Genetics Test #2 End of Chapter Questions

7.27 In the nematode Caenorhabditis elegans, the linked genes dpy (dumpy body) and unc (uncoordinated behavior) recombine with a frequency P. If a repulsion heterozygote carrying recessive mutations in these genes is self-fertilized, what fraction of the offspring will be both dumpy and uncoordinated?

(P/2)2

6.8 If nondisjunction of chromosome 21 occurs in the division of a secondary oocyte in a human female, what is the chance that a mature egg derived from this division will receive two number 21 chromosomes?

1/2

7.32 One of the metacentric chromosomes in a species of bean typically has one chiasma in each of its arms during meiosis. What is the genetic map length of this chromosome?

100cM

7.30 A chromosome is 120 cM long. On the average, how many chiasmata will occur when it goes through meiosis?

2.4 chiasmata

7.28 In the following testcross, genes a and b are 20 cM apart, and genes b and c are 10 cM apart: a + + b + × a b b c. If the coefficient of coincidence is 0.5 over this interval on the linkage map, how many triply homozygous recessive individuals are expected among 1000 progeny?

5

7.31 The total map length of the Drosophila genome is about 270 cM. On the average, how many chiasmata will occur in an oocyte going through meiosis?

5.4 Chiasmata

6.25 The following diagram shows two pairs of chromosomes in the karyotypes of a man, a woman, and their child. The man and the woman are phenotypically normal, but the child (a boy) suffers from a syndrome of abnormalities, including poor motor control and severe mental impairment. What is the genetic basis of the child's abnormal phenotype? Is the child hyperploid or hypoploid for a segment in one of his chromosomes? Pictures in book

6.25 The following diagram shows two pairs of chromosomes in the karyotypes of a man, a woman, and their child. The man and the woman are phenotypically normal, but the child (a boy) suffers from a syndrome of abnormalities, including poor motor control and severe mental impairment. What is the genetic basis of the child's abnormal phenotype? Is the child hyperploid or hypoploid for a segment in one of his chromosomes?

6.17 One chromosome in a plant has the sequence A B C D E F, and another has the sequence M N O P Q R. A reciprocal translocation between these chromosomes produced the following arrangement: A B C P Q R on one chromosome and M N O D E F on the other. Illustrate how these translocated chromosomes would pair with their normal counterparts in a heterozygous individual during meiosis.

ANS Online:

5.18 In Drosophila, the gene for bobbed bristles (recessive allele bb, bobbed bristles; wild-type allele , normal bristles) is located on the X chromosome and on a homologous segment of the Y chromosome. Give the genotypes and phenotypes of the offspring from the following crosses: (a) (b) (c) (d)

ANS: (a) 1/2 Xbb Xbb bobbed females, 1/2 Xbb Y+ wild-type males; (b) 1/2 X+ Xbb wild-type females, 1/2 Xbb Ybb bobbed males; (c) 1/4 X+ X+ wild-type females, 1/4 X+ Xbb wild-type females, 1/4 X+ Ybb wild-type males, 1/4 Xbb Ybb bobbed males; (d) 1/4 X+ Xbb wild-type females, 1/4 Xbb Xbb bobbed females, 1/4 X+ Y+ wild-type males, 1/4 Xbb Y+ wild-type males.

7.39 The order of three genes and the centromere on one chromosome of Neurospora is centromere — x —y — z. A cross between + + + and x y z produced an ascus with the following ordered array of ascospores (only one member of each spore pair is shown): (++z) (+y z) (x + +) (x y +). (a) Is this ascus most likely the result of a meiotic event in which 0, 1, 2, or 3 crossovers occurred? (b) In what interval(s) did the crossover(s) most likely occur? (c) If double or triple crossovers were involved, were they two-strand, three-strand, or four-strand multiple crossovers?

ANS: (a) 2 crossovers; (b) one exchange between x and y, the other between y and z; (c) 4-strand double crossover.

7.40 The following tetrads were produced by a cross of a Neurospora strain that had white spores (w) and a nutritional requirement for the amino acid arginine (arg) with a strain that had dark spores and no arginine requirement:

ANS: (a) 8 cM; (b) 21.5 cM; (c) the arg and w loci are located in different arms of a chromosome.

7.7 Genes a and b are 20 cM apart. An a+ b+/a+ b+ individual was mated with an a b/a b individual. (a) Diagram the cross and show the gametes produced by each parent and the genotype of the F1. (b) What gametes can the F1 produce, and in what proportions? (c) If the F1 was crossed to a b/a b individuals, what offspring would be expected, and in what proportions? (d) Is this an example of the coupling or repulsion linkage phase? (e) If the F1 were intercrossed, what offspring would be expected and in what proportions?

ANS: (a) Cross: a+ b+/a+ b+ a b/a b Gametes: a+ b+ from one parent, a b from the other F1: a+ b+/a b (b) 40% a+ b+, 40% a b, 10% a+ b, 10% a b+ (c) F2 from testcross: 40% a+ b+/a b, 40% a b/a b, 10% a+ b/a b, 10% a b+/a b (d) Coupling linkage phase (e) F2 from intercross: Summary of phenotypes: a+ and b+ 66% a+ and b 9% a and b+ 9% a and b 16%

7.8 Answer questions (a)-(e) in the preceding problem under the assumption that the original cross was a+ b/a+ b × a b+/a b+.

ANS: (a) Cross: a+ b/a+ b ab+/ab+ Gametes: a+ b from one parent, ab+ from the other F1: a+ b/ab+ (b) 40% a+ b, 40% ab+, 10% a+ b+, 10% ab (c) F2 from testcross: 40% a+ b/ab, 40% ab+/ab, 10% a+ b+/ab, 10% ab/ab (d) Repulsion linkage phase (e) F2 from intercross: Summary of phenotypes: a+ and b+ 51% a+ and b 24% a and b+ 24% a and b 1%

5.19 Predict the sex of Drosophila with the following chromosome compositions (A haploid set of autosomes): (a) 4X 4A; (b) 3X 4A; (c) 2X 3A; (d) 1X 3A; (e) 2X 2A; (f) 1X 2A.

ANS: (a) Female; (b) intersex; (c) intersex; (d) male: (e) female; (f) male.

6.4 The following table presents chromosome data on four species of plants and their F1 hybrids: (a) Deduce the chromosomal origin of species A. (b) How many bivalents and univalents would you expect to observe at meiotic metaphase I in a hybrid between species C and species B? (c) How many bivalents and univalents would you expect to observe at meiotic metaphase I in a hybrid between species D and species B?

ANS: (a) Species A is an allotetraploid with a genome from each of species C and species D; (b) 0 bivalents and 15 univalents; (c) 0 bivalents and 15 univalents.

