this case, in the Lac- mutant), the mutation rate in the original Lac+ culture is also at least 9 x 10-9. The qualifier at least is important because not all nonsense suppressors may suppress the mutation in the particular Lac- strain (that is, some suppressors may insert an amino acid that still results in a nonfunctional enzyme.)
13.13 (a) 5 x 10-5 is the probability of a mutation per generation, and so 1 - (5 x 10-5) = 0.99995 is the probability of no mutation per generation. The probability of no mutations in 10,000 generations equals (0.99995)10,000 = 0.61. (b) The average number of generations until a mutation occurs is 1/(5 x 10-5) = 20,000 generations.
13.14 At 1 Gy, the total mutation rate is double the spontaneous mutation rate. Because a dose of 1 Gy increases the rate by 100 percent, a dose of 0.5 Gy increases it by 50 percent and a dose of 0.1 Gy by 10 percent.
13.15 Amino acid replacements at many positions in the A protein do not eliminate its ability to function; only amino acid replacements at critical positions that do eliminate tryptophan synthase activity are detectable as Trp- mutants.
13.16 The possible lysine codons are AAA and AAG, and the possible glutamic acid codons are GAA and GAG. Therefore, the mutation results from an AT ' GC transition in the first position of the codon.
13.17 Six amino acid replacements are possible because the codons UAY (Y stands for either pyrimidine, C or U) can mutate in a single step to codons for Phe (UUY), Ser (UCY), Cys (UGY), His (CAY), Asn (AAY), or Asp (GAY).
13.18 Single nucleotide substitutions that can account for the amino acid replacements are (1) Met (AUG) to Leu (UUG), (2) Met (AUG) to Lys (AAG), (3) Leu (CUN) to Pro (CCN), (4) Pro (CCN) to Thr (ACN), and (5) Thr (ACR) to Arg (AGR), where R stands for any purine (A or G) and N for any nucleotide. Substitutions (1), (2), (4), and (5) are transversion, and substitution (3) is a transition. Because 5-bromouracil primarily induces transitions, the
Leu ' Pro replacement is expected to be the most frequent.
13.19 The sectors are explained by a nucleotide substitution in only one strand of the DNA duplex. Replication of this heteroduplex yields one daughter DNA molecule with the wildtype sequence and another with the mutant sequence. Cell division produces one Lac+ and one Lac- cell, located side by side. Because the cells do not move on the agar surface, the Lac+ cells form a purple half-colony, and the Lac- cells form a pink half-colony.
13.20 The a and b gene products may be subunits of a multimeric protein. A few proteins with compensatory amino acid replacements can interact to give wildtype activity, but most mutant proteins cannot interact in this way with each other or with wildtype.
13.21 Some of the mutagenized Pl particles acquire mutations in the a gene and become a- b+. Therefore, the next step is to select b+ transductants and screen among these for ones that are a-. The procedure depends on close linkage.
13.22 (a) All progeny would be able to grow in the absence of arginine. (b) The genotype of the diploid is Su+/Sw arg+/arg", where Su+ indicates the ability to suppress, and Sw indicates inability to suppress. The resulting haploid spores are 1/4 Su+ arg+, 1/4 Su+ arg-, 1/4 Sw arg+, and 1/4 Sw arg-. The first three are Arg+ and the last Arg-, so the ratio is 3 Arg+: 1 Arg-. c The diploid genotype is Su+ arg~/Sw arg+. The haploid spores are in the proportions 0.05 Su+ arg+, 0.45 Su+ arg-, 0.45 Sw arg+, and 0.05 Sw arg'. Therefore, 95 percent of the progeny are Arg+ and 5 percent are Arg-.
13.23 Asci C, D, and F might arise from gene conversion at the m locus as follows. Branch migration through m results in two regions of heteroduplex DNA. In ascus type C, one region is replicated before repair, leaving + and m alleles intact; the other region is repaired before replication (+ to m), yielding two m alleles (therefore a 3 : 5 ratio). In ascus type D, both heteroduplex regions are replicated before repair, leaving all alleles intact but changing their order in the ascus (therefore a 3 : 1 : 1 : 3 ratio). In ascus type F, both heteroduplex regions are repaired (+ to m) before replication, yielding four m alleles (therefore a 2 : 6 ratio).
14.1 Inheritance is strictly maternal, so yellow is probably inherited cytoplasmically. A reasonable hypothesis is that
the gene for yellow is contained in chloroplast DNA.
14.2 Pollen is not totally devoid of cytoplasm, and occasionally it contains a chloroplast that is transmitted to the zygote.
14.3 This trait is maternally inherited, so the phenotype is that of the mother. The progeny are (a) green, (b) white, (c) variegated, and (d) green.
14.4 With X-linked inheritance, half of the F2 males (1/4 of the total progeny) would be white. The observed result is that all of the F2 progeny are green.
14.5 If the daughter chloroplasts segregate at random, both of them will go to the same pole half the time. For all four products to contain a descendant chloroplast, the segregation has to be perfect in three divisions, which occurs with probability (1/2)3 = 1/8. Therefore, the probability that at least one cell lacks a descendant chloroplast is 1 - 1/8 = 7/8.
14.6 The chloroplast from the mt strain is preferentially lost, and the mt alleles segregate 1 : 1. The expected result is 1/2 a+ b- mt+ and 1/2 a+ b mt.
14.7 Petite, because neither parent contributes functional mitochondria.
14.8 Because every tetrad shows uniparental inheritance, it is likely that the trait is determined by factors transmitted through the cytoplasm, such as mitochondria. The 4 : 0 tetrads result from diploids containing only mutant factors, and the 0 : 4 tetrads result from diploids containing only wildtype factors.
14.9 In autogamy, one product of meiosis becomes a homozygous diploid nucleus, so half of the Aa cells that undergo autogamy become AA and half become aa.
14.10 The mating pairs are 1/4 AA x AA, 1/2 AA x aa, and 1/4 aa x aa. These produce only AA, Aa, and aa progeny, respectively, so the progeny are 1/4 AA, 1/2 Aa, and 1/4 aa. In autogamy, the Aa genotypes become homozygous for either A or a, so the expected result of autogamy is 1/2 AA and 1/2aa.
14.11 The only functional pollen is a B. All the progeny of the AA bb plants are Aa Bb, and those of the aa BB plants are aa BB.
14.12 The F1 females will be heterozygous for nuclear genes, and possibilities (1) and (2) imply that these females should have some viable sons. The fact that all progeny are female supports the hypothesis of cytoplasmic inheritance.
14.13 Because the progeny coiling is determined by the genotype of the mother, the rightward-coiling snail has genotype ss. To explain the rightward coiling, the mother of this parent snail would have had the + allele and must in fact have been +/s.
14.14 Among Africans, the average time traversed in tracing the ancestry from a present-day mitochondrial DNA back to a common ancestor and forward again to another present-day mitochondrial DNA is between 78 x 1500 years and 78 x 3000 years, or 117,000 to 234,000 years; the backward and foward times are equal, so the person having the common ancestor of the mitochondrial DNA lived between 117,000/2 = 58,500 years ago and 234,000/2 = 117,000 years ago. Among Asians, similar calculations result in a range of 43,500 to 87,000 years for the common ancestor of mitochondrial DNA. Among Caucasians, the calculated range is 30,000 to 60,000 years.
14.15 The accompanying table shows the percent of restriction sites shared by each pair of mitochondrial DNA molecules.
The entries in the table may be explained by example. Molecules 3 and 4 together have 9 restriction sites, of which 1 is shared, so the entry in the table is 1/9 = 11 percent. The diagonal entries of 100 mean that each molecule shares 100 percent of the restriction sites with itself. The two most closely related molecules are 2 and 5 (in fact, they are identical), so these also must be grouped together. The two next closest are 1 and 3, so these also must be grouped together. The 2-5 group shares an average of 10 percent restriction sites with molecule 4, and the 1-3 group shares an average of 10.5 percent restriction sites [that is, (10 + 11)/2] with molecule 4. Therefore, the 2-5 group and the 13 group are about equally distant from molecule 4. The diagram shows the inferred relationships among the species. Data of this type do not indicate where the ''root" should be positioned in the tree diagram.
15.1 Adults 2/3 Aa and 1/3 aa. It is not necessary to assume random mating, because the only fertile matings are Aa x Aa. These produce 1/4 AA, 1/2 Aa, and 1/4 aa zygotes, but the AA zygotes die, leaving 2/3 Aa and 1/3 aa adults.
15.2 The harmful recessive allele will have a smaller equilibrium frequency in the X-linked case, because the allele is exposed to selection in hemizygous males.
15.3 The 10 AA individuals have 20 A alleles, the 15 Aa have 15 A and 15 a, and the 4 aa have 8 a. Altogether there 35 A and 23 a alleles, for a total of 58. The allele frequency of A is 35/58 = 0.603, and that of a is 23/58 = 0.397.
15.4 The numbers of alleles of each type are A, 2 + 2 + 5 + 1 = 10; B, 5 + 12 + 12 + 10 = 39; C, 10 + 5 + 5 + 1 = 21. The total number is 70, so the allele frequencies are A, 10/70 = 0.14; B, 39/70 = 0.56; and C, 21/70 = 0.30.
15.5 (a) Two bands indicate heterozygosity for two different VNTR alleles. (b) A person who is homozygous for a VNTR allele will have a single VNTR band. (An alternative explanation is also possible: Some VNTR alleles are present in small DNA fragments that migrate so fast that they are lost from the gel; a heterozygote for one of these alleles will have only one band—the one corresponding to the larger DNA fragment—because the band corresponding to the smaller DNA fragment will not be detected. (c) VNTR alleles are codominant because both are detected in heterozygotes. (d) Because none of the bands in the four pairs of parents have the same electrophoretic mobility, the parents of each child can be identified as the persons who share a single VNTR band with the child. Child A has parents 7 and 2; child B has parents 4 and 1; child C has parents 6 and 8; and child D has parents 3 and 5.
15.6 Allele 5.7 frequency = (2 x9 + 21 + 15 + 6)/200 = 0.30; allele 6.0 frequency = (2 x 12 + 21 + 18 + 7)/200 = 0.35; allele 6.2 frequency = (2 x 6 + 15 + 18 + 5)/200 =
0.25; allele 6.5 frequency =(2 x 1 + 6 + 7 + 5) 200 = 0.10.
15.7 If the genotype frequencies satisfy the Hardy-Weinberg principle, they should be in the proportions p2, 2pq, and q2. In each case, p equals the frequency of AA plus one-half the frequency of Aa, and q = 1-p. The allele frequencies and expected genotype frequencies with random mating are (a) p = 0.50, expected: 0.25, 0.50, 0.25; (b) p = 0.635, expected: 0.403, 0.464, 0.133; (c) p = 0.7, expected : 0.49, 0.42, 0.09; (d) p = 0.775, expected: 0.601, 0.349, 0.051; (e) p = 0.5, expected: 0.25, 0.50, 0.25. Therefore, only (a) and (c) fit the Hardy-Weinberg principle.
15.8p2 = 0.09, and sop = (0.09)1/2 = 0.30. All other alleles have a combined frequency of 1 - 0.30 = 0.70.
15.9 The frequency of the recessive allele is q = (1/14,000)1/2 = 0.0085. Hencep = 1 - 0.0085 = 0.9915, and the frequency of heterozygotes is 2pq = 0.017 (about 1 in 60).
15.10 Because the frequency of homozygous recessives is 0.10, the frequency of the recessive allele q is (0.10)1/2 =0.316. This meansp = 1 -0.316 =0.684, so the frequency of heterozygotes is 2 x 0.316 x 0.684 = 0.43. Because the proportion 0.90 of the individuals are nondwarfs, the frequency of heterozygotes among nondwarfs is 0.43/0.90 = 0.48.
15.11 (a) Because the proportion 0.60 can germinate, the proportion 0.40 that cannot are the homozygous recessives. Thus the allele frequency q of the recessive is (0.4)1/2 = 0.63. This implies thatp = 1 0.63 = 0.37 is the frequency of the resistance allele. (b) The frequency of heterozygotes is 2pq = 2 x 0.37 x 0.63 = 0.46. The frequency of homozygous dominants is (0.37)2 = 0.14, and the proportion of the surviving genotypes that are homozygous is 0.14/0.60= 0.23.
15.12 The bald and nonbald alleles have frequencies of 0.3 and 0.7, respectively. Bald females are homozygous and occur in the frequency (0.3)2 = 0.09; nonbald females occur with frequency 0.91. Bald males are homozygous or heterozygous and occur with frequency (0.3)2 + 2(0.3)(0.7) = 0.51; nonbald males occur with frequency 0.49.
15.13 IAIA, (0.16)2 = 0.026; IAIO, 2(0.16)(0.74) = 0.236; IBIB, (0.10)2 = 0.01; IBIO, 2(0.10)(0.74) = 0.148;IOlO, (0.74)2 = 0.548; lAlB, 2(0.16)(0.10) = 0.032. The phenotype frequencies are O, 0.548; A, 0.026 + 0.236 = 0.262;B, 0.01 + 0.148 = 0.158; AB, 0.032.
15.14 (a) There are 17 A and 53 B alleles, and the allele frequencies are A, 17/70 = 0.24; B, 53/70 = 0.76. (b) The expected genotype frequencies are AA, (0.24)2 = 0.058;AB, 2(0.24)(0.76) = 0.365; BB, (0.76)2 = 0.578. The expected numbers are 2.0, 12.8, and 20.2, respectively.
15.15 The frequency in males implies that q = 0.2. The expected frequency of the trait in females is q2 = 0.0004. The expected frequency of heterozygous females is 2pq = 2(0.98)(0.02) = 0.0392.
15.16 The frequency of the yellow allele y is 0.2. In females, the expected frequencies of wildtype (+ + and y +) and yellow (yy) are (0.8)2 + 2(0.8)(0.2) = 0.96 and (0.2)2 = 0.04, respectively. Among 1000 females, there would be 960 wildtype and 40 yellow. The phenotype frequencies are very different in males. In males, the expected frequencies of wildtype (+ Y) and yellow (y Y) are 0.8 and 0.2, respectively, where Y represents the Y chromosome. Among 1000 males, there would be 800 wildtype and 200 yellow.
15.17 The frequency of heterozygotes is smaller by the amount 2pqF, in which F is the inbreeding coefficient.
15.18 Inbreeding is suggested by a deficiency of heterozygotes relative to the 2pq expected with random mating. The suggestive populations are (d) with expected heterozygote frequency 0.349 and (e) with expected heterozygote frequency 0.5.
15.19 The allele frequency is the square root of the frequency of affected offspring among unrelated individuals, or (8.5 x 10-6)1 2 = 2.9 x 10-3. With inbreeding, the expected frequency of homozygous recessives is q2(l - F) + qF, in which F is the inbreeding coefficient. When F = 1/16, the expected frequency of homozygous recessives is (2.9 x 10- 3)2(1 - 1/16) + (2.9 x 10-3)(1/16) = 1.9 x 10-4;when F = 1/64, the value is 5.3 x 10-5.
15.20 (a) Here F = 0.66, p = 0.43, and q = 0.57. The expected genotype frequencies are AA, (0.43)2(1 - 0.66) + (0.43) (0.66) = 0.347; AB, 2(0.43)(0.57)(1 0.66) = 0.167;BB (0.57)2(1 - 0.66) + (0.57)(0.66) = 0.487. (b) With random mating, the expected genotype frequencies are AA, (0.43)2 = 0.185; AB, 2(0.43)(0.57) = 0.490; BB, (0.57)2 = 0.325.
15.21 Use Equation (11) with n = 40, p0q0 = 1 (equal amounts inoculated), and p40q40 = 0.35/0.65. The solution for w
is w = 0.9846, which is the fitness of strain B relative to strain A, and the selection coefficient against B is s = 1 - 0.9846 = 0.0154.
15.22 (a) A fixed; (b) A lost; (c) stable equilibrium; (d) A lost.
15.23 An Aa female produces 1/2 A and 1/2 a gametes, and these yield heterozygous offspring if they are fertilized by a and by A, respectively. The overall probability of a heterozygous offspring is (1/2)q + (1/2)p = (1/2)(p + q) = 1/2.
15.24 The frequency of each allele is 1 n, and there are n possible homozygotes. Therefore, the total frequency of homozygotes is n x (1/n)2 = 1 /n, which implies that the total frequency of heterozygotes is 1 -1 /n.
15.25 The frequencies are calculated as in the HardyWeinberg principle except that, for apparently homozygous alleles, 2p is used instead ofp2. With regard to the genotype (6, 9.3) for HUMTHO1, (10, 11) for HUMFES, (6, 6) for
D12S67, and (18, 24) for D1S80, the calculation is 2 x 0.230 x 0.310 x 2 x 0.321 x 0.373 x 2 x 0.324 x 2 x 0.293 x 0.335 = 0.0043, or about one in 230 people. With regard to the genotype (8, 10) for HUMTHO1, (8, 13) for HUMFES, (1, 2) for D12S67, and (19, 19) for D1S80, the calculation is 2 x 0.105 x 0.002 x 2 x 0.007 x 0.066 x 2 x 0.015 x 0.005 x 2 x 0.011 = 1.28 x 10-12, or about one in 8 x 1011 people. The point is that DNA typing is at its most powerful in discriminating among people who carry rare alleles.
16.1 Genotype-environment interaction.
16.2 Since the variance equals the square of the standard deviation, the only possibilities are that the variance equals either 1 or 0.
16.3 (a) The mean or average. (b) The one with the greater variance is broader. (c) 68 percent within one standard deviation of the mean, 95 percent within two.