7.17 The Drosophila genes vg (vestigial wings) and cn (cinnabar eyes) are located at 67.0 and 57.0, respectively, on chromosome 2. A female from a homozygous strain of vestigial flies was crossed with a male from a homozygous strain of cinnabar flies. The F1 hybrids were phenotypically wild-type (long wings and dark red eyes). (a) How many different kinds of gametes could the F1 females produce, and in what proportions? (b) If these females are mated with cinnabar, vestigial males, what kinds of progeny would you expect, and in what proportions?

ANS: (a) The F1 females, which are cn vg+/cn+ vg, produce four types of gametes: 45% cn vg+, 45% cn+ vg, 5% cn+ vg+, 5% cn vg. (b) 45% cinnabar eyes, normal wings; 45% reddish-brown eyes, vestigial wings; 5% reddish-brown eyes, normal wings; 5% cinnabar eyes, vestigial wings.

5.3 In grasshoppers, rosy body color is caused by a recessive mutation; the wild-type body color is green. If the gene for body color is on the X chromosome, what kind of progeny would be obtained from a mating between a homozygous rosy female and a homozygous wild-type male? (In grasshoppers, females are XX and males are XO.)

ANS: All the daughters will be green and all the sons will be rosy.

7.15 In Drosophila, the genes sr (stripe thorax) and e (ebony body) are located at 62 and 70 cM, respectively, from the left end of chromosome 3. A striped female homozygous for e+ was mated with an ebony male homozygous for sr+. All the offspring were phenotypically wild-type (gray body and unstriped). (a) What kind of gametes will be produced by the F1 females, and in what proportions? (b) What kind of gametes will be produced by the F1 males, and in what proportions? (c) If the F1 females are mated with striped, ebony males, what offspring are expected, and in what proportions? (d) If the F1 males and females are intercrossed, what offspring would you expect from this intercross, and in what proportions?

ANS: (a) The F1 females, which are sr e+/sr+ e, produce four types of gametes: 46% sr e+, 46% sr+ e, 4% sr e, 4% sr+ e+. (b) The F1 males, which have the same genotype as the F1 females, produce two types of gametes: 50% sr e+, 50% sr+ e; remember, there is no crossing over in Drosophila males. (c) 46% striped, gray; 46% unstriped, ebony; 4% striped, ebony; 4% unstriped, gray. (d) The offspring from the intercross can be obtained from the following table. Summary of phenotypes striped, gray 0.25 unstriped, gray 0.50 striped, ebony 0 unstriped, ebony 0.25

5.10 A man who has color blindness and type O blood has children with a woman who has normal color vision and type AB blood. The woman's father had color blindness. Color blindness is determined by an X-linked gene, and blood type is determined by an autosomal gene. (a) What are the genotypes of the man and the woman? (b) What proportion of their children will have color blindness and type B blood? (c) What proportion of their children will have color blindness and type A blood? (d) What proportion of their children will be color blind and have type AB blood?

ANS: (a) The man is XcY ii; the woman is X+ Xc IA IB. (b) Probability color blind = 1/2; probability type B blood = 1/2; combined probability = (1/2) (1/2) = 1/4. (c) Probability color blind = 1/2; probability type A blood = 1/2; combined probability (1/2) (1/2) = 1/4. (d) 0.

7.23 Female Drosophila heterozygous for three recessive mutations e (ebony body), st (scarlet eyes), and ss (spineless bristles) were testcrossed, and the following progeny were obtained:

ANS: (a) Two of the classes (the parental types) vastly outnumber the other six classes (recombinant types); (b) st + +/+ ss e; (c) st—ss—e; (d) [(145 + 122) 1 + (18) 2]/1000 = 30.3 cM; (e) (122 + 18)/1000 = 14.0 cM; (f) (0.018)/ (0.163 0.140) = 0.789. (g) st + +/+ ss e females st ss e/st ss e males 2 parental classes and 6 recombinant classes.

5.22 What is the maximum number of Barr bodies in the nuclei of human cells with the following chromosome compositions: (a) XY; (b) XX; (c) XXY; (d) XXX; (e) XXXX; (f) XYY?

ANS: (a) Zero; (b) one; (c) one; (d) two; (e) three; (f) zero.

7.4 Mice have 19 autosomes in their genome, each about the same size. If two autosomal genes are chosen randomly, what is the chance that they will be on the same chromosome?

ANS: 1/19.

7.6 If two loci are 10 cM apart, what proportion of the cells in prophase of the first meiotic division will contain a single crossover in the region between them?

ANS: 20%.

5.14 In Drosophila, a recessive mutation called chocolate (c) causes the eyes to be darkly pigmented. The mutant phenotype is indistinguishable from that of an autosomal recessive mutation called brown (bw). A cross of chocolate-eyed females to homozygous brown males yielded wild-type females and darkly pigmented males. If the flies are intercrossed, what types of progeny are expected, and in what proportions? (Assume the double mutant combination has the same phenotype as either of the single mutants alone.)

ANS: 3/8 wild-type (red), 5/8 brown for both male and female F2 progeny.

6.2 In human beings, a cytologically abnormal chromosome 22, called the "Philadelphia" chromosome because of the city in which it was discovered, is associated with chronic leukemia. This chromosome is missing part of its long arm. How would you denote the karyotype of an individual who had 46 chromosomes in his somatic cells, including one normal 22 and one Philadelphia chromosome?

ANS: 46, XX, 22q- or 46, XY, 22q-, depending on the sex chromosome constitution.

6.20 Distinguish between a compound chromosome and a Robertsonian translocation.

ANS: A compound chromosome is composed of segments from the same pair of chromosomes, as when two X chromosomes become attached to each other. A Robertsonian translocation involves a fusion of segments from two different pairs of chromosomes. These segments fuse at or near the centromeres, usually with the loss of the short arms of each of the participating chromosomes.

6.28 In Drosophila, vestigial wing (vg), hairy body (h), and eyeless (ey) are recessive mutations on chromosomes 2, 3, and 4, respectively. Wild-type males that had been irradiated with X rays were crossed to triply homozygous recessive females. The F1 males (all phenotypically wild-type) were then testcrossed to triply homozygous recessive females. Most of the F1 males produced eight classes of progeny in approximately equal proportions, as would be expected if the vg, h, and ey genes assort independently. However, one F1 male produced only four classes of offspring, each approximately one-fourth of the total: (1) wild-type, (2) eyeless, (3) vestigial, hairy, and (4) vestigial, hairy, eyeless. What kind of chromosome aberration did the exceptional F1 male carry, and which chromosomes were involved?

ANS: A reciprocal translocation between chromosomes 2 and 3. One translocated chromosome carries the wild-type alleles of vg and h on chromosomes 2 and w, respectively, and the other carries the recessive mutant alleles of these genes. The chromosome that carries the ey gene (chromosome 4) is not involved in the rearrangement.

7.5 Two yeast strains differing in three linked genes were crossed: A B C × a b c. Among the tetrads that were analyzed, one contained the following spores: A B C, A b C, a B c, and a b c. How did this tetrad originate?

ANS: A two-strand double crossover must have occurred.

7.45 A Drosophila geneticist has identified a strain of flies with a large inversion in the left arm of chromosome 3. This inversion includes two mutations, e (ebony body) and cd (cardinal eyes), and is flanked by two other mutations, sr (stripe thorax) on the right and ro (rough eyes) on the left. The geneticist wishes to replace the e and cd mutations inside the inversion with their wild-type alleles; he plans to accomplish this by recombining the multiply mutant, inverted chromosome with a wild-type, inversion-free chromosome. What event is the geneticist counting on to achieve his objective? Explain.

ANS: A two-strand double crossover within the inversion; the exchange points of the double crossover must lie between the genetic markers and the inversion breakpoints.

6.21 A yellow-bodied Drosophila female with attached-X chromosomes was crossed to a white-eyed male. Both of the parental phenotypes are caused by X-linked recessive mutations. Predict the phenotypes of the progeny.

ANS: All the daughters will be yellow-bodied and all the sons will be white-eyed.

6.10 The Drosophila fourth chromosome is so small that flies monosomic or trisomic for it survive and are fertile. Several genes, including eyeless (ey), have been located on this chromosome. If a cytologically normal fly homozygous for a recessive eyeless mutation is crossed to a fly monosomic for a wild-type fourth chromosome, what kinds of progeny will be produced, and in what proportions?

ANS: Approximately half the progeny should be disomic ey/+ and half should be monosomic ey/O. The disomic progeny will be wild-type, and the monosomic progeny will be eyeless.

6.24 Each of six populations of Drosophila in different geographic regions had a specific arrangement of bands in one of the large autosomes: (a) 12345678 (b) 12263478 (c) 15432678 (d) 14322678 (e) 16223478 (f) 154322678 Assume that arrangement (a) is the original one. In what order did the other arrangements most likely arise, and what type of chromosomal aberration is responsible for each change?

ANS: Arrangement (a) produced (c) by inversion of segment 2345; (c) produced (f) by a duplication of band 2; (f) produced (d) by a deletion of band 5; (d) produced (e) by inversion of segment 43226; (e) produced (b) by inversion of segment 622.

7.16 In Drosophila, genes a and b are located at positions 22.0 and 42.0 on chromosome 2, and genes c and d are located at positions 10.0 and 25.0 on chromosome 3. A fly homozygous for the wild-type alleles of these four genes was crossed with a fly homozygous for the recessive alleles, and the F1 daughters were backcrossed to their quadruply recessive fathers. What offspring would you expect from this backcross, and in what proportions?

ANS: Because the two chromosomes assort independently, the genetic makeup of the gametes (and, therefore, of the backcross progeny) can be obtained from the following table.

5.24 A breeder of sun conures (a type of bird) has obtained two true-breeding strains, A and B, which have red eyes instead of the normal brown found in natural populations. In Cross 1, a male from strain A was mated to a female from strain B, and the male and female offspring all had brown eyes. In Cross 2, a female from strain A was mated to a male from strain B, and the male offspring had brown eyes and the female offspring had red eyes. When the birds from each cross were mated brother to sister, the breeder obtained the following results: Proportion in Proportion in Phenotype of Cross 1 of Cross 2 Brown male Red male Brown female Red female Provide a genetic explanation for these results.

ANS: Color is determined by an autosomal gene (alleles A and a) and a sex-linked gene (alleles B and b) on the Z chromosome (females are ZW and males are ZZ) and the recessive alleles are mutually epistatic—that is, aa, bb, or bW birds have red eyes, and A- B- or A- BW birds have brown eyes. Crosses Online

7.9 If the recombination frequency in the previous two problems were 40 percent instead of 20 percent, what change would occur in the proportions of gametes and testcross progeny?

ANS: Coupling heterozygotes a+ b+/a b would produce the following gametes: 30% a+ b+, 30% a b, 20% a+ b, 20% a b+; repulsion heterozygotes a+ b/a b+ would produce the following gametes: 30% a+ b, 30% a b+, 20% a+ b+, 20% a b. In each case, the frequencies of the testcross progeny would correspond to the frequencies of the gametes.

5.2 A male with singed bristles appeared in a culture of Drosophila. How would you determine if this unusual phenotype was due to an X-linked mutation?

ANS: Cross the singed male to wild-type females, and then intercross the offspring. If the singed bristle phenotype is due to an X-linked mutation, approximately half the F2 males, but none of the F2 females, will show it.

5.21 A Drosophila male carrying a recessive X-linked mutation for yellow body is mated to a homozygous wild-type female with gray body. The daughters of this mating all have uniformly gray bodies. Why aren't their bodies a mosaic of yellow and gray patches?

ANS: Drosophila does not achieve dosage compensation by inactivating one of the X chromosomes in females.

5.11 A Drosophila female homozygous for a recessive X-linked mutation that causes vermilion eyes is mated to a wild-type male with red eyes. Among their progeny, all the sons have vermilion eyes, and nearly all the daughters have red eyes; however, a few daughters have vermilion eyes. Explain the origin of these vermilion-eyed daughters.

ANS: Each of the rare vermilion daughters must have resulted from the union of an X(v) X(v) egg with a Y-bearing sperm. The diplo-X eggs must have originated through nondisjunction of the X chromosomes during oogenesis in the mother. However, we cannot determine if the nondisjunction occurred in the first or the second meiotic division.

5.13 A Drosophila female heterozygous for the recessive X-linked mutation w (for white eyes) and its wild-type allele is mated to a wild-type male with red eyes. Among the sons, half have white eyes and half have red eyes. Among the daughters, nearly all have red eyes; however, a few have white eyes. Explain the origin of these white-eyed daughters.

ANS: Each of the rare white-eyed daughters must have resulted from the union of an X(w) X(w) egg with a Y-bearing sperm. The rare diplo-X eggs must have originated through nondisjunction of the X chromosomes during the second meiotic division in the mother.

5.25 In 1908 F. M. Durham and D.C.E. Marryat reported the results of breeding experiments with canaries. Cinnamon canaries have pink eyes when they first hatch, whereas green canaries have black eyes. Durham and Marryat crossed cinnamon females with green males and observed that all the progeny had black eyes, just like those of the green strain. When the males were crossed to green females, all the male progeny had black eyes, whereas all the female progeny had either black or pink eyes, in about equal proportions. When the males were crossed to cinnamon females, four classes of progeny were obtained: females with black eyes, females with pink eyes, males with black eyes, and males with pink eyes—all in approximately equal proportions. Propose an explanation for these findings.