16.4 Broad-sense heritability is the proportion of the phenotypic variance attributable to differences in genotype, or H2 = / <J* , which includes dominance effects and interactions between alleles. Narrow-sense heritability is the
proportion of the phenotypic variance due only to additive effects, or <7'' , which is used to predict the resemblance between parents and offspring in artificial selection. If all allele frequencies equal 1 or 0, both heritabilities equal zero.
16.5 The F1 generation provides an estimate of the environmental variance. The variance of the F2 generation is determined by the genotypic variance and the environmental variance.
16.6 For fruit weight, <7|T /rr"- = 0.13 is given. Because fr= + <7|T , then H2 = /rr" = 1 - 0.13 = 0.87. Therefore, 87 percent is the broad-sense heritability of fruit weight. The values of H2 for soluble-solid content and acidity are 91 percent and 89 percent, respectively.
16.7 Estimates are as follows: mean, 104/10 = 10.4; variance, 24.4/(10 - 1) = 2.71; standard deviation (2.71)1 2 = 1.65.
16.8 An IQ of 130 is two standard deviations above the mean, so the proportion greater equals (1 - 0.95)/2 = 2.5 percent; 85 is one standard deviation below the mean, so the proportion smaller equals (1 - 0.68)/2 = 16 percent; the proportion above 85 equals 1 - 0.16 = 84 percent, which can be calculated alternatively as 0.50 + 0.68/2 = 0.84.
16.9 (a) To determine the mean, multiply each value for the pounds of milk produced (using the midpoint value) by the number of cows, and divide by 304 (the total number of cows) to obtain a value of the mean of 4032.9. The estimated variance is 693,633.8, and the estimated standard deviation, which is the square root of the variance, is 832.8. (b) The range that includes 68 percent of the cows is the mean minus the standard deviation to the mean plus the standard deviation, or 4033 - 833 = 3200 to 4033 + 833 = 4866. (c) Rounding to the nearest 500 yields 30005000. The observed number in this range is 239; the expected number is 0.68 x 304 = 207. (d) For 95 percent of the animals, the range is within two standard deviations, or 2367-5698. Rounding off to the nearest 500 yields 25006000. The observed number is 296, which compares well to the expected number of 0.95 x 304 = 289.
16.10 (a) Because 15 to 21 is the range of one standard deviation from the mean, the proportion is 68 percent. (b) Because 12 to 24 is the range of two standard deviations from the mean, the proportion is 95 percent. (c) A phenotype of 15 is one standard deviation below the mean; because a normal distribution is symmetrical about the mean, the proportion with phenotypes more than one standard deviation below the mean is (1 - 0.68)/2 =0.16. (d) Those with 24 leaves or greater are more than two standard deviations above the mean, so the proportion is (1 - 0.95)/2 = 0.025. (e) Those with 21 to 24 leaves are between one and two standard deviations above the mean, and the expected proportion is 0.16 - 0.025 = 0.135 (because 0.16 have more than 21 leaves, and 0.25 have more than 24 leaves).
16.11 (a) The genotypic variance is 0 because all F1 mice have the same genotype. (b) With additive alleles, the mean of the F1 generation will equal the average of the parental strains.
16.12 Among the F1, the total variance equals the environmental variance, which is given as 1.46. Among the F2, the total variance, given as 5.97, is the sum of the environmental variance (1.46) and the genotypic variance. Thus the
genotypic variance is 5.97 - 1.46 = 4.51. The broad-sense heritability is the genotypic variance (4.51) divided by the total variance (5.97), or 4.51/5.97 = 0.76.
16.13 From the F, data, <J|T = (2 f = 4; from the F2 data, + <7|T = (3)2 = 9; hence =9-4 = 5, and so H2 = 5/9 = 56 percent.
16.14 Using Equation (4), we find that the minimum number of genes affecting the trait is D2l%'Ji = (17.6)2/ (8x0.88)=44.
16.15 =0.0570 -0.0144 = 0.0426 and D= 1.689 -(- 0.137) = 1.826. The minimum number of genes is estimated as n = (1.826)2/(8 x 0.0426) = 9.8.
16.16 Use Equation (6) with M* - M = 50 grams in each generation of selection, with M initially 700 grams. The expected gain in each generation of selection is 0.8 x 50 = 40 grams, so after five generations, the expected mean weight is 700 + 5 x 40 = 900 grams.
16.17 Use Equation (6) with M = 10.8, M* = 5.8, andM = 8.8. The narrow-sense heritability is estimated as h2 = (8.8 - 10.8)/(5.8 - 10.8) = 0.40.
16.18 (a) In this case, M = 10.8, M* = 5.8, andM = 9.9. The estimate of h2 is (9.9 - 10.8)/(5.8 - 10.8) = 0.18. (b) The result is consistent with the earlier result because of the differences in the variance. In the previous case, the additive variance was 0.4 x 4 = 1.6. In the present case, the phenotypic variance is 9. If the additive variance did not change, then the expected heritability would be 1.6/9 = 0.18, which is that observed. This problem has been idealized somewhat;in real examples, the additive variance would also be expected to change with a change in environment.
16.19 The offspring are genetically related as half-siblings, and Table 16.3 says that the theoretical correlation coefficient is h2/4.
16.20 Table 16.3 gives the correlation coefficient between first cousins as h2/8, and so h2 = 8 x 0.09 = 72 percent.
16.21 The mean weaning weight of the 20 individuals is 81 pounds, and the standard deviation is 13.8 pounds. The mean of the upper half of the distribution can be calculated by using the relationship given in the problem, which says that the mean of the upper half of the population is 81 + 0.8 x 13.8 = 92 pounds. Using Equation (6), we find that the expected weaning weight of the offspring is 81 + 0.2(92 - 81) = 83.2 pounds.
16.22 One approach is to use Equation (4) with D = In- 0 = 2/7, from which n = (2n)2/8 , or fJii = n/2. A more direct approach uses the fact that with unlinked, additive genes, the total genetic variance is the summation of the genetic variance resulting from each gene. With 1 gene, the genetic variance equals 1/2, so with n genes, the total genetic variance equals n/2.
16.23 (a) This is a nonrecombinant class and so should have the genotype Q/q for the QTL affecting egg lay; the average should be (300 + 270)/2= 285 eggs per hen. (b) This is a crossover class of progeny. In view of the relative distances of the two intervals of 0.07 : 0.03, 7/10 of the crossovers should occur in the region Cx-Q and 3/10 in the region Q-Dx. The expected genotypes with respect to the QTL are 7/10 q/q (crossing-over in the Cx-Q region) and 3/10 Q / q (crossing-over in the Q-Dx region). The average phenotype is expected to be (7/10) x 270 + (3/10) x 285 = 274.5 eggs per hen. (c) This is also a crossover class, but it is the reciprocal of that in the previous part. In this case, the expected genotypes with respect to the QTL are 7/10 Q/q (crossing-over in the Cx-Q region) and 3/10 q/q (crossing-over in the Q-Dx region). The expected average egg lay is (7/10) x 285 + (3/10) x 270 = 280.5 eggs per hen. (d) These are again nonrecombinants with QTL genotype q/q, and the expected egg lay is 270 eggs per hen.
16.24 (a) The rare genetic disease is transmitted as a simple Mendelian dominant. Genotype Q/q is affected and q/q is not. Because the disease is rare, the frequency of Q/Q homozygous genotypes is negligible. (b) Male IV-2 has the genotype Q/q, so the probability that he has an affected daugher is 1/2. (c) Male IV-2 has the genotype H/H for gene 2; with random mating, the probability that his daughter will have the genotype H/F is equal to the probability that a randomly chosen egg contains the F allele, which is given in the population-frequency table as 0.05, or 5 percent. (d) The genotypes of the individuals in the pedigree are as shown in the accompanying table. At this stage in the analysis, we do not yet know anything about linkage, but we can, in some cases, specify the linkage phase in terms of which alleles were transmitted together in a single gamete. When known, the alleles transmitted in a single gamete are grouped together in parentheses. Also, the order in which the genes are listed is arbitrary. (The two columns on the right are referred to in the next part of the analysis.)
I-1 (q A E)/(q A E) I-2 Q/q ; C/D ; F/G II-1 (q A E)/(Q D F) NRC II-2 (q A H)/(q B H) II-3 (q A E)/(q C G) NRC
(q A G)/(q A H)
II-5 (q A E)/(Q D G) NRC
II-6 (q A H)/(q B H)
II-7 (q A E)/(Q D F) NRC II-8 (q A E)/(q C G) NRC II-9 q/q ; A/B ; E/G
III-1 (q B H)/(q B H)
III-2 (q B H)/(Q D E) NRC REC
III-3 (q A H)/(Q D E) NRC REC
III-4 (q A H)/(q A F) NRC REC
III-5 (q A G)/(q A E) NRC
III-6 (q A H)/(q A G) REC
III-7 (q B H)/(q A E) NRC NRC
III-8 (q A H)/(Q D G) NRC NRC
III-9 q/q ; B/C ; E/H
III-10 (q B H)/(q A G) NRC REC
III-11 (q A E)/(q A E) NRC
III-12 (q B G)/(q A G) REC
III-13 q/q ; B/C; E/G Undeterminable
IV-1 (q B H)/(Q D E) NRC NRC IV-2 (q B H)/(Q D H) NRC REC IV-3 (q B H)/(q B E) NRC REC IV-4 (q B E)/(Q D G) NRC NRC IV-5 (q C H)/(q A H) NRC NRC IV-6 (q C H)(Q D G) NRC NRC
e There is evidence of linkage between genes 1 and 3, In particular, the allele Q is always transmitted with allele D, starting with male 1-2 at the top of the pedigree. If we assume that this male has the genotype (Q D)/(q C), then throughout the pedigree, there are 17 individuals in which recombination between genes 1 and 3 could have been detected. These cases are identified in the third column of the foregoing table. All 17 are nonrecombinant, so the recombination fraction between genes 1 and 3 in these data is 0/17. As far as one can discern from the pedigree alone, the Q and D alleles could be identical. However, the population-frequency table gives the allele frequency of D as 5 percent, and the text says that the Q allele is rare. Hence most D alleles in the population must be on chromosomes carrying q, so that D and Q are separable. Turning to genes 1 and 2, there are 17 people in whom recombination could be detected. These are identified in the fourth column in the foregoing table. Among the 17 cases, one is undeterminable, 8 are recombinant, and 8 are nonrecombinant. Genes 1 and 2 are therefore unlinked. They are probably on different chromosomes, but they could also be very far apart on the same chromosome.
17.1 There are numerous chemoreceptors in E. coli, each of which interacts with a specific set of chemosensors for different small molecules. Mutations in a particular chemoreceptor affect the chemotactic response to all molecules whose chemosensors feed into the chemoreceptor. Hence cells with mutant chemosensors are multiply nonchemotactic.
17.2 The mutant would be unable to make P-CheA and, therefore, P-CheY, so the phenotype would be that of continuous swimming.
17.3 Extreme cheB mutants lack the CheB methylase and have high levels of chemoreceptor methylation; their phenotype is that of continuous tumbling, because the methylated chemoreceptors require high concentrations of attractant to be stimulated (which is why they swim in response to very strong attractant stimuli). Extreme cheR mutants cannot methylate the chemoreceptors and so are highly sensitive to stimuli; their phenotype is that of continuous swimming, but very strong repellent stimuli overcome the effects of undermethylation and result in tumbling.
17.4 Phosphate groups are transferred among several key components (CheA, CheY, and CheB) in the chemotaxis- signaling pathway. Stimulated chemoreceptors promote autophosphorylation of CheA, in which a phosphate group is transferred from ATP to a specific amino acid residue in cheA. The phosphate group can be transferred readily from P-CheA to CheB or CheY. The P-CheY promotes clockwise flagellar rotation and causes the bacteria to tumble. If phosphate transfers were inhibited, the lack of P-CheY would result in failure to switch from swimming behavior to tumbling behavior, and chemotaxis would be impossible. Because protein phosphorylation is common to all of the chemoreceptors, inhibition of phosphorylation would result in a generally nonchemotactic phenotype.
17.5 Male mating success is more likely to be affected by mutations in per because these mutations affect the courtship song, and it is the male who must sing acceptably to the female for successful courtship.
17.6 Mate a mutagen-treated male with an attached-X female and select the male progeny. Each male progeny carries a different, mutagen-treated X chromosome. Mate each of the male progeny with an attached-X female; the progeny of this mating consist of attached-X daughters and XY sons, and the X chromosome in each son is a replica of the single mutagen-treated X chromosome present in the father.
17.7 Either genetic or molecular tests would reveal whether the mutant allele is in per. The first step in genetic analysis is genetic mapping. If the mutant does not map near the chromosomal location of per, then it must be a different gene; if it does map near per, then a complementation test should be carried out with per0. A molecular analysis would entail determining the DNA sequence of all or part of the per gene and comparing it with wildtype. If the sequences are identical, then the mutation must be in a gene different from per; if the sequences differ, then separate tests must be carried out to demonstrate that the mutant phenotype results from the difference in DNA sequence. The new mutation could easily be in a gene different from per, because courtship behavior is a complex trait influenced by many different genes.
17.8 (a) A short-period rhythm of approximately 40 seconds.(b) A long-period rhythm of approximately 90 seconds. (c) Because perS shortens the period and perL lengthens it, the perS/perL heterozygote would be expected to exhibit an intermediate period of approximately (40 + 90)/2 = 65 seconds, which is very close to the period of the wildtype courtship song.
17.9 A fly with a mutation that eliminates only one of the wavelength preferences would be classified as specifically
nonphototactic. The mutation might affect the photopigments necessary for the detection of light of a particular wavelength, or it might affect the photoreceptor cells containing the particular visual pigment. A fly with a mutation that eliminates the phototactic response to both wavelengths is generally nonphototactic. The mutation might affect components of the visual pathway common to both 480-nm and 370-nm light—for example, a component common to all photoreceptor cells that is necessary for stimulation by light.
17.10 It is an example of sensory adaptation, because intense light of a given wavelength results in a temporarily decreased response to light of the same wavelength.
17.11 It is unlikely that an inbred strain would produce selected lines that differed significantly in maze-learning ability (or in any other trait), because genetic variation is essential for the success of artificial selection.
17.12 Variability in the sense of smell in the initial population might result in variability in the ability of the rats to detect the food at the end of the maze. Therefore, differences in
maze-learning ability may result from differences in the ability of the animals to detect the food used to motivate them to run through the maze. In this example, maze-learning ability is not a good measure of intelligence. You could control for variation in the sense of smell by using odorless food, masking the smell of the food, smothering the entire maze with the smell of the food, or using a nonfood goal.
17.13 It is an example of genotype-environment interaction, because the relative performances of the two populations are reversed merely by changing the environment in which the test is carried out.
17.14 The waltzer allele is recessive to the wildtype allele.
17.15 Under conditions in which behavioral choices are strongly influenced by the ability to distinguish among different colors—for example, the ability to distinguish red and green in obeying traffic lights.
17.16 The test would take advantage of the size differences between the wildtype and mutant alleles. The DNA of parents and child would be digested with a restriction enzyme that cleaves the DNA at sites flanking the CAG repeat. The resulting fragments would be separated by size with electrophoresis, and the fragments of interest would be identified by hybridization using a radioactive probe for the CAGcontaining fragment. The mutant allele should be longer than the wildtype allele because of the greater number of CAG repeats. Because the mutant allele is dominant, any mating between heterozygous mutant and homozygous wildtype is expected to result in 50 percent affected offspring.
17.17 This example illustrates some of the complexities of behavior genetics in human beings. The correlation between the MAO gene and the behavioral disorder in the large pedigree is suggestive. However, single pedigrees may be misleading, because a correlation with a genetic marker may result from chance. Even if the disorder in this example is genetically determined, the cause need not be MAO. The disorder may result from a different gene that is genetically linked to MAO, or the large pedigree may also contain other genes influencing the disorder, and the MAO mutation may be only a contributing factor. The finding that unrelated men with similar behavioral disorders may have abnormal levels of monoamine oxidase is also suggestive. However, without data on the distribution of enzyme levels among normal men and among those with the behavioral disorder, it is impossible to evaluate the strength of the evidence.
The Molecular Basis of Heredity and Variation
51.1 Classify each of the following statements as true or false.
(a) Each gene is responsible for only one visible trait.
(b) Every trait is potentially affected by many genes.
(c) The sequence of nucleotides in a gene specifies the sequence of amino acids in a protein encoded by the gene.
(d) There is one-to-one correspondence between the set of codons in the genetic code and the set of amino acids encoded.
51.2 Is the following statement correct? " Heredity is solely responsible for the manifestation of anemia." Explain.
51.3 When a human trait is genetic in origin (for example, a disorder caused by an inborn error of metabolism), does this mean that the trait cannot be influenced by the environment?
51.4 What is meant by the term "pleiotropy" and what is a "pleiotropic effect"? What is the relevance of sickle-cell anemia to this phenomenon?
51.5 What facts known prior to Avery, MacLeod, and McCarty's experiment indirectly suggested that DNA might play a role in the genetic material?
51.6 Prior to the experiments of Avery, MacLeod, and McCarty and of Hershey and Chase, why did most biologists and biochemists believe that proteins were probably the genetic material?
xS1.7 Each of the following statements pertains to the experiments of Avery, MacLeod, and McCarty on Streptococcus pneumoniae. Classify each statement as true or false. Which single statement is a conclusion that can be drawn from the experiments themselves?
(a) The polysaccharide capsule is required for virulence (the ability to infect and kill mice).
(b) The polysaccharide capsule is not important for virulence.