ANS: Eye color in canaries is due to a gene on the Z chromosome, which is present in two copies in males and one copy in females. The allele for pink color at hatching (p) is recessive to the allele for black color at hatching (P). There is no eye color gene on the other sex chromosome (W), which is present in one copy in females and absent in males. The parental birds were genotypically p/W (cinnamon females) and P/P (green males). Their F1 sons were genotypically p/P (with black eyes at hatching). When these sons were crossed to green females (genotype P/W), they produced F2 progeny that sorted into three categories: males with black eyes at hatching (P/-, half the total progeny), females with black eyes at hatching (P/W, a fourth of the total progeny), and females with pink eyes at hatching (p/W, a fourth of the total progeny). When these sons were crossed to cinnamon females (genotype p/W), they produced F2 progeny that sorted into four equally frequent categories: males with black eyes at hatching (genotype P/p), males with pink eyes at hatching (genotype p/p), females with black eyes at hatching (genotype P/W), and females with pink eyes at hatching (genotype p/W).

7.29 Drosophila females heterozygous for three recessive mutations, a, b, and c, were crossed to males homozygous for all three mutations. The cross yielded the following results:

ANS: From the parental classes, + + c and a b +, the heterozygous females must have had the genotype + + c/a b +. The missing classes, + b + and a + c, which would represent double crossovers, establish that the gene order is b—a—c. The distance between b and a is (96 + 110)/1000 = 20.6 cM and that between a and c is (65 + 75)/1000 = 14.0 cM. Thus, the genetic map is b—20.6 cM—a—14.0 cM—c.

7.37 Analysis of unordered yeast tetrads from the cross +++ × a b c yielded the following data:

ANS: Genes b and c are linked and are 17 cM apart. Gene a is located on a different chromosome.

7.41 The following pedigree, described in 1937 by C. L. Birch, shows the inheritance of X-linked color blindness and hemophilia in a family. What is the genotype of II-2? Do any of her children provide evidence for recombination between the genes for color blindness and hemophilia?

ANS: II-1 has the genotype C h/c H, that is, she is a repulsion heterozygote for the alleles for color blindness (c) and hemophilia (h). None of her children are recombinant for these alleles.

7.42 The following pedigree, described in 1938 by B. Rath, shows the inheritance of X-linked color blindness and hemophilia in a family. What are the possible genotypes of II-1? For each possible genotype, evaluate the children of II-1 for evidence of recombination between the color blindness and hemophilia genes.

ANS: II-1 is either (a) C h/c H or (b) c h/C H. Her four sons have the genotypes c h (1), C h (2), c H (3), and C H (4). If II-1 has the genotype C h/c H, sons 1 and 4 are recombinant and sons 2 and 3 are nonrecombinant. If II-1 has the genotype c h/C H, sons 2 and 3 are recombinant and sons 1 and 4 are nonrecombinant. Either way, the frequency of recombination is 0.5.

7.1 Mendel did not know of the existence of chromosomes. Had he known, what change might he have made in his Principle of Independent Assortment?

ANS: If Mendel had known of the existence of chromosomes, he would have realized that the number of factors determining traits exceeds the number of chromosomes, and he would have concluded that some factors must be linked on the same chromosome. Thus, Mendel would have revised the Principle of Independent Assortment to say that factors on different chromosomes (or far apart on the same chromosome) are inherited independently.

7.26 Assume that in Drosophila there are three genes x, y, and z, with each mutant allele recessive to the wild-type allele. A cross between females heterozygous for these three loci and wild-type males yielded the following progeny:

ANS: Ignore the female progeny and base the map on the male progeny. The parental types are + + z and x y +. The two missing classes (+ y + and x + z) must represent double crossovers; thus, the gene order is y—x—z. The distance between y and x is (32 + 27)/1000 = 5.9 cM and that between x and z is (31 + 39)/1000 = 7.0 cM. Thus, the map is y—5.9 cM—x—7.0 cM—z. The coefficient of coincidence is zero.

6.3 During meiosis, why do some tetraploids behave more regularly than triploids?

ANS: In allotetraploids, each member of the different sets of chromosomes can pair with a homologous partner during prophase I and then disjoin during anaphase I. In triploids, disjunction is irregular because homologous chromosomes associate during prophase I by forming either bivalents and univalents or by forming trivalents.

7.19 In maize, the genes Pl for purple leaves (dominant over pl for green leaves), sm for salmon silk (recessive to Sm for yellow silk), and py for pigmy plant (recessive to Py for normal-size plant) are on chromosome 6, with map positions as shown: pl-45, sm-55, py-65 Hybrids from the cross Pl sm sm py × pl Sm Sm Py were testcrossed with pl sm sm py plants. Predict the phenotypes of the offspring and their frequencies assuming (a) no interference and (b) complete interference.

ANS: In the enumeration below, classes 1 and 2 are parental types, classes 3 and 4 result from a single crossover between Pl and Sm, classes 5 and 6 result from a single crossover between Sm and Py, and classes 7 and 8 result from a double crossover, with one of the exchanges between Pl and Sm and the other between Sm and Py.

7.18 In Drosophila, the genes st (scarlet eyes), ss (spineless bristles), and e (ebony body) are located on chromosome 3, with map positions as indicated: st-44 ss-58 e-70 Each of these mutations is recessive to its wild-type allele (st+, dark red eyes; ss+, smooth bristles; e+, gray body). Pheno-typically wild-type females with the genotype st ss e+/ st+ ss+ e were crossed with triply recessive males. Predict the phenotypes of the progeny and the frequencies with which they will occur assuming (a) no interference and (b) complete interference.

ANS: In the enumeration below, classes 1 and 2 are parental types, classes 3 and 4 result from a single crossover between st and ss, classes 5 and 6 result from a single crossover between ss and e, and classes 7 and 8 result from a double crossover, with one of the exchanges between st and ss and the other between ss and e.

7.20. In maize, the genes Tu, j2, and gl3 are located on chromosome 4 at map positions 101, 106, and 112, respectively. If plants homozygous for the recessive alleles of these genes are crossed with plants homozygous for the dominant alleles, and the F1 plants are testcrossed to triply recessive plants, what genotypes would you expect, and in what proportions? Assume that interference is complete over this map interval.

ANS: In the enumeration below, classes 1 and 2 are parental types, classes 3 and 4 result from crossing over between Tu and J2, and classes 5 and 6 result from crossing over between J2 and Gl3; only the chromosome from the triply heterozygous F1 plant is shown. Because interference is complete, there are no double crossover progeny.

6.16 In plants translocation heterozygotes display about 50 percent pollen abortion. Why?

ANS: In translocation heterozygotes, only alternate segregation leads to euploid gametes, and the frequency of alternate segregation is typically around 5 percent.

7.44 Two strains of maize, M1 and M2, are homozygous for four recessive mutations, a, b, c, and d, on one of the large chromosomes in the genome. Strain W1 is homozygous for the dominant alleles of these mutations. Hybrids produced by crossing M1 and W1 yield many different classes of recombinants, whereas hybrids produced by crossing M2 and W1 do not yield any recombinants at all. What is the difference between M1 and M2?

ANS: M2 carries an inversion that suppresses recombination in the chromosome.

5.7 If a father and son both have defective color vision, is it likely that the son inherited the trait from his father?

ANS: No. Defective color vision is caused by an X-linked mutation. The son's X chromosome came from his mother, not his father.

7.3 If a is linked to b, and b to c, and c to d, does it follow that a recombination experiment would detect linkage between a and d? Explain.

ANS: No. The genes a and d could be very far apart on the same chromosome—so far apart that they recombine freely, that is, 50 percent of the time.

7.10 A homozygous variety of maize with red leaves and normal seeds was crossed with another homozygous variety with green leaves and tassel seeds. The hybrids were then backcrossed to the green, tassel-seeded variety, and the following offspring were obtained: red, normal 124; red, tassel 126; green, normal 125; green, tassel 123. Are the genes for plant color and seed type linked? Explain.

ANS: No. The leaf color and tassel seed traits assort independently.

6.11 A woman with X-linked color blindness and Turner syndrome had a color-blind father and a normal mother. In which of her parents did nondisjunction of the sex chromosomes occur?

ANS: Nondisjunction must have occurred in the mother. The color blind woman with Turner syndrome was produced by the union of an X-bearing sperm, which carried the mutant allele for color blindness, and a nullo-x egg.

5.12 In Drosophila, vermilion eye color is due to a recessive allele (v) located on the X chromosome. Curved wings is due to a recessive allele (cu) located on one autosome, and ebony body is due to a recessive allele (e) located on another autosome. A vermilion male is mated to a curved, ebony female, and the males are phenotypically wild-type. If these males were backcrossed to curved, ebony females, what proportion of the offspring will be wild-type males?

ANS: P(male) = 1/2 ; P(male transmits first WT autosome) = 1/2; P(male transmits other WT autosome) = 1/2; therefore, combined proportion, P(WT male) = 1/8

7.34 In yeast, the ad mutation requires adenine for growth, and the in mutation requires inositol for growth. The wild-type alleles of these two mutations are AD and IN. C. C. Lindegren analyzed 48 tetrads from the cross AD IN × ad in; among these tetrads, 22 were parental ditype, 3 were nonparental ditype, and 23 were tetratype. Are the AD and IN genes linked? If so, what is the distance between them?

ANS: PD >> NPD, so the genes are linked; the distance between the genes is estimated as [(1/2) x (23 + 3)]/48 = 30 cM.

7.14 In tomatoes, tall vine (D) is dominant over dwarf (d), and spherical fruit shape (P) is dominant over pear shape (p). The genes for vine height and fruit shape are linked with 20 percent recombination between them. One tall plant (I) with spherical fruit was crossed with a dwarf, pear-fruited plant. The cross produced the following results: tall, spherical 81; dwarf, pear 79; tall, pear 22; dwarf spherical 17. Another tall plant with spherical fruit (II) was crossed with the dwarf, pear-fruited plant, and the following results were obtained: tall, pear 21; dwarf, spherical 18; tall, spherical 5; dwarf, pear 4. Diagram these two crosses, showing the genetic markers on the chromosomes. If the two tall plants with spherical fruit were crossed with each other, that is, I × II, what phenotypic classes would you expect from the cross, and in what proportions?

ANS: Plant I has the genotype D P/d p, and when crossed to a d p/d p plant, produces four classes of progeny:.... look on website tall, spherical 0.54 tall, pear 0.21 dwarf, spherical 0.21 dwarf, pear 0.04

7.25 In Drosophila, the X-linked recessive mutations prune (pn) and garnet (g) recombine with a frequency of 0.4. Both of these mutations cause the eyes to be brown instead of dark red. Females homozygous for the pn mutation were crossed to males hemizygous for the g mutation, and the F1 daughters, all with dark red eyes, were crossed with their brown-eyed brothers. Predict the frequency of sons from this last cross that will have dark red eyes.

ANS: The F1 females are genotypically pn +/+ g. Among their sons, 40 percent will be recombinant for the two X-linked genes, and half of the recombinants will have the wild-type alleles of these genes. Thus the frequency of sons with dark red eyes will be 1/2 x 40% = 20%.

6.6 A plant species X with n = 5 was crossed with a related species Y with n = 7. The F1 hybrid produced only a few pollen grains, which were used to fertilize the ovules of species Y. A few plants were produced from this cross, and all had 19 chromosomes. Following self-fertilization, the F1 hybrids produced a few F2 plants, each with 24 chromosomes. These plants were phenotypically different from either of the original species and were highly fertile. Explain the sequence of events that produced these fertile F2 hybrids.

ANS: The F1 hybrid had 5 chromosomes from species X and 7 from species Y, for a total of 12. When this hybrid was backcrossed to species Y, the few progeny that were produced had 5 + 7 = 12 chromosomes from the hybrid and 7 from species Y, for a total of 19. This hybrid was therefore a triploid. Upon self-fertilization, a few F2 plants were formed, each with 24 chromosomes. Presumably the chromosomes in these plants consisted of 2 5 = 10 from species X and 2 7 = 14 from species Y. These vigorous and fertile F2 plants were therefore allotetraploids.

6.14 In a Drosophila salivary chromosome, the bands have a sequence of 1 2 3 4 5 6 7 8. The homologue with which this chromosome is synapsed has a sequence of 1 2 3 6 5 4 7 8. What kind of chromosome change has occurred? Draw the synapsed chromosomes.

ANS: The animal is heterozygous for an inversion:

7.38 In Neurospora, the mutations arg, thi, and leu block the synthesis of arginine, thiamine, and leucine, respectively. The following ordered tetrad data come from two crosses with these mutations:

ANS: The arg and thi loci are unlinked; however, the arg and leu loci are linked. The distance between arg and leu is [(1/2) x 44]/200 = 11 cM; the distance between arg and its centromere is (1/2) x (1/300) = 0.17 cM; thus, the genetic map for the chromosome that carries arg and leu is centromere----0.17 cM----arg----11 cM----leu. The thi gene is very tightly linked to its centromere.

6.19 A phenotypically normal boy has 45 chromosomes, but his sister, who has Down syndrome, has 46. Suggest an explanation for this paradox.