(c) DNA encodes the sequence of sugars in the polysaccharide capsule.
(d) DNA is the polysaccharide capsule.
(e) Proteins are not the molecules that carry the hereditary information for synthesis of the polysaccharide capsule.
51.8 What critical finding in Hershey and Chase's experiment enabled them to interpret the results as a proof that the genetic material in bacteriophage T2 is DNA?
51.9 Define the following terms: replication, transcription, translation, mutation, natural selection.
51.10 What can we infer about the function of DNA from its structure?
51.11 Is it correct to say that DNA is always the genetic material?
51.12 One of the early models for the structure of DNA was the so-called tetranucleotide hypothesis. According to this hypothesis, one of each of the four nucleotides containing adenine, thymine, guanine, and cytosine joined together to form a covalently linked planar unit (the "tetranucleotide"); these units were, in turn, linked together to yield a repeating polymer of the tetranucleotide. Why could a molecule with this structure never serve as the genetic material?
51.13 What are the major chemical components found in DNA?
51.14 The repeating unit of DNA is:
(a) a ribonucleotide.
(b) an amino acid.
(c) a nitrogenous base.
(d) a sugar, deoxyribose.
(e) none of the above.
51.15 The two strands of a duplex DNA molecule are said to be complementary in base sequence. Which of the following statements are implied by the complementary nature of the strands?
(a) The two strands have the same sequence of bases.
(b) The base A in one strand is paired with a T in the other strand.
(c) One strand consists of A's paired with T's; the other strand consists of G's paired with C's.
(d) The base G in one strand is paired with a C in the other strand.
(e) The two strands have the same sequence of bases when read in opposite directions.
51.16 In the following statements about double-stranded DNA, square brackets indicate the number of molecules. For example, [A] means the number of molecules of the base
adenine, and [deoxyribose] means the number of molecules of 2' deoxyribose. Classify each of the statements as true or false.
(a) [A] = [G]
(b) [A] = [C]
(c) [A] = [T]
(d) [A] + [G] = [T] + [C]
(e) [deoxyribose] = [phosphate]
51.17 When the base composition of double-stranded DNA from Micrococcus luteus was determined, 37.5 percent of the bases were found to be cytosine. What is the percentage of adenine in the DNA of this organism?
51.18 Which nucleic acid base is found only in DNA? Which is found only in RNA?
51.19 What are three principal structural differences between RNA and DNA?
51.20 What is the most likely reason for the participation of RNA intermediates in the process of information flow from DNA to protein?
51.21 What are the consequences of the Glu ' Val replacement at amino acid position 6 in the beta-globin chain?
51.22 Using the vermilion and cinnabar mutations affecting eye color in D. melanogaster as an example, explain what is meant by an epistatic interaction between genes.
51.23 Can all the existing diversity of life be explained as a result of adaptive evolution brought about by natural selection?
Principles of Genetic Transmission
52.1 Define the following terms: genotype, phenotype, allele, dominant, recessive.
52.2 How many different alleles of a particular gene may exist in a population of organisms? How many alleles of a particular gene can be present in an individual organism?
52.3 A breeder of Irish setters has a particularly valuable male show dog descended from a famous female named Rheona Didona, which was known to carry a harmful recessive gene for blindness caused by atrophy of the retina. Before she breeds the dog, she wants some assurance that it does not carry this harmful allele. How would you advise her to test the dog?
52.4 Any of a large number of different recessive mutations cause profound hereditary deafness. Children homozygous for such mutations are often found in small, isolated communities in which matings tend to be within the group. In these instances, a mutation that originates in one person can be transmitted to several members of the population in future generations and become homozygous through a marriage between group members who are carriers. A deaf man from one community marries a deaf woman from a different, unrelated community. Both of them have deaf parents. The marriage produces three children, all with normal hearing. How can you explain this result?
52.5 Consider a human family with four children, and remember that each birth is equally likely to result in a boy or a girl.
(a) What proportion of sibships will include at least one boy?
(b) What fraction of sibships will have the gender order GBGB?
52.6 Mendel's cross of homozygous round yellow with homozygous wrinkled green peas yielded the phenotypes 9/16 round yellow, 3/16 round green, 3/16 wrinkled yellow, and 1/16 wrinkled green progeny in the F2 generation. If
he had testcrossed the round green progeny individually, what proportion of them would have yielded some progeny with round yellow seeds?
52.7 Intestinal lactase deficiency (ILD) is a common inborn error of metabolism in humans; affected persons are homozygous for a recessive allele of a gene on chromosome 2. A woman and her husband, both phenotypically normal, had a daughter with ILD. She married a normal man and had three sons. One had ILD and two were normal. Recognizing that each succeeding child's genotype is not influenced by the genotypes of older siblings, determine the probability that the couple's next child will have ILD.
(d) 1.00 (e) 0.0625
52.8 Seborrheic keratosis is a rare hereditary skin condition due to an autosomal dominant mutation. Affected people have skin marked with numerous small, sharply margined, yellowish or brownish areas covered with a thin, greasy scale. A man with keratosis marries a normal woman, and they have three children. What is the chance that all three are normal? What is the chance that all three are affected?
52.9 In Drosophila, the dominant allele Cy (Curly) results in curly wings. The cross Cy + XCy + (where + represents the wildtype allele of Cy) results in a ratio of 2 curly : 1 wildtype F1 progeny. The cross between the curly F1 progeny also gives a ratio of 2 curly : 1 wildtype F2 progeny. How can this result be explained?
52.10 White Leghorn chickens are homozygous for a dominant allele, C, of a gene responsible for colored feathers, and
also for a dominant allele, I, of an independently segregating gene that prevents the expression of C. The White Wyandotte breed is homozygous recessive for both genes cc ii. What proportion of the F2 progeny obtained from mating White Leghorn x White Wyandotte F1 hybrids would be expected to have colored feathers?
52.11 The tailless trait in the Manx cat is determined by the alleles of a single gene. In the cross Manx x Manx, both tailless and tailed progeny are produced, in a ratio of 2 tailless : 1 tailed. All tailless progeny from this cross, when mated with tailed, produce a 1 : 1 ratio of tailless to tailed progeny.
(a) Is the allele for the tailless trait dominant or recessive?
(b) What genetic hypothesis can account for the 2 : 1 ratio of tailless: tailed and the result of the testcrosses with the tailless cats?
52.12 Absence of the enzyme alpha-1-antitrypsin is associated with a very high risk of developing emphysema and early death due to damaged lungs. Enzyme deficiency is usually due to homozygosity for an allele denoted PI-Z. A phenotypically normal man whose father died of the emphysema associated with homozygosity for PI-Z marries a phenotypically normal woman with a negative family history for the trait. They consider having a child.
(a) Draw the pedigree as described above.
(b) If the frequency of heterozygous PI-Z in the population is 1 in 30, what is the chance that the first child will be homozygous PI-Z?
(c) If the first child is homozygous PI-Z, what is the probability that the second child will be either homozygous or heterozygous for the normal allele?
52.13 Plants of the genotypes Aa Bb cc and aa Bb Cc are crossed. The three genes are on different chromosomes. What proportion of the progeny will have the phenotype A- B- C-, and what fraction of the A- B- C- plants will have the genotype Aa Bb Cc?
(a) 1/16; all
(b) 3/8; 2/3
(c) 3/16; 2/3
(d) 1/2; 1/2 (e) 1/2; all
52.14 In the cross Aa Bb Cc Dd xAa Bb Cc Dd, in which all four genes undergo independent assortment, what proportion of offspring are expected to be heterozygous for all four genes?
What is the answer if the cross is Aa Bb Cc Dd x Aa Bb cc dd?
52.15 In the cross AA Bb Cc x Aa Bb cc, where all three genes assort independently and each uppercase allele is dominant, what proportion of the progeny are expected to have the phenotype A- bb cc?
(d) 1/8 (e) 1/4
52.16 In the cross Aa Bb xAa Bb, what fraction of the progeny will be homozygous for one gene and heterozygous for the other if the two genes assort independently?
S2.17 The pedigree in the accompanying illustration shows the inheritance of coat color in a group of cocker spaniels. The coat colors and genotypes are indicated in the key.
(a) Specify in as much detail as possible the genotype of each dog in the pedigree.
(b) What are the possible genotypes of animal III-4, and what is the probability of each genotype?
(c) If a single pup is produced from the mating of III-4 x III-3, what is the probability that the pup will be lemon?
52.18 In terms of the type of epistasis indicated, would you say that a 27 : 37 ratio has more in common with a 9 : 7 ratio or with a 15 : 1 ratio? Explain your argument.
52.19 A trithybrid cross AA BB rr x aa bb RR is made in a plant species in which A and B are dominant to their respective alleles but there is no dominance between R and r. Assuming independent assortment, consider the F2 progeny from this cross.
(a) How many phenotypic classes are expected?
(b) What is the probability of the parental aa bb RR genotype?
(c) What proportion of the progeny would be expected to be homozygous for all three genes?
52.20 Four babies accidentally become mixed up in a maternity ward. The ABO types of the four babies are known to be O, A, B, and AB, and those of the four sets of parents are shown below. One of the attendants insists that DNA typing will be necessary to straighten out the mess, but another claims that the ABO blood types alone are sufficient. To
determine who is correct, try to match each baby with its correct set of parents.
(a) AB x O
(b) A x O
(c) A x AB
(d) O x O
52.21 For each of the following F2 ratios, some indicating epistasis, what phenotypic ratio would you expect in a testcross of the F1 dihybrid?
52.22 In plants of the genus Primula, the K locus controls synthesis of a compound called malvidin. Two plants heterozygous at the K locus were crossed, producing the following distribution of progeny:
1010 Make malvidin 345 Do not make malvidin
You wish to determine if the D locus (located on a different chromosome from K), affects production of malvidin as well. True-breeding plants that make malvidin (KKdd) were crossed to true-breeding plants that do not make malvidin (kk DD). All F1 plants failed to produce malvidin. F1 heterozygotes were self-fertilized, and the F2 progeny, assayed for malvidin synthesis, yielded the following distribution:
522 Make malvidin 2270 Do not make malvidin
(a) Write the genotypes and the corresponding phenotypes of all the F2 progeny obtained.
(b) Does the D locus affect malvidin synthesis? What is the basis of your conclusion?
52.23 From the complementation data in the table below, assign the ten mutations al through a10 to complementation groups. Use the complementation groups to deduce the expected results of the complementation tests indicated as missing data (denoted by the question marks).
Genes and Chromosomes
S3.1 Classify each statement as true or false as it applies to mitosis.
(a) DNA synthesis takes place in prophase; chromosome distribution to daughter cells takes place in metaphase.
(b) DNA synthesis takes place in metaphase; chromosome distribution to daughter cells takes place in anaphase.
(c) Chromosome condensation takes place in prophase; chromosome distribution to daughter cells takes place in anaphase.
(d) Chromosomes pair in metaphase.
53.2 Which of the following statements are true?
(a) A bivalent contains one centromere, two sister chromatids, and one homolog.
(b) A bivalent contains two centromeres, four sister chromatids, and two homologs.
(c) A bivalent contains two linear, duplex DNA molecules.
(d) A bivalent contains four linear, duplex DNA molecules.
(e) A bivalent contains eight linear, duplex DNA molecules.
53.3 Which of the following statements correctly describes a difference between mitosis and meiosis?
(a) Meiosis includes two nuclear divisions, mitosis only one.
(b) Bivalents appear in mitosis but not in meiosis.
(c) Sister chromatids appear in meiosis but not in mitosis.
(d) Centromeres do not divide in meiosis; they do in mitosis.
Complementation Data Problem S2.23 a1 a2 a3 a4 a5 a6 a7 a8 a9 a10
a1 - + + + - + + + ? +
a2 - + ? + + + -? +
a3 - + + -? + + +
a4 - + + ?- + +
a5 _ + + + _ +
a6 - - + + +
a7 - + + +
a8 - + +
a9 - +
S3.4 In which of the following situations can the result of segregation be observed phenotypically?
(a) In the haploid products of meiosis or the cells derived from them by mitosis, provided that these can grow and divide (as in certain fungi).
(b) In the testcross progeny of a heterozygous diploid organism.
(c) In the F2 progeny of genetically different diploid parents.
(d) Statements (a), (b), and (c).
(e) Statements (a) and (b) only.
53.5 Classify each of the following statements as true or false as it applies to the segregation of alleles for a genetic trait in a diploid organism.
(a) There is a 1 : 1 ratio of meiotic products.
(b) There is a 3 : 1 ratio of meiotic products.
(c) There is a 1 : 2 : 1 ratio of meiotic products.
(d) There is a 3 : 1 ratio of F2 genotypes.
(e) All of the above.
53.6 What is a chiasma? In what stage of meiosis are chiasmata observed?
53.7 When in meiosis does the physical manifestation of independent assortment take place?
53.8 In the life cycle of corn, Zea mays, how many haploid chromosome sets are there in cells of the following tissues?
(c) Root tip
(e) Embryo (e) Microspore
53.9 Somatic cells of the red fox, Vulpes vulpes, normally have 38 chromosomes. What is the number of chromosomes present in the nucleus in a cell of this organism in each of the following stages? (For purposes of this problem, count each chromatid as a chromosome in its own right.)
(a) Metaphase of mitosis
(b) Metaphase I of meiosis
(c) Telophase I of meiosis
(d) Telophase II of meiosis
S3.10 The terms ''reductional division" and "equational division" are literally correct in describing the two meiotic divisions with respect to division of the centromeres. To assess the applicability of the terms to alleles along a chromosome, consider a chromosome arm in which exactly one exchange event takes place in prophase I. Denote as A the region between the centromere and the position of the exchange and as B the region between the position of the exchange and the tip of the chromosome arm. If "reductional" is defined as resulting in products that are
genetically different and "equational" is defined as resulting in products that are genetically identical, what can you say about the "reductional" or "equational" nature of the two meiotic divisions with respect to the regions A and B?
53.11 What differences are there between the sexes in the pattern of transmission of genes located on the autosomes versus those on the sex chromosomes? (Assume that females are the homogametic sex and males the heterogametic sex.)
53.12 What observations about Drosophila provided proof of the chromosomal theory of heredity?
53.13 People with the chromosome constitution 47, XXY are phenotypically males. A normal woman whose father had hemophilia mates with a normal man and produces an XXY son who also has hemophilia. What kind of nondisjunction can explain this result?
53.14 Which of the following types of inheritance have the feature that an affected male has all affected daughters but no affected sons?
(a) Autosomal recessive
(b) Autosomal dominant
(d) X-linked recessive
(e) X-linked dominant
53.15 Mendel studied the inheritance of phenotypic characters determined by seven pairs of alleles. It is an interesting coincidence that the pea plant also has seven pairs of chromosomes. What is the probability that, if seven loci are chosen at random in an organism that has seven chromosomes, each locus is in a different chromosome? (Note: Mendel's seven genes are actually located in only four chromosomes, but only two of the genes—tall versus short plant and smooth versus constricted pod—are close enough together that independent assortment would not be observed; he apparently never examined these two traits for independent assortment.)
53.16 Drosophila montana has a somatic chromosome number of 12. Say the centromeres of the six homologous pairs are designated as A/a, B/b, C/c, D/d, E/e, and F/f.
(a) How many different combinations of centromeres can be produced in meiosis?
(b) What is the probability that a gamete will contain only centromeres designated by capital letters?
53.17 How many different genotypes are possible for
(a) an autosomal gene with four alleles?
(b) an X-linked gene with four alleles?
53.18 For an autosomal gene with m alleles, how many different genotypes are there?
53.19 Consider the mating Aa x Aa.
(a) What is the probability that a sibship of size 8 will contain no aa offspring?
(b) What is the probability that a sibship of size 8 will contain a perfect Mendelian ratio of 3 A-: 1 aa?
53.20 Matings between two types of guinea pigs with normal phenotypes produce 64 offspring, 11 of which showed a hyperactive behavioral abnormality. The other 53 offspring were behaviorally normal. Assume that the parents were heterozygous.
(a) Do these data fit the model that hyperactivity is caused by a recessive allele of a single gene?
(b) Do they fit the model that hyperactivity is caused by simultaneous homozygosity for recessive alleles of each of two unlinked genes?
53.21 In a X2 test with 1 degree of freedom, the value of X2 that results in significance at the 5 percent level (P = 0.05) is approximately 4. If a genetic hypothesis predicts a 1 : 1 ratio of two progeny types, calculate what deviations from the expected 1 : 1 ratio yield P = 0.05 (and rejection of the genetic hypothesis) when
(a) the total number of progeny equals 50
(b) the total number of progeny equals 100 Chapter 4—
Gene Linkage and Chromosome Mapping
54.1 After a large number of genes have been mapped, how many linkage groups will be found in each of the following?
(a) The bacterial cell Salmonella typhimurium, with a single, circular DNA molecule
(b) The haploid yeast Saccharomyces cerevisiae, with 16 chromosomes
(c) The common wheat Triticum vulgare, with 42 chromosomes per somatic cell
54.2 Does physical distance always correlate with map distance?
54.3 In genetic linkage studies, what is the definition of a map unit?
54.4 Distinguish between the terms "chromosome interference" and "chromatid interference." Which is measured by the coincidence?
54.5 What does the coefficient of coincidence measure?
(a) The frequency of double crossovers
(b) The ratio of double crossovers to single crossovers
(c) The ratio of single crossovers in two adjacent regions
(d) The ratio of double crossovers to the number expected if there were no interference
(e) The ratio of recombinant to parental progeny
54.6 A coefficient of coincidence of 0.25 means that:
(a) the frequency of double crossovers is 1/4.
(b) the frequency of double crossovers is 1/4 of the number expected if there were no interference.
(c) there were four times as many single crossovers as double crossovers.
(d) there were four times as many single crossovers in one region as there were in an adjacent region.
(e) there were four times as many parental as recombinant progeny.
54.7 In wheat, normal plant height requires the presence of either or both of two dominant alleles, A and B. Plants
homozygous for both recessive alleles (a b/a b) are dwarfed but otherwise normal. The two genes are on the same chromosome and recombine with a frequency of 16 percent. From the cross A b/a B x A B/a b, what is the expected frequency of dwarfed plants among the progeny?
S4.8 Two yeast strains were mated; one was red with mating type a; the other was red+ with mating type a. The resulting diploids were induced to undergo meiosis, and the meiotic products of the diploid were as follows (total progeny = 100):
9 red+ a
7 red a
How many map units separate the red gene from that for mating type?
54.9 In meiosis, under what conditions does seconddivision segregation of a pair of alleles take place?
54.10 Explain why the most accurate measurement of map distance between two genes is obtained by summing shorter distances between the genes.
54.11 In D. melanogaster, the recessive allele ap (apterous) drastically reduces the size of the wings and halteres (flight balancers), and the recessive allele cl (clot) produces dark maroon eye color. Both genes are located in the second chromosome. You have a friend taking a genetics laboratory course. To obtain the genetic distance between these genes, she performs a cross of a double-heterozygous female and a maroon-eyed, wingless male. She examines 25 flies in the F1 generation and calculates that the frequency of recombination is 56 percent. What is wrong with this number? What is a plausible explanation for the results obtained?
54.12 Your friend in Problem S4.11 logs onto FlyBase and learns that the ap-cl distance is 36 map units. She therefore decides to repeat the cross and to examine more F1 flies. Because both genes are autosomal, she does not expect any difference between reciprocal crosses. Hence, in setting up the cross this time, she mates a double- heterozygous male with a maroon-eyed, wingless female. To her astonishment, among 1000 F1 progeny she was unable to find even one recombinant progeny. How can this anomaly be explained?
54.13 The genetic map of an X chromosome of Drosophila melanogaster has a length of 73.1 map units. The X chromosome of the related species, Drosophila virilis, is even longer—170.5 map units. In view of the fact that the frequency of recombination between two genes cannot exceed 50 percent, how is it possible for the map distance between genes at opposite ends of a chromosome to exceed 50 map units?
54.14 In D. virilis, the mutations dusky body color (dy), cut wings (ct), and white eyes (w) are all recessive alleles located in the X chromosome. Females heterozygous for all three mutations were crossed to dy ct w males, and the following phenotypes were observed among the offspring.
cut, white 150
dusky, cut 72
dusky, cut, white 22
dusky, white 3
(a) What are the genotypes of two maternal X chromosomes?
(b) What is the map order of the genes?
(c) What are the two-factor recombination frequencies between each pair of genes?
(d) Are there as many observed double-crossover gametes as would be expected if there were no interference?
(e) What are the coefficient of coincidence and the interference across this region of the X chromosome?
54.15 The genes considered in Problem S4.14 are also located on the X chromosome in D. melanogaster. However, their order and the distances between them are completely different because of chromosome rearrangements that have happened since the divergence of two species approximately 40 million years ago. In D.melanogaster, the genetic map of this region is dy-16.2-ct-18.5-w. If you performed a cross identical to that in Problem S4.14 with D. melanogaster and observed 500 progeny, what types of progeny would be expected and in what numbers? Assume that the interference between the genes is the same in D.melanogaster as in D. virilis, and round the expected number of progeny to the nearest whole number. For convenience, list the progeny types in the same order as in Problem S4.14.
54.16 What experiment in Drosophila demonstrated that crossing-over takes place in the four-strand stage? How would the result of the experiment have been different if crossing-over took place in the two-strand stage?
54.17 Neurospora crassa is a haploid ascomycete fungus. A mutant strain that requires arginine for growth (arg-) is crossed with a strain that requires tyrosine and phenylalanine (try phe'). Each requirement results from a single mutation. Tests on 200 randomly collected spores from the cross gave the following results, in which the + in any column indicates the presence of the wildtype allele of the gene.
- + + 42 + - + 43
- + - 42 + + - 8
- - + 9 + + + 6
- - - 7
(a) What are the three possible two-gene recombination frequencies?
(b) Is any linkage suggested by these data?
54.18 In the case of independently assorting genes, why are parental ditype (PD) and nonparental ditype (NPD) tetrads formed in equal numbers? Why are nonparental ditype tetrads rare for linked genes?
54.19 The yeast Saccharomyces cerevisiae has unordered tetrads. In a cross made to study the linkage relationship among three genes, the following tetrads were obtained. The cross was between a strain of genotype + b c and one of genotypea + +.
Genotypes of spores in tetrads
Number of tetrads
a + +
+ b c
+ b c
a b +
+ b c
a b c
a + c
a b c
a b c
+ ++ Total
(a) From these data, determine which (if any) of the genes are linked.
(b) For any linked genes, determine the map distances.
S4.20 Consider a strain of Neurospora that exhibits complete chromosomal interference, even across the centromere. Two genes, A and B, are located in the same chromosome but on different sides of the centromere. Gene A is 10 map units from the centromere, and gene B is 5 map units from the centromere. Calculate the expected percentage of each of the following types of asci.
(a) Parental ditype
(b) Nonparental ditype
(d) First-division segregation for A, first for B
(e) First-division segregation for A, second for B
(f) Second-division segregation for A, first for B
(g) Second-division segregation for A, second for B
S4.21 You are a geneticist writing a set of test questions and wish to invent plausible numbers of the possible types of
progeny in a three-point testcross to yield a frequency of recombination of 0.2 in "region I," a frequency of recombination of 0.3 in "region II," and an interference value of 2/3. Among 1000 total progeny, how many should be:
(b) recombinant in region I only?
(c) recombinant in region II only?
(d) recombinant in both regions (double crossovers)?
(e)In each class of progeny above, how should the total number be allocated between the two reciprocal products of recombination?
54.22 What is the answer to Problem S4.21 in the general case when the frequency of recombination in region I is r1, that in region II is r2, and the coincidence equals c?
54.23 In a testcross of a parent heterozygous for each of four linked genes, in which the order of the genes is known, the following numbers of progeny were obtained:
Recombinant in region I only 61
Recombinant in region II only 143
Recombinant in region III only 226
Recombinant in regions I and II only 10
Recombinant in regions II and III only 45
Recombinant in regions I and III only 27
Recombinant in regions I and II and III 2
(a) What are the map distances in centimorgans of regions I, II, and III? (Hint: Solve the problem as in a three-point cross, but remember that recombinants in two or three regions are recombinant for each region separately.)
(b) Calculate the values of the interference for regions I + II, II + III, and I + III. Does the ranking of the interference values make intuitive sense?
54.24 In a fungal organism with ordered tetrads, deduce whether one observes first-division segregation or second- division segregation when the region between a gene and the centromere undergoes
(a) a two-strand double crossover
(b) either of the two types of three-strand double crossover
(c) a four-strand double crossover
If the four types of double crossover are equally frequent, what proportion of asci formed from double crossing-over will show second-division segregation?
54.25 Show that each of the following statements is true in a fungal organism with ordered tetrads,
(a) If a configuration of multiple crossovers will result in first-division segregation, then an additional crossover (distal to the existing ones) will convert the ascus to second-division segregation with probability 1.
(b) If a configuration of multiple crossovers will result in second-division segregation, then an additional crossover (distal to the existing ones) will convert the ascus to first-division segregation with probability 1/2.
The Molecular Structure and Replication of the Genetic Material
55.1 You have determined that one strand of a DNA double helix has the sequence 5'-AGCCTAG-3'. What is the sequence of the complementary strand?
55.2 In the replication of a bacterial chromosome, does replication begin at a random point?
55.3 The haploid genome of the wall cress, Arabidopsis thaliana, contains 100,000 kb and has five chromosomes. If a particular chromosome contains 10 percent of the DNA in the haploid genome, what is the approximate length of its DNA molecule in micrometers?
55.4 For the chromosome of A. thaliana in Problem S5.3, estimate the time of replication, assuming that there is only one origin of replication (exactly in the middle), that replication is bidirectional, and that the rate of DNA synthesis is
(a) 1500 nucleotide pairs per second (typical of bacterial cells)
(b) 50 nucleotide pairs per second (typical of eukaryotic cells)
55.5 In terms of the mechanism of DNA replication, explain why a linear chromosome without telomeres is expected to become progressively shorter in each generation.
55.6 A friend brings you three samples of nucleic acid and asks you to determine each sample's chemical identity (whether DNA or RNA) and whether the molecules are double-stranded or single-stranded. You use powerful nucleases to degrade each sample to its constituent nucleotides, and then you determine the approximate relative proportions of nucleotides. The results of your assays are presented below. What can you tell your friend about the nature of these samples?
Sample 1: dGTP 14% Sample 2: dGTP 12% Sample 3: GTP 22%
Assay results Problem S5.6
dCTP 15% dATP
dCTP 36% dATP
CTP 47% ATP
36% dTTP 35% 47% dTTP 5% 17% UTP 14%
55.7 What is meant by the statement that the DNA replication fork is asymmetrical?
55.8 What is the role of each of the following proteins in DNA replication?
(a) DNA helicase
(b) Single-stranded DNA-binding protein
(c) DNA polymerase
(d) DNA ligase
55.9 Which enzyme in DNA replication
(a) alters the helical winding of double-stranded DNA by causing a single-strand break, exchanging the relative position of the strands, and sealing the break?
(b) uses ribonucleoside triphosphates to synthesize short RNA primers?
55.10 Describe why topoisomerase is necessary during replication of the E. coli chromosome.
55.11 The accompanying drawing is of a replication fork. The ends of each strand are labeled with the letters a through h.
(a) Label the leading and lagging strands and, by the addition of arrowheads, show the direction of movement of the replication fork and the direction of synthesis of each DNA strand or fragment.
(b) Specify which of the letters a through h label 5' ends and which label 3' ends.
S5.12 Some "thermophilic" bacteria can live in water so hot that it approaches the boiling point. Which of the following DNA sequences would be more likely to be found in such an organism? Explain.
(a) 5'- TTATAAAATATATTTTTTATAT - 3' 3'-AATATTTTATATAAAAATATA-5'
(b) 5'-CCCCCGCGCGGCCGGGCGCGCG-3' 3'-GGGGGCGCGCCGGCCCGCGCGC-5'
55.13 A 3.1-kilobase linear fragment of DNA was digested with PstI and produced a 2-kb and a 1.1 -kb fragment. When the same 3.1-kb fragment was cut with HindIII, it yielded a 1.5-kb fragment, a 1.3-kb fragment, and a 0.3-kb fragment. When the 3.1-kb molecule was cut with a mixture of the two enzymes, fragments of 1.5, 0.8, 0.5, and 0.3 kb resulted. Draw a map of the original 3.1-kb fragment, and label restriction sites and the distances between the sites.
55.14 The X-linked recessive mutation dusky in Drosophila causes small, dark wings. In a stock of wildtype flies, you find a single male that has the dusky phenotype. In wildtype flies, the dusky gene is contained within an 8-kb XhoI restriction fragment. When you digest genomic DNA from the mutant male with XhoI and probe with the 8-kb restriction fragment on a Southern blot, the size of the labeled fragment is 10 kb. You clone the 10-kb fragment and
use it as a probe for a polytene chromosome in situ hybridization in a number of different wildtype strains, and you notice that this fragment hybridizes to multiple locations along the polytene chromosomes. Each wildtype strain has a different pattern of hybridization. What do these data suggest about the origin of the dusky mutation that you isolated?
S5.15 A short region of an mRNA molecule is used as a primer in dideoxy sequencing. After transfer to a filter and autoradiography, the resulting film looks as follows, where the orientation is with the samples loaded at the top and the letters indicate the dideoxynucleotide present during synthesis.
(a) Write the sequence implied by the bands in the gel from the 5' end to the 3' end.
(b) Is this the sequence of the elongated RNA strand or the sequence of its complementary DNA strand?
S5.16 For the following restriction endonucleases, calculate the average distance between restriction sites in an organism whose DNA has a random sequence and equal proportions of all four nucleotides. The symbol R means any purine (A or G) and Y means any pyrimidine (T or C), but an R-Y pair must be either A-T or G-C.
TaqI 5' -TCGA- 3
BamHI 5'-GGATCC-3' 3'-CCTAGG-5'
HaeII 5' -RGCGCY-3' 3'-YCGCGR-5'
55.17 For each of the restriction enzymes in Problem S5.16, calculate the expected number of restriction sites in the genome of a nematode Caenorhabditis elegans, assuming the sequence is approximately random. The genome size of this organism is 100,000 kb.
55.18 In early studies of the properties of Okazaki fragments, it was shown that these fragments could hybridize to both strands of E. coli DNA. This was taken as an evidence that they are synthesized on both branches of the replication fork. However, one feature of the replication of E. coli DNA that was not known at the time invalidates this conclusion. What is this characteristic?
55.19 In experiments that confirmed that DNA replication is semiconservative, E. coli DNA was labeled by growing cells for a number of generations in a medium containing "heavy" 15N isotope and then observing the densities of the molecules during subsequent generations of growth in a medium containing a great excess of "light" 14N. If DNA replication were conservative, what pattern of molecular densities would have been observed after one generation and after two generations of growth in the 14N-containing medium?
55.20 A covalently closed circular single-stranded DNA molecule is hybridized to a linear complementary strand that has a segment of 100 nucleotides missing. This duplex is exhaustively treated with various nucleases. List the final reaction products for each enzyme.
(a) An endonuclease that cuts only single-stranded DNA
(b) An exonuclease that degrades only single-stranded DNA
(c) An exonuclease that attacks only double-stranded DNA and degrades it in the 3' ' 5' direction (removing the shorter strand from a paired duplex when the partners are of unequal length)
(d) An endonuclease that attacks and degrades only double-stranded DNA
(e) A mixture of enzymes (b) and (d)
(f) A mixture of enzymes (a) and (c)
(g) A mixture of enzymes (c) and (d) Chapter 6—
The Molecular Organization of Chromosomes
56.1 What are the principal molecular constituents of the nucleosome?
56.2 Is it possible for a gene of average size to be present in a single nucleosome?
56.3 A geneticist wishes to isolate mutations that affect the amino acid sequence of any of the histone genes H2, H3A, H3B, and H4. What is the likely phenotype of such a mutation?
56.4 In spermatogenesis in many male animals, the normal somatic-cell histones are removed from the DNA and replaced with a still more highly basic (arginine-rich and lysine-rich) set of sperm-specific histones. Suggest one possible explanation.
56.5 What are principal structural levels of chromosome organization?
56.6 Why are centromeres and telomeres required for chromosomes to be genetically stable?
56.7 The nematode C. elegans has holocentric chromosomes. If the chromosomes in a cell are fragmented into many pieces by treatment with x rays, how would you expect each fragment to behave in cell division?
56.8 Name three features that distinguish euchromatin from heterochromatin.
56.9 Explain why topoisomerase is necessary during replication of the E. coli chromosome.
56.10 What is one possible reason why there is usually a single gene in the genome coding for a particular protein, whereas genes for rRNA and tRNA are repeated many times?
56.11 The X-linked recessive mutation singed (sn) causes gnarled bristles in Drosophila. In a stock of wildtype flies, you find a single male that has the singed phenotype.
(a) Diagram a crossing scheme that you will use to generate a true-breeding line of sn flies. Note that, other than the single singed male, you have only wildtype flies available.
(b) In wildtype flies, the sn gene is contained within an 8-kb BamHI restriction fragment. When you digest genomic DNA from the mutant line with BamHI and probe with the 8-kb restriction fragment on a Southern blot, the size of the labeled fragment on the filter is 13 kb. What possible explanations could account for this finding?
56.12 A genetically engineered strain of Drosophila virilis (strain 1) is homozygous for the recessive mutation y (yellow), an X-linked mutation that causes yellow body color. This strain also carries, at a site in the X chromosome, a genetically engineered transposable element called Hermes into which a copy of the y+ gene was inserted; the engineered Hermes element is unable to produce the transposase protein needed for transposition of the element. Another genetically engineered strain, strain 2, is also homozygous y but contains a copy of Hermes that can produce the transposase in each homolog of chromosome 3 . A female of strain 1 is crossed with a male of strain 2. The phenotypically wildtype F1 males are crossed with yellow females. About 5 percent of the F2 progeny males have wildtype body color. Give two possible explanations of how these males might arise.
56.13 The term "satellite DNA" was originally coined for highly repetitive simple sequences that, in equilibrium density-gradient centrifugation in CsCl, formed a distinct band (a "satellite" band) different from the bulk of the
genomic DNA, because of a difference in density of the DNA fractions. You analyze the genome organization of a new species by the kinetics of DNA renaturation and find that 10 percent of DNA is represented by a highly repetitive simple sequence. Does this mean that, after equilibrium density-gradient centrifugation in CsCl, you will find a satellite band?
56.14 DNA from species A that is labeled with 14N and randomly fragmented is renatured with an equal concentration of DNA from species B that is labeled with 15N and also randomly fragmented. The resulting mixture of molecules is centrifuged to equilibrium in CsCl. Ten percent of the total renatured DNA has a hybrid density. What fraction of the base sequences in the two species will be similar enough to renature?