ANS: The boy carries a translocation between chromosome 21 and another chromosome, say chromosome 14. He also carries a normal chromosome 21 and a normal chromosome 14. The boy's sister carries the translocation, one normal chromosome 14, and two normal copies of chromosome 21.

7.2 From a cross between individuals with the genotypes Cc Dd Ee × cc dd ee, 1000 offspring were produced. The class that was C- D- ee included 351 individuals. Are the genes c, d, and e on the same or different chromosomes? Explain.

ANS: The class represented by 351 offspring indicates that at least two of the three genes are linked.

5.4 In the mosquito Anopheles culicifacies, golden body (go) is a recessive X-linked mutation, and brown eyes (bw) is a recessive autosomal mutation. A homozygous XX female with golden body is mated to a homozygous XY male with brown eyes. Predict the phenotypes of their offspring. If the progeny are intercrossed, what kinds of progeny will appear in the and in what proportions?

ANS: The cross is go/go +/+ female +/Y bw/bw male --> F1: go/+ bw/+ females (wild-type eyes and body) and go/Y bw/+ males (golden body, wild-type eyes). An intercross of the F1 offspring yields the following F2 phenotypes in both sexes. Body Eyes Genotype Proportion golden brown go/go or Y bw/bw (1/2) (1/4) = 1/8 golden wild-type go/go or Y +/bw or + (1/2) (3/4) = 3/8 wild-type brown +/go or Y bw/bw (1/2) (1/4) = 1/8 wild-type wild-type +/go or Y +/bw or + (1/2) (3/4) = 3/8

6.12 In humans, Hunter syndrome is known to be an X-linked trait with complete penetrance. In family A, two phenotypically normal parents have produced a normal son, a daughter with Hunter and Turner syndromes, and a son with Hunter syndrome. In family B, two phenotypically normal parents have produced two phenotypically normal daughters and a son with Hunter and Klinefelter syndromes. In family C, two phenotypically normal parents have produced a phenotypically normal daughter, a daughter with Hunter syndrome, and a son with Hunter syndrome. For each family, explain the origin of the child indicated in italics.

ANS: The daughter with Turner and Hunter syndrome in family A must have received her single X chromosome from her mother, who is heterozygous for the mutation causing Hunter syndrome. The daughter did not receive a sex chromosome from her father because sex chromosome nondisjunction must have occurred during meiosis in his germline. The son with Klinefelter syndrome in family B is karyotypically XXY, and both of his X chromosomes carry the mutant allele for Hunter syndrome. This individual must have received two mutant X chromosomes from his heterozygous mother due to X chromosome nondisjunction during the second meiotic division in her germline. The daughter with Hunter syndrome in family C is karyotypically XX, and both of her X chromosomes carry the mutant allele for Hunter syndrome. This individual received the two mutant X chromosomes from her heterozygous mother through nondisjunction during the second meiotic division in the mother's germline. Furthermore, because the daughter did not receive a sex chromosome from her father, sex chromosome nondisjunction must have occurred during meiosis in his germline too.

7.36 The following tetrad data were obtained from the cross A B × a b in Neurospora:

ANS: The distance between a and b is [(1/2) x (220 + 14)]/2000 = 5.85 cM; the distance between a and the centromere is (1/2) x (220/2000) = 5.5 cM; the distance between b and the centromere is (1/2) x (14/2000) = 0.35 cM. Thus, the genetic map is a----5.5 cM----centromere----0.35 cM----b.

7.35 A geneticist obtained the following ordered tetrad data from a cross with Neurospora:

ANS: The distance is half the frequency of second division segregation asci: (1/2) x (84/200) = 21 cM.

7.21 A Drosophila geneticist made a cross between females homozygous for three X-linked recessive mutations (y, yellow body; ec, echinus eye shape; w, white eye color) and wild-type males. He then mated the F1 females to triply mutant males and obtained the following results:Determine the order of the three loci y, ec, and w, and estimate the distances between them on the linkage map of the X chromosome. Look in Book

ANS: The double crossover classes, which are the two that were not observed, establish that the gene order is y—w—ec. Thus, the F1 females had the genotype y w ec/+ + +. The distance between y and w is estimated by the frequency of recombination between these two genes: (8 + 7)/1000 = 0.015; similarly, the distance between w and ec is (18 + 23)/1000 = 0.041. Thus, the genetic map for this segment of the X chromosome is y—1.5 cM—w—4.1 cM—ec.

6.18 In Drosophila, the genes bw and st are located on chromosomes 2 and 3, respectively. Flies homozygous for bw mutations have brown eyes, flies homozygous for st mutations have scarlet eyes, and flies homozygous for bw and st mutations have white eyes. Doubly heterozygous males were mated individually to homozygous bw; st females. All but one of the matings produced four classes of progeny: wild-type, and brown-, scarlet- and white-eyed. The single exception produced only wild-type and white-eyed progeny. Explain the nature of this exception.

ANS: The exceptional male, whose genotype is bw/+ st/+, is heterozygous for a translocation between chromosomes 2 and 3. It is not possible to determine whether the translocation is between the two mutant chromosomes or between the two wild-type chromosomes, that is, whether it is T(bw; st) or T(+; +); however it clearly is not between a mutant chromosome and a wild-type chromosome, that is, T(bw; +) or T(+; st). If it were, the progeny would be either brown or scarlet, not either wild-type or white.

The recessive alleles w and dor cause mutant eye colors (white and deep orange, respectively). However, w is epistatic over dor; that is, the genotypes w and w dor have white eyes. If there is 40 percent recombination between w and dor, what proportion of the sons from this heterozygous female will show a mutant phenotype? What proportion will have either red or deep orange eyes?

ANS: The female will produce four kinds of gametes: 30% w +, 30% + dor, 20% w dor, and 20% + +; thus, 80% of the progeny will be mutant (either white or deep orange), and 50% will be pigmented (either red or deep orange).

6.5 A plant species A, which has seven chromosomes in its gametes, was crossed with a related species B, which has nine. The hybrids were sterile, and microscopic observation of their pollen mother cells showed no chromosome pairing. A section from one of the hybrids that grew vigorously was propagated vegetatively, producing a plant with 32 chromosomes in its somatic cells. This plant was fertile. Explain.

ANS: The fertile plant is an allotetraploid with 7 pairs of chromosomes from species A and 9 pairs of chromosomes from species B; the total number of chromosomes is (2 7) + (2 9) = 32.

6.9 A Drosophila female homozygous for a recessive X-linked mutation causing yellow body was crossed to a wild-type male. Among the progeny, one fly had sectors of yellow pigment in an otherwise gray body. These yellow sectors were distinctly male, whereas the gray areas were female. Explain the peculiar phenotype of this fly.

ANS: The fly is a gynandromorph, that is, a sexual mosaic. The yellow tissue is X(y)/O and the gray tissue is X(y)/X(+). This mosaicism must have arisen through loss of the X chromosome that carried the wild-type allele, presumably during one of the early embryonic cleavage divisions.

7.22 A Drosophila geneticist crossed females homozygous for three X-linked mutations (y, yellow body; B, bar eye shape; v, vermilion eye color) to wild-type males. The F1 females, which had gray bodies and bar eyes with dark red pigment, were then ..Determine the order of these three loci on the X chromosome and estimate the distances between them.... Book

ANS: The last two classes, consisting of yellow, bar flies and vermilion flies, with a total of 50 progeny, result from double crossovers. Thus, the order of the genes is y—v—B, and the F1 females had the genotype y v B/+ + +. The distance between y and v is the average number of crossovers between them: (244 + 50)/1000 = 29.4 cM; likewise, the distance between v and B is (160 +50)/1000 = 21.0 cM. Thus, the genetic map is y--29.4 cM--v--21.0 cM--B.

7.33 A Drosophila second chromosome that carried a recessive lethal mutation, l(2)g14, was maintained in a stock with a balancer chromosome marked with a dominant mutation for curly wings. This latter mutation, denoted Cy, is also associated with a recessive lethal effect—but this effect is different from that of l(2)g14. Thus, flies survive, and they have curly wings. Flies without the Cy mutation have straight wings. A researcher crossed females to males that carried second chromosomes with different deletions (all homozygous lethal) balanced over the Cy chromosome (genotype ). Each cross was scored for the presence or absence of progeny with straight wings.

ANS: The lethal mutation resides in band 7.

5.1 What are the genetic differences between male- and female-determining sperm in animals with heterogametic males?

ANS: The male-determining sperm carries a Y chromosome; the female-determining sperm carries an X chromosome.

6.26 A male mouse that is heterozygous for a reciprocal translocation between the X chromosome and an autosome is crossed to a female mouse with a normal karyotype. The autosome involved in the translocation carries a gene responsible for coloration of the fur. The allele on the male's translocated autosome is wild-type, and the allele on its nontranslocated autosome is mutant; however, because the wild-type allele is dominant to the mutant allele, the male's fur is wild-type (dark in color). The female mouse has light color in her fur because she is homozygous for the mutant allele of the color-determining gene. When the offspring of the cross are examined, all the males have light fur and all the females have patches of light and dark fur. Explain these peculiar results.

ANS: The phenotype in the female offspring is mosaic because one of the X chromosomes is inactivated in each of their cells. If the translocated X is inactivated, the autosome attached to it could also be partially inactivated by a spreading of the inactivation process across the translocation breakpoint. This spreading could therefore inactivate the color-determining gene on the translocated autosome and cause patches of tissue to be phenotypically mutant.

5.9 A man with X-linked color blindness marries a woman with no history of color blindness in her family. The daughter of this couple marries a normal man, and their daughter also marries a normal man. What is the chance that this last couple will have a child with color blindness? If this couple has already had a child with color blindness, what is the chance that their next child will be color blind?

ANS: The risk for the child is P(mother is C/c) P(mother transmits c) P(child is male) = (1/2) (1/2) (1/2) = 1/8; if the couple has already had a child with color blindness, P(mother is C/c) = 1, and the risk for each subsequent child is 1/4.

5.8 A normal woman, whose father had hemophilia, marries a normal man. What is the chance that their first child will have hemophilia?

ANS: The risk for the child is P(woman transmits mutant allele) P(child is male) = (1/2) (1/2) = 1/4.

6.27 In Drosophila, the autosomal genes cinnabar (cn) and brown (bw) control the production of brown and red eye pigments, respectively. Flies homozygous for cinnabar mutations have bright red eyes, flies homozygous for brown mutations have brown eyes, and flies homozygous for mutations in both of these genes have white eyes. A male homozygous for mutations in the cn and bw genes has bright red eyes because a small duplication that carries the wild-type allele of bw (bw+) is attached to the Y chromosome. If this male is mated to a karyotypically normal female that is homozygous for the cn and bw mutations, what types of progeny will be produced?

ANS: The sons will have bright red eyes because they will inherit the Y chromosome with the bw+ allele from their father. The daughters will have white eyes because they will inherit an X chromosome from their father.

6.23 Analysis of the polytene chromosomes of three populations of Drosophila has revealed three different banding sequences in a region of the second chromosome: Explain the evolutionary relationships among these populations. (look in book)

ANS: The three populations are related by a series of inversions: ---insert art from problem 6.23 in 4e---

7.43 A normal woman with a color-blind father married a normal man, and their first child, a boy, had hemophilia. Both color blindness and hemophilia are due to X-linked recessive mutations, and the relevant genes are separated by 10 cM. This couple plans to have a second child. What is the probability that it will have hemophilia? color blindness? both hemophilia and color blindness? neither hemophilia nor color blindness?

ANS: The woman is a repulsion heterozygote for the alleles for color blindness and hemophilia—that is, she is C h/c H. If the woman has a boy, the chance that he will have hemophilia is 0.5 and the chance that he will have color blindness is 0.5. If we specify that the boy have only one of these two conditions, then the chance that he will have color blindness is 0.45. The reason is that the boy will inherit a nonrecombinant X chromosome with a probability of 0.9, and half the nonrecombinant X chromosomes will carrya the mutant allele for color blindness and the other half will carry the mutant allele for hemophilia. The chance that the boy will have both conditions is 0.05 and the chance that he will have neither condition is 0.05. The reason is that the boy will inherit a recombinant X chromosome with a probability of 0.1, and half the recombinant X chromosomes will carry both mutant alleles and the other half will carry neither mutant allele.

5.16 A woman carries the testicular feminization mutation (tfm) on one of her X chromosomes; the other X carries the wild-type allele (Tfm). If the woman marries a normal man, what fraction of her children will be phenotypically female? Of these, what fraction will be fertile?

ANS: Three-fourths will be phenotypically female (genotypically tfm/Tfm, Tfm/Tfm, or tfm/Y). Among the females, 2/3 (tfm/Tfmm and Tfm/Tfm) will be fertile; the tfm/Y females will be sterile.

ANS: Eye color in canaries is due to a gene on the Z chromosome, which is present in two copies in males and one copy in females. The allele for pink color at hatching (p) is recessive to the allele for black color at hatching (P). There is no eye color gene on the other sex chromosome (W), which is present in one copy in females and absent in males. The parental birds were genotypically p/W (cinnamon females) and P/P (green males). Their F1 sons were genotypically p/P (with black eyes at hatching). When these sons were crossed to green females (genotype P/W), they produced F2 progeny that sorted into three categories: males with black eyes at hatching (P/-, half the total progeny), females with black eyes at hatching (P/W, a fourth of the total progeny), and females with pink eyes at hatching (p/W, a fourth of the total progeny). When these sons were crossed to cinnamon females (genotype p/W), they produced F2 progeny that sorted into four equally frequent categories: males with black eyes at hatching (genotype P/p), males with pink eyes at hatching (genotype p/p), females with black eyes at hatching (genotype P/W), and females with pink eyes at hatching (genotype p/W).