56.15 You are studying the denaturation of two unusual DNA molecules, both consisting of 99 percent AT base pairs. Both molecules have the same melting temperature. However, observations in the electron microscope indicate that the strands of one of the molecules do not separate completely until raised to a temperature 12°C higher than that required for complete separation of the strands of the other molecule. Suggest an explanation.
56.16 A molecule of 5000 nucleotide pairs consists of the repeating sequence 5'-ACGTAG-3'. How would the melting temperature of this molecule compare to that of another molecule of the same length and base composition (50 percent G + C) but with a random sequence?
56.17 A Cot analysis is carried out with DNA obtained from a particular plant species. The chloroplast DNA is readily detected as a rapidly reannealing fraction because of its abundance in the DNA sample. Plants of the same species are maintained in the dark for several weeks until they lose all their chlorophyll and the leaves become white. DNA from these plants has the same Cot curve as observed in plants maintained in the light. What does this result tell you about the loss of chlorophyll?
Variation in Chromosome Number and Structure
57.1 Drosophila and human beings have very different mechanisms of sex determination. Which of the following statements correctly describes the phenotype of an XXY organism in both species?
(a) Fertile male in Drosophila, sterile female in humans.
(b) Sterile female in Drosophila, fertile male in humans.
(c) Fertile female in Drosophila, fertile male in humans.
(d) Fertile female in Drosophila, sterile male in humans.
(e) Fertile female in Drosophila, fertile female in humans.
57.2 What are the consequences of X-chromosome inactivation in mammals?
(a) The formation of Barr bodies
(b) A mutation
(c) A mosaic expression of different phenotypes in different cells of heterozygous X-linked genes in females
(d) A mosaic expression of heterozygous autosomal genes in females
(e) Turner syndrome
57.3 Nondisjunction of chromosomes in meiosis may take place at the first or the second meiotic division. If nondisjunction of the XY pair of chromosomes took place at the first division in the formation of sperm, which of the following chromosomal types of sperm could be formed?
57.4 Which of the following human disorders can result from nondisjunction during the formation of the egg?
(a) Down syndrome
(b) Klinefelter syndrome
(c) XYY male
57.5 Why are abnormalities in the number of sex chromosomes in mammals usually less severe in their phenotypic effects than abnormalities in the number of autosomes?
57.6 What are the consequences of a single crossover within the inverted region of a pair of homologous chromosomes with the gene order is A B C D in one and a c b d in the other, in each of the following cases?
(a) The centromere is not included within the inversion.
(b) The centromere is included within the inversion.
57.7 What role might inversions play in evolution?
57.8 In human beings, trisomy 21 is the cause of Down syndrome. Ordinarily, only one child with Down syndrome is seen in a sibship. Occasionally, a sibship with multiple affected children is observed. In one such family, a pair of normal parents had seven children. Four were normal, and three had an atypical form of Down syndrome. In this atypical form, the slowed growth, mental handicap, transverse crease on the palm of the hand, and short broad hands were present, but the ear and heart defects often observed with Down syndrome were not present. One of the normal children gave rise to five offspring, two of whom had the same atypical form of Down syndrome. Provide a genetic explanation for this atypical form of Down syndrome and for its inheritance pattern.
57.9 Colchicine is used to cause a doubling of chromosome number in many plant species. Starting with a diploid strain of watermelon (Citrullus lanatus), outline a series of crosses you would use to create a pentaploid strain of this species.
57.10 Explain why, in Drosophila, triploid flies are much more likely to survive than trisomics.
57.11 In a certain fertile tetraploid plant, a locus with alleles A and a is situated very near the centromere of its chromosome. If a tetraploid plant of genotype AAaa is crossed with a diploid plant of genotype aa, what genotypic ratios are expected? What phenotypic ratios are expected if A is dominant to a? (Assume that homologous chromosomes in the tetraploid form pairs at random and that all gametes produced by the tetraploid are diploid. Assume also that the dominant phenotype is expressed whenever at least one A allele is present.)
57.12 Explain the apparent paradox that familial retinoblastoma is regarded as an example of a recessive oncogene (tumor suppressor gene) even though it is inherited in pedigrees as an autosomal dominant. Familial retinoblastomas are generally bilateral (that is, there are multiple independent tumors in both eyes of affected individuals), whereas sporadic retinoblastoma is generally limited to one eye and to fewer independent tumors. Account for these observations.
57.13 Two strains of a plant called shepherd's purse (Capsella bursa-pastoris), a widespread lawn and roadside weed of the mustard family, are compared. In strain 1, the recombination frequency between genes A and B is 6 percent, whereas in strain 2, it is 50 percent. The F1 progeny from a cross between these strains produce gametes with the following properties: About half are viable and contain only non-recombinant chromosomes, and about half are nonviable. How can these results be explained?
57.14 True-breeding tetrasomic organisms are much easier to create than true-breeding trisomic organisms. Suggest a reason for this observation.
57.15 In plants, a cross between a diploid organism and a tetraploid organism of the same species is usually more fertile than that between a diploid organism and a triploid organism of the same species. Suggest a possible explanation.
57.16 In Drosophila melanogaster, the genes for brown eyes (bw) and humpy thorax (hy) are about 12 map units distant on the same arm of chromosome 2. A paracentric inversion spans about one-third of the region but does not include these genes. What recombinant frequency between bw and hy would you expect in each of the following?
(a) Females that are homozygous for the inversion
(b) Females that are heterozygous for the inversion
57.17 The sequence of genes in a normal human chromosome is 123 • 456789, where the raised dot (•) represents the centromere. Aberrant chromosomes with the following arrangements of genes were observed.
I. 123 •476589
II. 123 •46789
III. 1654 •32789
IV. 123 •4566789
(a) Identify each chromosome as a paracentric inversion, pericentric inversion, deletion, or duplication, and diagram how each rearrangement would pair with the normal homolog in meiosis.
(b) Draw the meiotic products of a crossover between genes 5 and 6 when chromosome I pairs with a wildtype chromosome.
(c) Draw the products of a crossover between genes 5 and 6 when chromosome III pairs with a wildtype chromosome.
S7.18 Semisterile tomato plants heterozygous for a reciprocal translocation between chromosomes 5 and 11 were crossed with chromosomally normal plants homozygous for the recessive mutant broad leaf on chromosome 11. When semisterile F1 plants were crossed with the plants of broad-leaf parental type, the following phenotypes were found in the backcross progeny:
semisterile broad leaf 38
fertile broad leaf 242
semisterile normal leaf 282
fertile normal leaf 33
(a) What is the recombination frequency between the broad-leaf gene and the translocation breakpoint in chromosome 11?
(b) What ratio of phenotypes in the backcross progeny would have been expected if the broad-leaf gene had not been on the chromosome involved in the translocation?
S7.19 Semisterile tomato plants heterozygous for the same translocation referred to in the previous problem were crossed with chromosomally normal plants homozygous for the recessive mutants wilty and leafy in chromosome 5. When semisterile F1 plants were crossed with plants of wilty, leafy parental genotype, the following phenotypes were found in the backcross progeny.
wildtype 333 19 352
wilty 17 6 23
leafy 1 8 9
wilty and leafy 25 273 298
What is the map distance between the two genes and the map distance between each of the genes and the translocation breakpoint in chromosome 5?
S7.20 The accompanying diagrams depict wildtype chromosomes 2 and 3 from a certain plant. The filled circles represent the respective centromeres.
(a) Show how these chromosomes would pair in meiosis in an individual heterozygous for an inversion that
includes genes E, F, and G as well as heterozygous for a reciprocal translocation between chromosomes 2 and 3 with breakpoints between B and C in 2 and between K and L in 3.
(b) Suppose that the breakpoints of the above inversion are sufficiently close to genes D and H that recombination between the genes is completely eliminated when the inversion is heterozygous. A plant heterozygous for the inversion and the translocation and of genotype Dd Hh is crossed with a plant with normal chromosomes and of genotype dd hh. The most frequent classes of offspring are dd Hh semisterile and Dd hh ertile. The rarest types are Dd hh semisterile and dd Hh fertile, which together have a combined frequency of 0.15. What is the recombination frequency between D and the translocation breakpoint?
S7.21 Six mutations of bacteriophage T4 (a throughf) are isolated and crossed in all pairwise combinations to three deletion mutations (numbered 1-3) with the results shown below. In the table, R means that wildtype recombinants were observed and 0 means that there were no wildtype recombinants.
Point mutations a b c d e f
Deletions 1 0 0 R R R R
Deletions 2 0 R 0 0 0 0
Deletions 3 0 0 R 0 0 0
In parallel with the above experiments, the mutations were tested in all possible pairwise combinations for complementation. The results are tabulated below, where + indicates complementation and - indicates no complementation.
a b c d e f
a - + + - + +
b +- + + + +
c + + - + - +
d - + + - + +
e - + + - + +
f + + + + + -
Assuming that complementation indicates that two mutations are in different genes, combine the deletion mapping and the complementation data to determine the most likely relative order of the mutations. Group the mutations into allelic classes, and determine the position and extent of the three deletions.
The Genetics of Bacteria and Viruses S8.1 Define the following terms:
(e) Temperate phage
58.2 Describe the differences between phage l and F-plasmid integration into bacterial chromosome.
58.3 Is it possible for a plasmid without any genes to exist?
58.4 What process is responsible for the conversion of an F+ bacterial cell into an Hfr cell?
58.5 What is the major difference in bacteriophage between the lytic cycle and the lysogenic cycle?
58.6 Why are specialized transducing particles of phage l generated only when a lysogen is induced to produce phage rather than in the process of lytic infection?
58.7 Why is generalized transduction not the preferred method for mapping two unknown mutants relative to one another on the bacterial chromosome?
58.8 A temperate phage has the gene order a b c d e f g h, whereas the prophage of the same phage has the gene order g h a b c d e f. What information does this circular permutation give you about the location of the phage attachment site?
58.9 Recombination in phage l crosses reveals the order of certain genes to be
where cos represents the cohesive ends and att is the phage attachment site. What order of the genes would be found if the l prophage were mapped by generalized P1 transduction of an E. coli strain lysogenic for phage l?
58.10 You are studying a biochemical pathway in Neurospora that leads to the production of substance A. You isolate a set of mutations, each of which is unable to grow on minimal medium unless it is supplemented with A. By performing appropriate matings, you group all the mutants into four complementation groups (genes) designated a1, a2, a3, and a4. You know beforehand that the biochemical pathway for the production of A includes four intermediates: B, C, D, and E. You test the nutritional requirements of your mutants by growing them on minimal medium supplemented with each of these intermediates in turn. The results are summarized in the accompanying table, where the plus signs indicate growth and the minus signs indicate failure to grow.
A B C D ti
Determine in what order the substances A, B, C, D, and E are most likely to participate in the biochemical pathway, and
indicate the enzymatic steps by arrows. Label each arrow with the name of the gene that codes for the corresponding enzyme.
S8.11 A experiment was carried out in E. coli to map five genes around the chromosome using each of three different Hfr strains. The genetic markers were bio, met, phe, his, and trp. The Hfr strains were found to transfer the genetic markers at the times indicated in the accompanying table. Construct a genetic map of the E. coli chromosome that includes all five genetic markers, the genetic distances in minutes between adjacent gene pairs, and the origin and direction of transfer of each Hfr. Complete the missing entries in the table, indicated by the question marks.
Hfr1 markers bio met phe
Time of entry 26
Time of entry ?
Hfr3 markers phe his
Hfr2 markers phe met bio
trp 75 bio
Time of entry 6
58.12 A first-year graduate student carried out transformation of an E. coli strain of genotype trp' azi' phe' gal' pur his' ser' using a plasmid with the selectable markers amp-r kan-r trp+ azi+ bio+ phe+ pur+ tet-r. After transformation, he plated the cells on minimal medium containing galactose as a carbon source along with histidine and serine and added ampicillin and kanamycin to select for the transformants. After incubating the plates overnight, the student returned to the lab to discover that no colonies had formed! Then he realized that he had made a stupid mistake. What was it?
58.13 The genes A, B, G, H, I, and T were tested in all possible pairs for cotransduction with bacteriophage P1. Only the following pairs were found able to be cotransduced: G and H, G and I, T and A, I and B, A and H. What is the order of the genes along the chromosome? Explain your logic.
58.14 Bacteriophage P1 was grown on a wildtype strain of E. coli, and the resulting progeny phage were used to infect a strain with three mutations, arg", pro', and his', each of which results in a requirement for an amino acid: arginine, proline, and histidine, respectively. Equal volumes of the infected culture were spread on plates containing minimal medium supplemented with various combinations of the amino acids, and the colonies appearing on each plate were counted. The results are shown here. In each row, the + signs indicate the amino acids present in the medium.
1 + + 0 1000
2 + 0 + 1000
3 0 + + 1000
4 + 0 0 200
5 0 + 0 100
6 0 0 +0
(a) For each of the six types of medium, list the donor markers that must be present to allow the transduced cell to form colonies.
(b) Using these results, construct a genetic map of the arg, pro, his region showing the relative order of the genes and the cotransduction frequencies between the markers.
58.15 The order of genes in the l phage virion is A B C D E att int xis N cI O P Q S R
(a) Given that the bacterial attachment site, att, is between gal and bio in the bacterial chromosome, what is the prophage gene order?
(b) A new phage is discovered whose gene order is identical in the virion and in the prophage. What does this say about the location of the att site with respect to the termini of the phage chromosome?
(c) A wildtype l lysogen is infected with another l phage carrying a genetic marker, Z, located between E and att. The superinfection gives rise to a rare, doubly lysogenic E. coli strain that carries both l and l-Z prophage. Assuming that the second phage also entered the chromosome at an att site, diagram two possible arrangements of the prophages in the bacterial chromosome, and indicate the locations of the bacterial genes gal and bio.
58.16 Four Hfr strains of E. coli transfer their genetic material in the following order. Construct a genetic map of the E. coli chromosome that includes all the markers listed here and the distances in minutes between adjacent gene pairs.
Hfr1 markers mal met thi thr Time of entry 10 17 22 33
Hfr2 markers his phe
Time of entry 18 23
Hfr3 markers phe his bio azi thr thi Time of entry 6 11 33 48 49 60
Hfr4 markers Time of entry
S8.17 An F' factor consisting of the E. coli sex factor, F, and a segment of the bacterial chromosome that includes the wildtype tryptophan synthetase A gene (trpA+) was isolated. The F' trpA factor was introduced by conjugation into an F- strain with a single nucleotide mutation in the trpA gene, and the resulting exconjugants were F' trpA+/trpA' partial diploids. These cells were streaked to obtain single colonies, and each colony was inoculated into liquid nutrient medium. When mixed with female cells, most of the resulting cultures transferred only the original F' trpA factor. However, a very small percentage of the cultures transferred a variable length of the bacterial chromosome, starting with the trpA gene. These cultures fell into two classes. Class 1 transferred trpA+ early, whereas class 2 transferred trpA' early.
(a) Draw a diagram of the interaction between the F' factor and the bacterial chromosome, showing how the cells of class 1 are likely to have arisen.
(b) Draw a similar diagram showing the probable origin of cells of class 2. Chapter 9—
Genetic Engineering and Genome Analysis
59.1 Define the following terms:
(b) Restriction site
(d) Cohesive end
(e) Blunt end
(f) Complementary DNA
(g) Insertional inactivation
(h) Colony hybridization assay
(i) Transgenic organism
59.2 What are some of the principal practical applications of genetic engineering?
59.3 Reverse transcriptase, like most enzymes that make DNA, requires a primer. Explain why, when cDNA is to be made for the purpose of cloning a eukaryotic gene, a convenient primer is a short sequence of poly(dT)? Why does this method not work with a prokaryotic messenger RNA?
59.4 A DNA molecule has 23 occurrences of the sequence 5'-AATT-3' along one strand. How many times does the same sequence occur along the other strand?
59.5 After doing a restriction digest with the enzyme SseI, which has the recognition site 5'-CCTGCA/GG-3' (the slash indicates the position of the cleavage), you wish to separate the fragments in an agarose gel. In order to choose the proper concentration of agarose, you need to know the expected size of the fragments. Assuming equivalent amounts of each of the four nucleotides in the target DNA, what average-size fragment would you expect?
59.6 How many clones are needed to establish a library of DNA from a species of monkey with a diploid genome size of 6 x 109 base pairs if (1) fragments whose sizes average 2 x 104 base pairs are used, and (2) one wishes 99 percent of the genomic sequences to be in the library? (Hint: If the genome is cloned at random with x-fold coverage, the probability that a particular sequence will be missing is e'x.)
59.7 The yeast Saccharomyces cerevisiae has 16 chromosomes. All of them can be separated by pulse field gel electrophoresis. Strains with the following chromosome rearrangements were examined by pulse field electrophoresis along with DNA from a wildtype control:
(a) A reciprocal translocation in which about one-third of chromosome IV (the largest chromosome) was exchanged with about one-third of chromosome I (the smallest chromosome).
(b) An insertional translocation in which two-thirds of one chromosome was inserted into another chromosome.
(c) A pericentric inversion in chromosome IV.
(d) A paracentric inversion in chromosome IV.
What band pattern would you expect to see in each case?
59.8 As a laboratory exercise, a friend is required to express a cloned Drosophila gene in bacteria to obtain large amounts of the recombinant protein. She is given a recombinant phage containing a 12-kb fragment of Drosophila genomic DNA, which, when introduced into the germline of a mutant fly, is able to rescue the mutant phenotype.
Hence, the 12-kb fragment contains the entire gene. The fragment also hybridizes to a 4.7-kb RNA on Northern blots. (A Northern blot is one in which RNA is separated by electrophoresis and the bands are transferred to a filter and probed with a labeled complementary sequence). The protein encoded by the gene is approximately 110 amino acids in length and can be isolated by standard methods. Your friend excises the 12-kb fragment out of the phage and inserts it into a bacterial expression vector (a plasmid designed to allow transcription of inserts). Although molecular analysis indicates that the DNA was inserted correctly and that the insert is being transcribed, she is unable to isolate any of the gene product. She turns to you for advice. What explanations would you suggest and what would you recommend she do to obtain the protein?
59.9 Aliquots of a 7.8-kilobase (kb) linear piece of DNA are digested with the restriction enzymes PvuII, HincII, ClaI, and BanII, alone and in pairs. The digestion products are separated by gel electrophoresis, and the size of each fragment is determined by comparison to size standards. The fragment sizes obtained, in kilobase pairs, are given below. Draw a diagram of the 7.8-kb fragment, showing the location of the restriction sites for each enzyme.
(a) Digestion with PvuII: 1.3 kb, 6.5 kb
(b) Digestion with HincII: 0.5 kb, 3.3 kb, 4 kb
(c) Digestion with ClaI: 1 kb, 6.8 kb
(d) Digestion with BanII: 1.3 kb, 6.5 kb
(e) Digestion with PvuII and ClaI: 1 kb, 5.5 kb, 1.3 kb
(f) Digestion with PvuII and HincII: 0.5 kb, 0.8 kb, 2.5 kb, 4 kb
(g) Digestion with HincII and ClaI: 0.5 kb, 1 kb, 3 kb, 3.3 kb
(h) Digestion with HincII and BanII: 0.5 kb, 0.8 kb, 2.5 kb, 4 kb
(i) Digestion with ClaI and BanII: 1 kb, 1.3 kb, 5.5 kb
59.10 A 3.3-kb size fragment created after the HincII digestion in the previous problem is inserted into a unique HincII site of the plasmid pUC18 and transformed into E. coli. Individual transformed colonies are isolated, and the recombinant plasmids from each colony are purified and mapped with the restriction enzyme PvuII. To the surprise of a novice molecular geneticist, the plasmids do not all give the same pattern of restriction fragments on a gel. The plasmids fall into two distinct groups. Both groups yield the same number of bands in the gel, the aggregate size of which is the same, but two of the fragment sizes are different. Explain the reason for this difference.
S9.11 You have cloned a region of the genome of the nematode C. elegans that contains a mutation in a gene involved in controlling biorhythms. You make a restriction map of the cloned region and find that the following HindIII restriction map can be constructed:
Now you perform a HindIII digest of genomic DNA from individuals homozygous for the mutation and also from those that are homozygous for the wildtype allele. You subject both of these digests to electrophoresis in an agarose gel and transfer the DNA to a nylon membrane. When the membrane is probed with fragment A, B, or D, the probe hybridizes to fragments of 4, 2, and 6 kb, respectively, in both the mutant and the wildtype strain. When the membrane is probed with fragment C, the probe hybridizes to 11 different fragments in the mutant genome (including one fragment of 5 kb) and to 10 fragments in the wildtype genome (but not one of 5 kb). What could account for both the mutation and the pattern of bands seen with the C probe?
59.12 You are given a plasmid containing part of a gene of D. melanogaster. The gene fragment is 303 base pairs long. You would like to amplify it using the polymerase chain reaction (PCR). You design oligonucleotide primers 19 nucleotides in length that are complementary to the plasmid sequences immediately adjacent to both ends of the cloning site. What would be the exact size of the resulting PCR product?
59.13 Suppose that you digest the genomic DNA of a particular organism with Sau3A (/GATC). Then you ligate the resulting fragments into a unique BamHI (G/GATCC) cloning site of a plasmid vector. Would it be possible to isolate the cloned fragments from the vector using BamHI? From what proportion of clones would it be possible?
59.14 Digestion of a DNA molecule with HindIII yields two fragments of 2.2 kb and 2.8 kb. EcoRI cuts the molecule, creating 1.8 kb and 3.2 kb fragments. When treated with both enzymes, the same DNA molecule produces four fragments of 0.8 kb, 1.0 kb, 1.2 kb, and 2.0 kb. Draw a restriction map of this molecule.
59.15 What are the main differences between cDNA and genomic DNA libraries?
59.16 You wish to amplify a unique 100-bp sequence from the human genome (3 x 109 bp) using the polymerase chain reaction. At the end of the amplification, you want the amount of amplified DNA to represent not less than 99 percent of all the DNA present. Assume that the amplification is exponential.
(a) How many cycles of amplification are necessary to achieve the 99 percent goal?
(b) How would the number of cycles change if you were to amplify a fragment of 1000 bp?
59.17 You are given a DNA fragment containing the Drosophila developmental gene fushi tarazu (ftz) along with the restriction map shown below. The restriction sites are HindIII (5'-A/AGCTT-3'), BclI (5'-T/GATCA-3'), EcoRI (5'-G/AATTC-3'), BamHI (5'-G/GATCC-3').
You are instructed to subclone the gene into a 3.8-kb plasmid vector having relevant restriction sites PstI (5'- CTGCA/G-3'), BglII (5'-A/GATCT-3'), and BamHI (5'-G/GATCC-3') positioned as follows:
(0) Pft I (50)
With what restriction enzymes should you cleave the ftz-containing region before ligating it with plasmid DNA that has been linearized with BglII?
59.18 Once you have subcloned the ftz region as in the previous problem, you want to determine the orientation in which the fragment is inserted in the resulting plasmid. You cleave the ftz-containing plasmid with BamHI (5'- G/GATCC-3'), PstI (5'-CTGCA/G-3'), and HindIII (5'-A/AGCTT-3'), and observe fragments of sizes approximately 3 kb and 1 kb.
(a) Can you determine the orientation of the inserted fragment from this result?
(b) Would your answer change if the cloning site used in the plasmid were BamHI rather than BglII? Explain.
59.19 You are given two strains of E. coli containing plasmids whose restriction maps are
You are asked to subclone the gene for alcohol dehydrogenase (Adh) from plasmid B into plasmid A. Unsure of which
cloning procedure would be the best, you try two different approaches.
(a) You digest both plasmids to completion with PpuI (A/TGCAT) and Sno (G/TGCAC), isolate the 4-kb PpuI-SnoI Adh gene fragment and the 3.5-kb PpuI-SnoI vector A fragment, carry out a ligation reaction containing both isolated DNA fragments, transform E. coli with the resulting DNA, and plate the transformed cells on medium containing tetracycline. The plates also contain Xgal and IPTG, which makes it possible to identify cells that contain the original plasmid A, because they turn blue as a result of the functional lacZ gene. Colonies with an insertion that interrupts the lacZ gene are white because of insertional inactivation. You follow the same procedure except that you use HindIII (A/AGCTT) instead of SnoI. Using approach (a), you obtain many more blue colonies than using approach (b). Explain why this is so.
S9.20 You purify plasmid DNA from white colonies obtained in both approaches in the previous problem. Predict the sizes of the resulting fragments when the isolated plasmids are digested with PstI.
Chapter 10— Gene Expression
510.1 Describe the main features of the genetic code.
510.2 If the triplet codons in the genetic code were read in an overlapping fashion, would there be any constraints on the amino acid sequences found in proteins?
510.3 If DNA consisted of only two nucleotides (say, A and T) in any sequence, what is the minimal number of adjacent nucleotides that would be needed to specify uniquely each of the 20 amino acids?
510.4 If DNA consisted of three possible base pairs (say, A-T, G-C, and X-Y), what is the minimal number of adjacent nucleotides that would be needed to specify uniquely each of the 20 amino acids?
510.5 A DNA strand consists of any sequence of four kinds of nucleotides. Suppose there were only 16 different amino acids instead of 20. Which of the following statements would be correct descriptions of the minimal number of nucleotides necessary to create a genetic code?
(b) 2, provided that chain termination does not require a special codon
(c) 3, provided that chain termination does require a special codon
(d) 2, no matter how chain termination is accomplished
(e) statements (b) and (c)
510.6 In the standard genetic code, synonymous codons often have the feature that the third nucleotide in the codon is either of the pyrimidines (T or C) or either of the purines (A or G) because of ''wobble" in the base pairing at the third position of the codon. If this rule were followed consistently for all codons, would it be possible to have a triplet code that specifies 20 amino acids and one stop codon? Would there be any room for degeneracy?
510.7 Which of the following statements is true of polypeptide chains and nucleotide chains?
(a) Both have a sugar-phosphate backbone.
(b) Both are chains consisting of 20 types of repeating units.
(c) Both types of molecules contain unambiguous and interconvertible triplet codes.
(d) Both types of polymers are linear and unbranched.
(e) None of the above.
510.8 The notion that a strand of DNA serves as a template for transcription of an RNA that is translated into a polypeptide is known as the "central dogma" of gene expression. All three types of molecules have a polarity. In the DNA template and the RNA transcript, the polarity is determined by the free 3' and 5' groups at opposite ends of the
polynucleotide chains; in a polypeptide, the polarity is determined by the free amino group (N terminal) and carboxyl group (C terminal) at opposite ends. Each statement below describes one possible polarity of the DNA template, the RNA transcript, and the polypeptide chain, respectively, in temporal order of use as a template or in synthesis. Which is correct?
(a) 5' to 3' DNA; 3' to 5' RNA; N terminal to C terminal.
(b) 3' to 5' DNA; 3' to 5' RNA; N terminal to C terminal.
(c) 3' to 5' DNA; 3' to 5' RNA; C terminal to N terminal.
(d) 3' to 5' DNA; 5' to 3' RNA; N terminal to C terminal.
(e) 5' to 3' DNA; 5' to 3' RNA; C terminal to N terminal.
510.9 Which of the following features of a protein structure is most directly determined by the genetic code?
(a) Its shape
(b) Its secondary structure
(c) Its catalytic or structural role
(d) Its amino acid sequence
(e) Its subunit composition (quaternary structure)
510.10 The structure below is that of a dipeptide composed of alanine and glycine.
(a) Which end, left or right, is the amino terminal end?
(b) Which amino acid is at the amino terminal end?
(c) Which end, left or right, is the carboxyl terminal end?
(d) Which amino acid is at the carboxyl terminal end?
(e) Which bond is the peptide bond?
510.11 Summarize the studies in phage T4 that provided the first genetic evidence for a triplet code.
510.12 An alternating polymer AGAG is used as a messenger RNA in an in vitro protein-synthesizing system that does not need a start codon. What types of polypeptide chain are expected?
510.13 In a segment of DNA that contains overlapping genes, would you expect the two proteins to terminate at the same site or have the same number of amino acids? Explain.
510.14 What unique feature of the 5' end of a eukaryotic mRNA distinguishes it from a prokaryotic mRNA?
510.15 Consider the two codons AGA and AGG, which code for the amino acid arginine. A, C, U, G, or I is allowed in the 5' position of the anticodon.
(a) What is the minimum number of tRNAArg types needed to translate these codons?
(b) For each of the tRNAArg types in part (a), what is the anticodon?
510.16 A wildtype protein has glutamic acid (Glu) at amino acid position 56. A mutation is found in which glutamic acid is replaced with a different amino acid. Among the reverse mutations that restore protein function, some have thrreonine (Thr) at the position, some have isoleucine (Ile), and some have glutamic acid. Assuming that the original mutation and all the revertants result from single nucleotide substitutions, determine the wildtype codon for position 56, that of the original mutation, and those of the revertants.
510.17 What amino acid sequence would correspond to the following mRNA sequence?
(If you use the conventional single-letter abbreviations for the amino acids, shown below, you will find a secret.)
Ala A Gly G Pro P
Arg R His H Ser S
Asn N Ile I Thr T
Asp D Leu L Trp W
Cys C Lys K Tyr Y
Gln Q Met M Val V
Glu E Phe F
S10.18 If the following double-stranded DNA molecule is transcribed from the bottom strand, does transcription start at the left or right end? What is the mRNA sequence? If translation starts at the first AUG, what is the polypeptide sequence?
5'-CGCTAGCATGGAGAACGACGAATTGAGCCCAGAAGCGTCATGAGG-3' 3' -GCCATCGTACCTCTTGCTGCTTAACTCGGGTCTTCGCAGTACTCC-5'
S10.19 The following table shows matching regions of the DNA, mRNA, tRNA, and amino acids encoded in a particular gene. The mRNA is shown with its 5' end at the left, and the tRNA anticodon is shown with its 3' end at the left. The vertical lines define the reading frame.
(a) Complete the nucleic acid sequences, assuming Watson-Crick pairing between each codon and anticodon.
(b) Is the DNA strand transcribed the top or the bottom strand?
(c) Translate the mRNA in all three reading frames.
(d) Specify the nucleic acid strand(s) whose sequence could be used as a probe:
(i) In a Southern blot hybridization, in which the hybridization is carried out against genomic DNA.
(ii) In a Northern blot hybridization, in which the hybridization is carried out against mRNA.
S10.20 Part of a DNA molecule containing a single short intron has the sequence
and the corresponding polypeptide has the sequence
Find the intron.
S10.21A double mutation produced by recombination contains two single nucleotide frameshifts separated by about 20 base pairs. The first is an insertion and the second a deletion. The amino acid sequence of the wildtype and that of the mutant polypeptide in this part of the protein are
Wildtype: Lys-Lys-Tyr-His-Gln-Trp-Thr-Cys-Asn Mutant: Lys-Gln-Ile-Pro-Pro-Val-Asp-Met-Asn
What are the original and the double-mutant mRNA sequences? Which nucleotide in the wildtype sequence is the frameshift addition? Which nucleotide in the double-mutant sequence is the frameshift deletion? (In working this problem, it will be convenient to use the conventional symbols Y for unknown pyrimidine, R for unknown purine, N for unknown nucleotide, and H for A, C, T.)
S10.22 The following RNA sequence is a fragment derived from a polycistronic messenger RNA including two genes.
5'-CUUAUGGAAGUAAACCCGAGC-3' This fragment is known to include the initiation codon of one protein and the termination codon of the other. (a) Determine the first four amino acids of the protein whose initiation codon is in the fragment.
(b) Determine the last two amino acids of the protein whose translation terminates in the fragment.
(c) What is unusual about these genes?
(d) What would be the consequence of a frameshift mutation that deleted the G from the initiation codon?
510.23 A strain of E. coli carries a mutation that completely inactivates the enzyme encoded in the gene. Several revertants with partly or fully restored activity were selected and the amino acid sequence of the enzyme determined. The only differences found were at position 10 in the polypeptide chain.
Revertant 1 had Thr. Revertant 2 had Glu. Revertant 3 had Met. Revertant 4 had Arg.
Assume that the initial mutation itself, as well as each revertant, resulted from a single nucleotide substitution.
(a) What amino acid is present at position 10 in the mutant protein?
(b) What codon in the messenger RNA would encode this amino acid?
510.24 Make a sketch of a mature eukaryotic messenger RNA molecule hybridized to the transcribed strand of DNA of a gene that contains two introns, oriented with the promoter region of the DNA at the left. Clearly label the DNA and the mRNA. Use the following letters to label the location and/or boundaries of each segment. Some letters may be used several times, as appropriate; some, which are not applicable, may not be used at all.
(a) 5' end
(b) 3' end
(c) Promoter region
(g) Polyadenylation signal
(h) Leader region
(i) Shine-Dalgarno sequence (j) Translation start codon (k) Translation stop codon (l) 5' cap
(m) Poly-A tail Chapter 11—
The Regulation of Gene Activity S11.1 Define the following terms:
(c) Cis-dominant mutation
(e) Combinatorial control
(f) Polycistronic mRNA
(g) Transcriptional activation by recruitment
511.2 How do gene and genome organization differ in eukaryotes and prokaryotes?
511.3 At what different levels can gene expression be regulated?
511.4 Give some examples of gene regulation by alterations in the DNA.
511.5 The regulation of transcription by attenuation cannot take place in eukaryotes. Why not?
511.6 Why is the lac operon of E. coli not inducible in the presence of glucose?
511.7 In eukaryotic organisms, the amounts of some proteins found in the cell are known to vary substantially with environmental conditions, whereas the amount of mRNA for these same proteins remains relatively constant. What mechanisms of regulation might be responsible for this phenomenon?
511.8 Two genotypes of E. coli are grown and assayed for levels of the enzymes of the lac operon. Using the information provided here for these two genotypes, predict the enzyme levels for the other genotypes listed in (a) through (d). (The levels of activity are expressed in arbitrary units relative to those observed under the induced conditions.)
I+ o+ Z+ Y+
I+ oc Z+ Y+
(a) I' O+ Z+ Y+
(b) F' I+ O+ Z- Y~H~ O+ Z+ Y+
(c) F' I+ O+ Z- Y"/I+ Oc Z+ Y+
(d) F' I+ O+ Z- Y'/I' Oc Z+ Y+
511.9 Imagine a bacterial species in which the methionine operon is regulated only by an attenuator and there is no repressor. In its mode of operation, the methionine attenuator is exactly analogous to the trp attenuator of E. coli. The relevant portion of the attenuator sequence is
The translation start site is located upstream from this sequence, and the AUG codons in the region shown cannot be used for translation initiation. What phenotype (constitutive, wildtype, or met-) would you expect of each of the following types of mutations? Explain your reasoning.
(a) The boldface A is deleted.
(b) Both the boldface and the underlined A are deleted.
(c) The first three As in the sequence are deleted.
511.10 How many proteins are bound to the trp operon when:
(a) Tryptophan and glucose are present?
(b) Tryptophan and glucose are absent?
(c) Tryptophan is present and glucose is absent?