ANS: Use one of the banding techniques.

6.7 Identify the sexual phenotypes of the following genotypes in human beings: XX, XY, XO, XXX, XXY, XYY

ANS: XX is female, XY is male, XO is female (but sterile), XXX is female, XXY is male (but sterile), XYY is male.

5.5 What are the sexual phenotypes of the following genotypes in Drosophila: XX, XY, XXY, XXX, XO?

ANS: XX is female, XY is male, XXY is female, XXX is female (but barely viable), XO is male (but sterile).

6.29 Cytological examination of the sex chromosomes in a man has revealed that he carries an insertional translocation. A small segment has been deleted from the Y chromosome and inserted into the short arm of the X chromosome; this segment contains the gene responsible for male differentiation (SRY). If this man marries a karyotypically normal woman, what types of progeny will the couple produce?

ANS: XX zygotes will develop into males because one of their X chromosomes carries the SRY gene that was translocated from the Y chromosome. XY zygotes will develop into females because their Y chromosome has lost the SRY gene.

6.13 Although XYY men are phenotypically normal, would they be expected to produce more children with sex chromosome abnormalities than XY men? Explain.

ANS: XYY men would produce more children with sex chromosome abnormalities because their three sex chromosomes will disjoin irregularly during meiosis. This irregular disjunction will produce a variety of aneuploid gametes, including the XY, YY, XYY, and nullo sex chromosome constitutions.

7.12 Another phenotypically wild-type female fruit fly heterozygous for the two genes mentioned in the previous problem was crossed to a homozygous black, vestigial male. The cross produced the following progeny: gray body, normal wings 23; gray body, vestigial wings 127; black body, normal wings 124; black body, vestigial wings 26. Do these data indicate linkage? What is the frequency of recombination? Diagram the cross, showing the arrangement of the genetic markers on the chromosomes.

ANS: Yes. Recombination frequency = (23 + 26)/(23 + 127 +124 +26) = 0.163. Cross:

7.11 A phenotypically wild-type female fruit fly that was hetero-zygous for genes controlling body color and wing length was crossed to a homozygous mutant male with black body (allele b) and vestigial wings (allele vg). The cross produced the following progeny: gray body, normal wings 126; gray body, vestigial wings 24; black body, normal wings 26; black body, vestigial wings 124. Do these data indicate linkage between the genes for body color and wing length? What is the frequency of recombination? Diagram the cross, showing the arrangement of the genetic markers on the chromosomes.

ANS: Yes. Recombination frequency = (24 + 26)/(126 + 24 + 26 + 124) = 0.167. Cross:

7.13 In rabbits, the dominant allele C is required for colored fur; the recessive allele c makes the fur colorless (albino). In the presence of at least one C allele, another gene determines whether the fur is black (B, dominant) or brown (b, recessive). A homozygous strain of brown rabbits was crossed with a homo-zygous strain of albinos. The F1 were then crossed to homo-zygous double recessive rabbits, yielding the following results: black 34; brown 66; albino 100. Are the genes b and c linked? What is the frequency of recombination? Diagram the crosses, showing the arrangement of the genetic markers on the chromosomes.

ANS: Yes. Recombination frequency is estimated by the frequency of black offspring among the colored offspring: 34/(66 + 34) = 0.34. Cross:

5.20 In chickens, the absence of barred feathers is due to a recessive allele. A barred rooster was mated with a nonbarred hen, and all the offspring were barred. These chickens were intercrossed to produce progeny, among which all the males were barred; half the females were barred and half were nonbarred. Are these results consistent with the hypothesis that the gene for barred feathers is located on one of the sex chromosomes?

ANS: Yes. The gene for feather patterning is on the Z chromosome. If we denote the allele for barred feathers as B and the allele for nonbarred feathers as b, the crosses are: B/B (barred) male b/W (nonbarred) female --> F1: B/b (barred) males and B/W (barred) females. Intercrossing the F1 produces B/B (barred) males, B/b (barred) males, B/W (barred) females, and b/W (nonbarred) females, all in equal proportions.

6.22 A man has attached chromosomes 21. If his wife is cytologically normal, what is the chance their first child will have Down syndrome?

ANS: Zygotes produced by this couple will be either trisomic or monosomic for chromosome 21. Thus, 100% of their viable children will develop Down syndrome.

5.15 Suppose that a mutation occurred in the SRY gene on the human Y chromosome, knocking out its ability to produce the testis-determining factor. Predict the phenotype of an individual who carried this mutation and a normal X chromosome.

Female

5.17 Would a human with two X chromosomes and a Y chromosome be male or female?

Male

5.6 In human beings, a recessive X-linked mutation, g, causes green-defective color vision; the wild-type allele, G, causes normal color vision. A man (a) and a woman (b), both with normal vision, have three children, all married to people with normal vision: a color-defective son (c), who has a daughter with normal vision (f); a daughter with normal vision (d), who has one color-defective son (g) and two normal sons (h); and a daughter with normal vision (e), who has six normal sons (i). Give the most likely genotypes for the individuals (a to i) in this family.

Pedigree Online (a) XGY; (b) XGXg; (c) XgY; (d) XGXg; (e) XGXG; (f) XGXg; (g) XgY; (h) XGY; (i) XGY

5.23 Males in a certain species of deer have two nonhomologous X chromosomes, denoted X1 and X2, and a Y chromosome. Each X chromosome is about half as large as the Y chromosome, and its centromere is located near one of the ends; the centromere of the Y chromosome is located in the middle. Females in this species have two copies of each of the X chromosomes and lack a Y chromosome. How would you predict the X and Y chromosomes to pair and disjoin during spermato-genesis to produce equal numbers of male- and female-determining sperm?

Picture Online: Since the centromere is at the end of each small X chromosome but in the middle of the larger Y, X1 and X2 both pair at the centromere of the Y chromosome during metaphase so that the two X chromosomes disjoin together and segregate from the Y chromosome during anaphase.

6.15 Other chromosomes have sequences as follows: (a) 1 2 5 6 7 8; (b) 1 2 3 4 4 5 6 7 8; (c) 1 2 3 4 5 8 7 6. What kind of chromosome change is present in each? Illustrate how these chromosomes would pair with a chromosome whose sequence is 1 2 3 4 5 6 7 8.

a. Deletion b. duplication c. a terminal inversion (Pictures Online for example)