511.11 Why are mutations of the lac operator often called cis- dominant? Why are some constitutive mutations of the lac repressor (lacI) called trans-recessive? Can you think of a way in which a noninducible mutation in the lacI gene might be trans-dominant?
511.12 An E. coli mutant strain synthesizes b-galactosidase whether or not the inducer is present. What genetic defect(s) might be responsible for this phenotype?
511.13 Temperature-sensitive mutations in the lacI gene of E. coli render the repressor nonfunctional (unable to bind the operator) at 42°C but leave it fully functional at 30°C. Under which of the following conditions would b- galactosidase be produced?
(a) In the presence of lactose at 30°C
(b) In the presence of lactose at 42°C
(c) In the absence of lactose at 30°C
(d) In the absence of lactose at 42°C
511.14 Many metabolic pathways are regulated by so-called feedback inhibition, in which the first enzyme in the pathway becomes inhibited by interacting with the final product of the entire pathway if the final product is present in a sufficient amount. Explain why a metabolic pathway should evolve in such a way that the first enzyme in the pathway is inhibited rather than the last enzyme in the pathway.
511.15 Among cells of a lacI+ lacO+ lacZ+ lacY+ strain of E. coli, a mutation was found in which the lacY gene underwent a precise inversion without any change in the coding region. However, the strain was found to be defective in the production of the permease. Why?
511.16 Cristina, a young immunologist, told her friend Jorge, who studied molecular biology, that there are 108 different antibodies produced by human B cells. To her astonishment, Jorge strongly disagreed, saying that there are not enough genes in the human genome to code for such a great variety of immunoglobulins. Who is right? Is there any way to reconcile their views?
511.17 You are engaged in the study of a gene, the transcription of which is activated in a particular tissue by a steroid hormone. From cultured cells isolated from this tissue, you are able to isolate mutant cells that no longer respond to the hormone. What are some possible explanations?
511.18 A laboratory strain of E. coli produces b-galactosidase constitutively and permease inducibly, despite the fact that these two proteins are co-regulated in the lac operon in wildtype strains. What is the genotype of the lac operon in this strain?
The Genetic Control of Development
S12.1 Define the following terms:
(a) Embryonic induction
(b) Gain-of-function mutation
(c) Fate map
(d) Pair-rule gene
(e) Segment-polarity gene
(g) Homeotic mutation
512.2 What is the relationship between maternal-effect genes and zygotic genes?
512.3 In the early development of C. elegans, what is the role of the anchor cell in the differentiation of the vulva?
512.4 When in the development of most metazoan animals is the polarity of the embryo determined?
512.5 The same transmembrane receptor protein encoded by the lin-12 gene is used in the determination of different developmental fates. What is the principal difference between the two types of target cells that develop ifferently in response to Lin-12 stimulation?
512.6 Programmed cell death (apoptosis) is responsible in part for shaping many organs and tissues in normal development. If a group of cells in the duck leg primordium that are destined to die are transplanted from their normal leg site to another part of the embryo just prior to the time they would normally die, they still die on schedule. The same operation performed a few hours earlier rescues the cells, and they do not die. How can you explain this observation?
512.7 Classify each of the following mutations:
(a) A mutation in C. elegans in which a cell that normally produces two distinct daughter cells gives rise to two identical cells.
(b) A mutation in Drosophila that causes an antenna to appear at the normal site of a leg.
(c) A lethal mutation in Drosophila that is responsible for abnormal gene expression in alternating segments of the embryo
512.8 What causes induction of the morphogenic events in the Drosophila imaginal disks? When does this induction happen?
512.9 If a small amount of cytoplasm is removed from the posterior end of the Drosophila embryo before blastoderm formation, the mature fly that develops from this embryo is normal in every way except it does not have germ cells. How can you explain this finding?
512.10 How would you prove that a newly discovered mutation has a maternal effect?
512.11 A recessive mutation smooth is found in a species of flower in which hairs (trichomes) normally present on the leaf surface are missing. Another mutation in the same species, called hairy, results in the presence of trichomes on the flower petals, where they are not normally formed. Cloning and sequencing indicate that these mutations are different alleles of the same gene. Molecular analysis reveals that smooth is a frameshift mutation near the start of the coding sequence and that hairy has an upstream insertion of a transposable element that allows transcription of the gene in petal tissue. What features of these observations suggest that the gene in question is an important gene for the regulation of trichome formation?
512.12 Two classes of genes involved in segmentation of the Drosophila embryo are gap genes, which are expressed in one region of the developing embryo, and pair-rule genes, which are expressed in seven stripes. Homozygotes for mutations in gap genes lack a continuous block of larval segments; homozygous mutations in pair-rule genes lack alternating segments. You examine gene expression by mRNA in situ hybridization and find that (1) the embryonic expression pattern of gap genes is normal in all pair-rule mutant homozygotes, and (2) the pair-rule gene expression pattern is abnormal in all gap gene mutant homozygotes. What do these observations tell you about the temporal hierarchy of gap genes and pair-rule genes in the developmental pathway of segmentation?
512.13 A paracentric inversion in Drosophila causes a dominant mutation that produces small wings. You mutagenize flies homozygous for this inversion with x rays and cross them to normal mates. Most progeny that arise have small wings, but a few rare progeny have normal wings. These rare progeny still have the original inversion but also contain a deletion spanning the more proximal of the two inversion breakpoints. (Proximal means closer to the centromere.) You correctly conclude from these observations that the dominant mutation is due to a genetic "gain of function" rather than to a reduction in gene activity. Explain how these results support this conclusion, and suggest an interpretation of how the gain-of-function phenotype arises.
512.14 One approach to the identification of important developmental genes in the mouse is to focus on genes whose DNA sequences are similar to those that have significant developmental roles in other organisms. Suppose that you want to study the mouse homolog of the lin-3 gene of C. elegans, which is important in vulva formation in the worm. Homozygotes for a null mutation of lin-3 lack a vulva. You propose that a homolog of lin-3 exists in the mouse, and you want to examine the phenotype of a mutant mouse that lacks a functional form of this gene. Arrange the following steps in the order in which you would perform them to generate a mouse homozygous for a null mutation of the gene.
(a) Identify an open reading frame in a mouse cDNA that is similar to the C. elegans lin-3 open reading frame.
(b) Create a deletion in a mouse genomic clone, and insert a gene resistant to the drug G418.
(c) Breed chimeric mice to donor strains to identify heterozygotes for a null mutation.
(d) Inject embryonic stem cells into differentially marked mouse blastocysts.
(e) Probe a mouse wildtype genomic library with a labeled mouse cDNA.
(f) Probe a mouse cDNA library with a labeled C. elegans lin-3 cDNA.
(g) Select for embryonic stem cells that can grow on G418.
(h) Intercross null mutant heterozygotes to obtain a homozygous null mouse.
(i) Identify chimeric mice.
(j) Inject the DNA of an engineered mutation in the mouse lin-3 homolog into embryonic stem cells. (k) Identify embryonic stem cells that have a null mutation in the mouse lin-3 homolog.
512.15 Edward B. Lewis, Christianne Nusslein-Volhard, and Eric Weichaus shared a 1995 Nobel Prize in Physiology or Medicine for their work on the developmental genetics of Drosophila. In their screen for developmental genes, NussleinVolhard and Weichaus initially identified 20 lines bearing maternal-effect mutations that produced embryos lacking anterior structures but having the posterior structures duplicated. When Nusslein- Volhard mentioned this result to a colleague, he was astonished to hear that mutations in 20 genes could give rise to
this phenotype. Explain why his astonishment was completely unfounded.
S12.16 The gene bicoid exemplifies a maternal-effect gene that, when mutated, produces embryos that lack anterior structures but have the posterior structures duplicated. The Bicoid protein is a morphogen that establishes the domain of expression of the gap gene hunchback in a concentration dependent manner. The Bicoid protein is usually present in an anterior-to-posterior gradient with the highest concentration at the anterior end of the early embryo. Diagrammed below is the major domain of expression of hunchback in an embryo from a wildtype mother.
hunchback egression psttem
What are the expected domains of hunchback expression in embryos from mothers with the following genotypes. Express your answers in the form of diagrams similar to the one here.
(a) Homozygous mutant for bicoid
(b) Homozygous for a duplication of a wildtype copy of bicoid
(c) Homozygous for transgene that causes bicoid to be localized at both poles of the embryo
Mutation, DNA Repair, and Recombination
513.1 Define the following terms:
(a) Gene conversion
(b) Mismatch repair
(c) Silent mutation
(e) Transversion mutation
513.2 Prior to the early 1940s, some bacteriologists maintained that the exposure of a bacterial population to an antibiotic or other selective agent induced mutations that enabled the bacteria to survive. Which of the following statements describes what Joshua and Esther Lederberg demonstrated with the replica plating technique?
(a) Streptomycin caused the formation of streptomycinresistant bacteria.
(b) Streptomycin revealed the presence of streptomycinresistant bacteria.
(c) Mutations are usually beneficial.
(d) Mutations are usually deleterious.
(e) None of the above.
513.3 If base substitutions were random, what ratio of transversions to transitions would be expected?
513.4 In the real world, spontaneous base substitutions are biased in favor of transitions. What is the consequence of this bias for base substitutions in the third position of codons?
513.5 How many different mutations can result from a single base substitution in the unique tryptophan codon TGG? Classify each as silent, missense, or nonsense.
513.6 What is the minimum number of single-nucleotide substitutions that would be necessary for each of the following amino acid replacements?
(a) Trp ' Lys
(b) Tyr ' Gly
(c) Met ' His
(d) Ala ' Asp
513.7 Every human gamete contains, very approximately, 50,000 genes. If a mutation rate is between 10-5 and 10-6 new mutations per gene per generation, what percentage of all gametes will carry a gene that has undergone a spontaneous mutation?
513.8 If a mutation rate is between 10-5 and 10-6 per gene per generation, what does this imply in terms of the number of mutations of a particular gene per million gametes per generation?
513.9 If the spontaneous mutation rate of a gene is 10-6 mutations per generation, and a dose of ionizing radiation of 10 grays doubles the rate of mutation, what mutation rate would be expected with 1 gray of radiation? Assume that the rate of induced mutation is proportional to the radiation dose.
513.10 In theory, deamination of both cytosine and 5-methyl-cytosine results in a mutation. However, nucleotide substitutions produced by loss of the amino group from cytosine are rarely observed, although those produced by
loss of the amino group from 5-methylcytosine are relatively frequent. What is a possible explanation for this phenomenon?
513.11 The mismatch repair system can detect mismatches that occur in DNA replication and can correct them after replication. In E. coli, how does the system use DNA methylation to determine which base in a pair is the incorrect one?
513.12 If one of the many copies of the tRNA gene for Glu underwent a mutation in which the anticodon became CUA instead of CUC, what would be the result? (Hint: The key is that only one of many tRNA copies is mutated.)
513.13 A strain of E. coli contains a mutant Leu tRNA. Instead of recognizing the codon 5'-CUG-3', as it does in nonmutant cells, the mutant tRNA now recognizes the codon 5'-GUG-3'. A missense mutation of another gene, affecting amino acid number 28 along the chain, is suppressed in cells with the mutant Leu tRNA.
(a) What are the anticodons of the wildtype and mutant Leu tRNA molecules?
(b) What kind of mutation is present in the mutant Leu tRNA gene?
(c) What amino acid would be inserted at position 28 of the mutant polypeptide chain if the missense mutation were not suppressed?
(d) What amino acid is inserted at position 28 when the missense mutation is suppressed?
513.14 A mutation is selected after treatment with a particular mutagen, and organisms containing the mutation are then mutagenized to find revertants.
(a) Can a mutation caused by hydroxylamine be induced to revert to the wildtype sequence by re-exposure to hydroxylamine?
(b) Can a mutation caused by hydroxylamine be induced to revert by nitrous acid?
(Hydroxylamine reacts specifically with cytosine and causes GC R AT transitions; nitrous acid causes deamination of A and C).
513.15 The tryptophan synthase of E. coli is a tetrameric enzyme formed by the polypeptide products of trpB and trpA genes. The wildtype subunit encoded by trpB has a glycine in position 38, whereas two mutants, B13 and B32, have Arg and Glu at this position, respectively. Plating these mutants on minimal medium sometimes yields spontaneous revertants to prototrophy. Four independent revertants of B13 have Ile, Thr, Ser, and Gly in position 38, and three independent revertants of B32 have Gly, Ala, and Val in the same position. What codon is in position 38 in the wildtype trpB gene?
513.16 In the rIIB gene of bacteriophage T4, there is a small coding region near the beginning of the gene in which the
particular amino acid sequence that is present is not essential for protein function. (Single-base insertions and deletions in this region were used to prove the triplet nature of the genetic code.) Under what condition would a + 1 frameshift mutation at the beginning of this region not be suppressed by a - 1 frameshift mutation at the end of the region?
513.17 E. coli mutation lacZ-1 was induced by acridine treatment, whereas lacZ-2 was induced by 5-bromouracil. What kind of mutations are they likely to be? What sort of mutant b-galactosidase will these mutations produce?
513.18 Is it correct to say that a Holliday structure and a chiasma are the same?
513.19 When two mutations are closely linked (for example, two nucleotide substitutions in the same gene), a single gene conversion event may include both sites. This phenomenon is called co-conversion. How would you expect the frequency of co-conversion to depend on the distance between mutant nucleotide sites?
513.20 In spite of gene conversion, among gametes chosen at random, one still observes the Mendelian segregation ratio of 1 : 1 in crosses. Why?
513.21 You carry out a large-scale cross of genotypes A m B x a + b of two strains of the mold Neurospora crassa and observe a number of aberrant asci, some of which are shown below.
The collegue who provided the strains insists that they are both deficient in the same gene in the DNA mismatch repair pathway. The results of your cross seem to contradict this assertion.
(a) Which of the asci depicted above would you exhibit as evidence that your colleague is incorrect?
(b) Which ascus (or asci) would you exhibit as definitive evidence that DNA mismatch repair in these strains is not 100 percent efficient?
(c) Among all asci resulting from meioses in which heteroduplexes form across the m + region, what proportion of acsi will be tetratype for the A and B markers and also show 4 : 4 segregation of m: +, under the assumption that 30 percent of heteroduplexes in the region are not corrected, 40 percent are corrected to m. and 30 percent are corrected to +?
S13.22 In molecular evolutionary studies, biologists often regard serine as though it were two different amino acids, one type of "serine" corresponding to the fourfold-degenerate UCN codon and the other type of "serine" corresponding to the twofold-degenerate AGY codon. Why does it make sense to regard serine in this way but not the other amino acids encoded by six codons: leucine (UUR and CUN) and arginine (AGR and CGN)? (In this symbolism for the nucleotides, Y is any pyrimidine, R any purine, and N any nucleotide.)
Chapter 14— Extranuclear Inheritance
S14.1 Define the following terms:
(a) Maternal inheritance
(b) Maternal effect
(c) Cytoplasmic male sterility
(e) RNA editing
(f) Phylogenetic tree
514.2 A woman with the mild form of a mitochondrial disease, Leber's hereditary optic neuropathy (LHON), has four children: two with severe disease expression, one unaffected, and one with the same level of expression as the mother. Explain this pattern of inheritance.
514.3 How do mitochondrial and nuclear genomes interact at the functional level?
514.4 A dwarfed corn plant is crossed as a female parent to a normal plant, and the F1 progeny are found to be dwarfed. The F1 plants are self-fertilized, and the F2 plants are found to be normal. Each F2 plant is self-fertilized, and 3/4 of F3 generation are found to be normal whereas 1/4 of the plants are dwarfed. Which of the following hypotheses are consistent with these results? Explain your reasoning.
(a) Cytoplasmic inheritance
(b) Nuclear inheritance
(c) Maternal effect
(d) Cytoplasmic inheritance plus mutation in a nuclear gene
514.5 A strain of yeast with a mutation ery-r conferring resistance to the antibiotic erycetin is crossed with a strain of opposite mating type with the genotype ery-s. After meiosis and sporulation, three tetrads are isolated. The spores have the following genotypes:
a ery-r a ery-s a ery-r
a ery-r a ery-s a ery-r
a ery-r a ery-s a ery-r
a ery-r a ery-s a ery-r
(a) What genetic hypothesis can explain these results?
(b) If an ery-r strain were used to generate petites, would you expect some of the petites to be sensitive to erycetin? Explain.
S14.6 In a fungal organism closely related to Saccharomyces cerevisiae, a strain resistant to three antibiotics is used to induce petite colonies. The antibiotics are erycetin, chlorpromanin, and paralysin, and the resistance genes are denoted ery-r. ch-r, andpar-r, respectively. Each of the resistance genes is known to be present in mitochondrial DNA. Each row in the accompanying table is the result of an experiment in which petite colonies selected for loss of genetic marker A (one of the antibiotic resistances) were tested for simultaneous loss of genetic marker B (another of the antibiotic resistances).
ery-r chl-r 170 19
ery-r par-r 54 81
chl-r par-r 116 8
From these data, deduce the order of the genes and the relative distances between them in the mitochondrial DNA.
514.7 Two haploid strains of Neurospora are isolated that have genetic defects in the ATPase complex. The mutants are designated A1' and A2'. The crosses AT x A1+ and A2' x A2+ were carried out, and one ascus was analyzed from each cross. The cross A1 x A1+ yielded a 4 : 4 ratio of A1+ : A1 ascospores, whereas the cross A2' x A2+ yielded an 8 : 0 ratio of A2+ : A2' ascospores.
(a) What difference between the mutations can explain these discordant results?
(b) What other types of tetrads would one expect from these crosses?
514.8 Mothers addicted to cocaine who use the drug during pregnancy can give birth to babies who are also addicted. Suggest a mechanism for this maternal effect.
514.9 If organelles were originally symbionts, why don't we call them symbionts instead of organelles?
514.10 How would you prove the existence of RNA editing in mitochondria?
514.11 What is controversial about the hypothesis of a mitochondrial Eve who lived in Africa? Is the controversial part the conclusion that there is a common ancestor of human mitochondrial DNA, or is it the inference of the geographical location of the common ancestor? Explain.
515.1 Define the following terms:
(b) Natural selection
(c) Random genetic drift
(d) Polymorphic gene
(e) Selectively neutral mutation
(f) Selection-mutation balance
515.2 In a large, randomly mating herd of cattle, 16 percent of the newborn calves have a certain type of dwarfism because of a homozygous recessive allele. Assume Hardy-Weinberg frequencies.
(a) What is the frequency of the recessive dwarfing allele?
(b) What is the frequency of heterozygous carriers in the herd?
(c) Among the cattle that are nondwarfs, what is the frequency of carriers?
515.3 A population of fish includes 1 percent albinos resulting from a homozygous recessive allele that eliminates the normal pigmentation. Assuming Hardy-Weinberg proportions, determine the frequency of heterozygous genotypes.
515.4 Hereditary hemochromatosis, a defect in iron metabolism resulting in impaired liver function, arthritis, and other symptoms, is one of the most common genetic diseases in northern Europe. The disease is relatively mild and easily treated when recognized. About one person in 400 is affected. Assuming Hardy-Weinberg proportions, determine the expected frequency of carriers.
515.5 An inherited defect in a certain receptor protein confers resistance to infection with HIV1 (human immunodeficiency virus 1, the cause of AIDS). Resistant people are homozygous recessive and are found at a frequency of 1 percent in the population. Assuming Hardy-Weinberg proportions, determine the frequency of heterozygous genotypes.
515.6 Organisms were sampled from three different populations polymorphic for the alleles A and a and yielded these observed numbers of each of the possible genotypes:
(a) AA: 5 Aa: 60 aa: 75 Total: 140
(b) AA: 10 Aa: 50 aa: 80 Total: 140
(c) AA: 15 Aa: 40 aa: 85 Total: 140
In terms of the magnitude of the chi-square obtained in a test of goodness of fit to Hardy-Weinberg proportions, which of these samples is closest to Hardy-Weinberg proportions? Do any of the samples deviate significantly from Hardy-Weinberg proportions?
515.7 In a population in Hardy-Weinberg equilibrium for a recessive allele at frequency q, what value of q is necessary for the ratio of carriers to affected organisms to equal each of the following?
515.8 A man is known to be a carrier of the cystic fibrosis allele. He marries a phenotypically normal woman. In the general population, the incidence of cystic fibrosis at birth is approximately one in 1700. Assume Hardy-Weinberg proportions.
(a) What is the probability that the wife is also a carrier?
(b) What is the probability that their first child will be affected?
515.9 A man with normal parents whose brother has phenylketonuria marries a phenotypically normal woman. In the general population, the incidence of phenylketonuria at birth is approximately one in 10,000. Assume Hardy- Weinberg proportions.
(a) What is the probability that the man is a carrier?
(b) What is the probability that the wife is also a carrier?
(c) What is the probability that their first child will be affected?
515.10 What genotype frequencies constitute the Hardy-Weinberg principle for a gene in a large, randomly mating allotetraploid population with two alleles at frequencies p and q?
515.11 A large population includes 10 alleles of a gene. The ratio of the allele frequencies forms the arithmetic progression 1 : 2 : 3 : . . .: 8 : 9 : 10. The genotypes are in Hardy-Weinberg proportions.
(a) What ratio is formed by the frequencies of the homozygous genotypes?
(b) Among all heterozygous genotypes that carry the most common allele, what ratio is formed by the frequencies of the heterozygous genotypes?
515.12 For an X-linked gene with two alleles in a large, randomly mating population, the frequency of carrier females equals one-half of the frequency of the males carrying the recessive allele. What are the allele frequencies?
515.13 A trait due to a harmful recessive X-linked allele in a large, randomly mating population affects one male in 50. What is the frequency of carrier females? What is the expected frequency of affected females?
515.14 In DNA typing, how many genes must be examined to prove that evidence left at a crime scene comes from a suspect? How many genes must be examined to prove that the evidence does not come from a suspect?
515.15 The DNA type of a suspect and that of blood left at the scene of a crime match for each of n genes, and each gene is heterozygous and so yields two bands. The frequency of each of the bands in the general population is 0.1. Use the product rule to calculate:
(a) The number of genes for which the probability of a match by chance is less than one in a million.
(b) The number of genes for which the probability of a match by chance is less than one in ten billion.
515.16 Two strains of bacteria, A and B, are placed into direct competition in a chemostat. A is favored over B. What is the value of the selection coefficient (s) if, after one generation, the ratio of the number of A cells to the number of B cells:
(a) Increases by 10 percent?
(b) Increases by 90 percent?
(c) Increases by a factor of 2?
515.17 Two strains of bacteria, A and B, are placed into direct competition in a chemostat. A is favored over B. If the selection coefficient per generation is constant, what is its value if, in an interval of 100 generations:
(a) The ratio of A cells to B cells increases by 10 percent?
(b) The ratio of A cells to B cells increases by 90 percent?
(c) The ratio of A cells to B cells increases by a factor of 2?
515.18 A strain of pathogenic bacteria undergoes cell division exactly two times as often as a nonpathogenic mutant strain of the same bacteria. What is the relative fitness of the pathogenic strain compared to the nonpathogenic strain? You may regard the generation time as the length of time for one division of a nonpathogenic cell.
515.19 An allele A undergoes mutation to the allele a at the rate of 10-5 per generation. If a very large population is fixed for A (generation 0), what is the expected frequency of A in the following generation (generation 1)? What is the expected frequency of A in generation 2? Deduce the rule for the frequency of A in generation n.
515.20 Large samples are taken from a single population of each of four related species of the flowering plant Phlox. The genotype frequencies for a gene with two alleles are estimated from each of the samples, and this yields the genotype frequencies in the accompanying table. For each species, calculate the allele frequencies and the genotype frequencies expected using the Hardy-Weinberg principle. Identify which of the populations give evidence of inbreeding, and for each apparently inbreeding population, calculate the inbreeding coefficient.
S15.21 The considerations in this problem illustrate the principle that for a rare, harmful, recessive allele, matings between relatives can account for a majority of affected individuals even though they account for a small proportion of all matings. Suppose that the frequency of the rare homozygous recessive genotype among the progeny of nonrelatives is one in 4 million and that first-cousin matings constitute 1 percent of all matings in the population.
(a) What is the frequency of the homozygous recessive among the offspring of first-cousin matings?
(a) Among all homozygous recessive genotypes, what proportion come from first-cousin matings?
515.22 A diploid population consists of 50 breeding organisms, of which one organism is heterozygous for a new mutation. If the population is maintained at a constant size of 50, what is the probability that the new mutation will be lost by chance in the first generation as a result of random genetic drift? (You may assume that the species is monoecious.) What is the answer in the more general case in which the breeding population consists of N diploid organisms?
515.23 A population segregating for the alleles A and a is sampled, and genotypes are found in the proportions 1/3 AA, 11/3 Aa, and 1/3 aa. These are not Hardy-Weinberg proportions. How big would the size of the sample have to be for the resulting P value to equal 0.05 in a chi-square test for goodness of fit?
515.24 An observed sample of a gene with two alleles from a population conforms almost exactly to the genotype frequencies expected for a population with an inbreeding coefficient F and allele frequencies 0.2 and 0.8. What is the smallest value of F that will lead to rejection of HardyWeinberg proportions at the 5 percent level of significance in each of the following cases? (Hint: Set chi-square equal to 4.0.)
(a) A sample size of 200 organisms.
(b) A sample size of 500 organisms.
(c) A sample size of 1000 organisms.
Chapter 16— Quantitative Genetics
516.1 Define the following terms:
(a) Multifactorial trait
(b) Quantitative-trait locus (QTL)
(c) Threshold trait
(d) Individual selection
(e) Broad-sense heritability
(f) Genotype-environment association
516.2 A population of sweet clover includes 20 percent with two leaves, 70 percent with three leaves, and 10 percent with four leaves. What are the mean and variance in leaf number in this population?
516.3 In an experimental population of the flour beetle Tribolium castaneum, the pupal weight is distributed normally with a mean of 2.0 mg and a standard deviation of 0.2 mg. What proportion of the population is expected to have a pupal weight between 1.8 and 2.2 mg? Between 1.6 and 2.4 mg? Would you expect to find an occasional pupa weighing 3.0 mg or more? Explain your answer.
516.4 Tabulated below are the numbers of eggs laid by 50 hens over a 2-month period. The hens were selected at random from a much larger population. Estimate the mean, variance, and standard deviation of the distribution of egg number in the entire population from which the sample was drawn.
48 50 51 47 54 45 50 38 40 52
58 47 55 53 54 41 59 48 53 49
51 37 31 47 55 46 49 48 43
59 51 52 66 54 37 46 55 59 45
44 44 57 51 50 57 50 40 63 33
516.5 Suppose that the 50 hens in the previous problem constituted the entire population of hens present in a flock instead of a sample from a larger flock. What are the mean, variance, and standard deviation of egg number in the population? Why are some of these numbers different from those calculated in the previous problem?
516.6 Two homozygous genotypes of Drosophila differ in the number of abdominal bristles. In genotype AA, the mean bristle number is 20 with a standard deviation of 2. In genotype aa, the mean bristle number is 23 with a standard deviation of 3. Both distributions conform to the normal curve, in which the proportions of the population that have a phenotype within an interval defined by the mean ±1, ±1.5, ± 2, and ± 3 standard deviations are 68, 87, 95, and 99.7 percent, respectively.
(a) In genotype AA, what is the proportion of flies with a bristle number between 20 and 23?
(b) In genotype aa, what is the proportion of flies with a bristle number between 20 and 23?
(c) What proportion of AA flies have a bristle number greater than the mean of aa flies?
(d) What proportion of aa flies have a bristle number greater than the mean of AA flies?
516.7 If the genotypes in the previous problem were crossed and the F1 flies mated to produce an F2 generation, would you expect the distribution of bristle number among F2 flies to show segregation of the A and a alleles? Explain your answer.
516.8 A meristic trait is determined by a pair of alleles at each of four unlinked loci: A, a and B, b and C, c and D, d. The alleles act in an additive fashion such that each capital letter adds one unit to the phenotypic score. Thus the phenotype ranges from a score of 0 (genotype aa bb cc dd) to 8 (genotype AA BB CC DD). What is the expected distribution of phenotype score in the F2 generation of the cross aa bb cc dd x AA BB CC DD? Draw a histogram of the distribution.
516.9 With respect to the trait in the previous problem, what is the expected distribution of phenotypes in a randomly mating population in which the frequency of each allele indicated with a capital letter is 0.8 and in which the alleles of the different loci are combined at random to produce the genotypes?
516.10 The F1 generation of a cross between two inbred strains of maize has a standard deviation in mature plant height of 15 cm. When grown under the same conditions, the F2 generation produced by the F1 x F1 cross has a stan-
dard deviation in plant height of 25 cm. What are the genotypic variance and the broad-sense heritability of mature plant height in the F2 population?
516.11 Two inbred strains of Drosophila are crossed and the F1 generation is brother-sister mated to produce a large, genetically heterogeneous F2 population that is split into two parts. One part is reared at a constant temperature of 25°C, and the other part is reared under conditions in which the temperature fluctuates among 18°C, 25°C, and 29°C. The broadsense heritability of body size at the constant temperature is 45 percent, but under fluctuating temperatures it is 15 percent. If the genotypic variance is the same under both conditions, by what factor does the temperature fluctuation increase the environmental variance?
516.12 The narrow-sense heritability of withers height in a population of quarterhorses is 30 percent. (Withers height is the height at the highest point of the back, between the shoulder blades.) The average withers height in the population is 17 hands. (A ''hand" is a traditional measure equal to the breadth of the human hand, now taken to equal 4 inches.) From this population, studs and mares with an average withers height of 16 hands are selected and mated at random. What is the expected withers height of the progeny? How does the value of the narrow-sense heritability change if withers height is measured in meters rather than hands?
516.13 The narrow-sense heritability of adult stature in people is about 25 percent. What does this value imply about the correlation coefficient in adult height between father and son?
516.14 Suppose that the narrow-sense heritability of adult stature in people equals 25 percent. A man whose height is 5 cm above the average height for men marries a woman whose average height equals that for women. Among their progeny, what is the expected average deviation from the mean adult height in the entire population?
516.15 From a population of Drosophila with an average abdominal bristle number of 23.2, flies with an average bristle number of 27.6 were selected and mated at random to produce the next generation. The average bristle number among the progeny was 25.4. What is the realized heritability? Is this a narrow-sense or a broad-sense heritability? Explain your answer.
516.16 In human beings, there is about a 90 percent correlation coefficient between identical twins in the total number of raised skin ridges forming the fingerprints. What is the broad-sense heritability of total fingerprint ridge count? What is the maximum value of the narrow-sense heritability?
516.17 In human beings, the correlation coefficient between first cousins in the total fingerprint ridge count is 10 percent. On the basis of this value, what is the narrow-sense heritability of this trait?
516.18 The narrow-sense heritability of 6-week body weight in mice is 25 percent. From a population whose average 6-week weight is 20 grams, individual mice weighing an average of 24 grams are selected and mated at random. What is the expected weight of the progeny mice in the next generation?
516.19 In the mouse population in the problem above, the selection for greater 6-week weight is carried out for a total of five generations, and in each generation, parents are chosen with an average 6-week weight that is, on average, 4 grams heavier than the population mean in their generation. What is the expected average 6-week weight after the five generations of selection? Make a graph of average 6week weight against generation number.
516.20 Suppose that the selection for 6-week weight in the mouse population above were carried out in both directions for five generations. In each generation, the parents in the "up" selection are, on average, 4 grams heavier than the population mean, and the parents in the "down direction" are, on average, 4 grams lighter than the population mean.
(a) In each generation, how different are the expected means of the "up" and "down" populations?
(b) Draw a graph illustrating the expected course of selection in both populations.
(c) In the "down" population, is the realized heritability of 6-week weight expected to remain 25 percent indefinitely?
516.21 Two inbred lines of maize differ in ear length. In one, the average length of ear is 30 cm, and in the other it is 15 cm. In both lines the variance in ear length is 2 cm2. In the F2 generation of a cross between the lines, the average ear length is 22.5 cm with a variance of 5 cm2. What is the minimum number of genes required to account for these
S16.22 A threshold trait affects 2.5 percent of the organisms in a population. The underlying liability is a multifactorial trait that is normally distributed with mean 0 and variance 1.
(a) What is the value of the threshold, a liability above which causes the appearance of the condition?
(b) If the underlying liability were normally distributed with a mean of 1 and a variance of 4, what would be the corresponding value of the threshold?
(c) Is there any real difference between these two situations? Chapter 17—
Genetics of Biorhythms and Behavior S17.1 Define the following terms:
(a) Specific nonchemotaxis
(b) Multiple nonchemotaxis
(c) General nonchemotaxis
(d) Sensory adaptation
(e) Circadian rhythm
(f) Genotype-enviroment interaction
517.2 What is the target of the molecular pathway common to all forms of chemotaxis?
517.3 What is the role of methylation in sensory adaptation? What is the expected phenotype of a mutant in cheR that cannot transfer methyl groups?
517.4 What period of courtship song would you expect of D. melanogaster males hemizygous for the per0 allele?
517.5 What kind of an experiment would enable you to demonstrate that per controls the expression of other genes?
517.6 How does the circadian system of Drosophila become reset each day so that the fly's intrinsic period becomes synchronized to the 24-hour day?
517.7 You are studying two groups of laboratory mice raised in similar environments. One is a highly inbred strain, and the other is derived from a large randomly mating population. Which group would you expect to have a higher value of the broad-sense heritability of a particular behavioral trait? Explain your reasoning.
517.8 Suppose that the broad-sense heritability of IQ score is 0.5. Can you conclude that genes are responsible for 50 percent of the IQ score of a particular person? Why or why not?
517.9 Explain why twins and adoption studies are useful in genetic research on behavioral traits in human beings. Why are ordinary family studies not sufficient?
517.10 For a given actual population size, how would the magnitude of random genetic drift differ in two animal populations, one with a strong social hierarchy and another lacking an established social structure?
517.11 It has been reported that males of Drosophila with a genotype that is uncommon in the population have greater reproductive success than males with more common genotypes. What effect would a rare-male mating advantage have on the genetic composition of a population of flies?
517.12 Humans have changed from a species living in small hunting populations to a species many members of which live in large and highly organized communities. Have those changes also led to changes in the relative contribution of different evolutionary factors (such as, for instance, random genetic drift) to the shaping of human genotypic structure?
517.13 Consider a complex behavioral trait in which the environmental variance consists solely of the variance that results from errors in measurement. In other words, the behavioral trait is completely determined genetically but is difficult to measure because of random fluctuations in measurement. For a single measurement of the trait, let the
environmental variance be represented as <7'' If errors in measurement are the only cause of environmental variance,
then it can be shown that the environmental variance for the mean of n independent measurements is <J|' /n For such
a trait with a broad-sense heritability of H2 = /(fJ^ + ) for a single measurement, show that the broad-sense
heritability of the mean of n measurements, H2, equals nH2/[1 + (n - 1)H2]. (Hint: Find an expression for<J' in terms of H2.)