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Header 1 Sub-Header Paragraph (for help, see below) Chapter 4: Chromosomes and Sex-Chromosome Inheritance Chapter Summary The chromosomes in somatic cells of higher plants and animals are present in pairs of homologous chromosomes. Homologous chromosomes are usually identical in appearance, whereas nonhomologous chromosomes often show differences in size and structural detail that make them visibly distinct from each other. A cell whose nucleus contains two sets of homologous chromosomes is diploid. One set of chromosomes comes from the maternal parent and the other from the paternal parent. Gametes are haploid. A gamete contains only one set of chromosomes, consisting of one member of each pair of homologs. Mitosis is the process of nuclear division that maintains the chromosome number when a somatic cell divides. Before mitosis, each chromosome replicates, forming a two-part structure consisting of two sister chromatids joined at the centromere (kinetochore). The chromosomes become visible at the onset of mitosis and, at metaphase, become aligned on the metaphase plate perpendicular to the spindle. At anaphase, the centromere of each chromosome divides, and the sister chromatids are pulled by spindle fibers to opposite poles of the cell. The separated sets of chromosomes present in telophase nuclei are genetically identical. Meiosis is the type of nuclear division that takes place in germ cells, and it reduces the diploid number of chromosomes to the haploid number. The genetic material is replicated before the onset of meiosis, so each chromosome consists of two sister chromatids. The first meiotic division is the reductional division, which reduces the chromosome number by half. The homologous chromosomes first pair (synapsis) and then, at anaphase I, separate. The resulting products contain chromosomes that consist of two chromatids attached to a common centromere. However, as a result of crossing-over, which takes place in prophase I, the chromatids may not be genetically identical along their entire length. In the second meiotic (equational) division, the centromeres divide and the homologous chromatids separate. The end result of meiosis is the formation of four genetically different haploid nuclei. 64 65 A distinctive feature of meiosis is the synapsis, or side-by-side pairing, of homologous chromosomes in the zygotene substage of prophase I. During the pachytene substage, the paired chromosomes become connected by chiasmata (the physical manifestations of crossing-over) and do not separate until anaphase I. This separation is called disjunction (unjoining), and failure of chromosomes to separate is called nondisjunction. Nondisjunction results in a gamete that contains either two copies or no copies of a particular chromosome. Meiosis is the physical basis of the segregation and independent assortment of genes. In Drosophila, an unexpected pattern of inheritance of the X-linked white gene was shown to be accompanied by nondisjunction of the X chromosome; these observations gave experimental proof of the chromosome theory of heredity. Unlike other chromosome pairs, the X and Y sex chromosomes are visibly different and contain different genes. In mammals and in many insects and other animals, as well as in some flowering plants, the female contains two X chromosomes (XX) and hence is homogametic, and the male contains one X chromosome and one Y chromosome (XY) and hence is heterogametic. In birds, moths, butterflies, and some reptiles, the situation is the reverse: Females are the heterogametic sex (WZ) and males the homogametic sex (ZZ). The Y chromosome in many species contains only a few genes. In human beings and other mammals, the Y chromosome includes a male-determining factor. In Drosophila, sex is determined by a male-specific or female-specific pattern of gene expression that is regulated by the ratio of the number of X chromosomes to the number of sets of autosomes. In most organisms, the X chromosome contains many genes unrelated to sexual differentiation. These X-linked genes show a characteristic pattern of inheritance that is due to their location in the X chromosome. The progeny of genetic crosses often conform to the theoretical predictions of the binomial probability formula. The degree to which the observed numbers of different genetic classes of progeny fit theoretically expected numbers is usually found with a chi-square (2) test. On the basis of the criterion of the chi-square test, Mendels data fit the expectations somewhat more closely than chance would dictate. However, the bias in the data is relatively small and is likely to be due to recounting or to repeating certain experiments whose results were regarded as unsatisfactory. 66 anaphase anaphase I anaphase II autosome bivalent cell cycle checkpoints centromere chiasma chi-square chromatid chromosomes chromosome complement crossing-over cytokinesis degrees of freedom diakinesis diploid diplotene equational division G1 period G2 period germ cell goodness of fit haploid heterogametic highly significant homogametic interphase kinetochore leptotene M period meiosis metaphase metaphase plate metaphase I metaphase II mitosis mitotic spindle nondisjunction nucleoli pachytene Pascals triangle prophase prophase I prophase II reductional division S phase sister chromatids sex chromosome somatic cell statistically significant synapsis telophase telophase I telophase II tetrad X chromosome X-linked gene Y chromosome zygotene Keywords 67 Fill-ins 1. ________ The period during the cell cycle when DNA synthesis takes place. 2. ________ This type of anaphase is associated with the reductional division. 3. ________ Failure of homologous chromosomes or chromatids to separate during cell division, resulting in extra or missing copies in the daughter cells. 4. ________ First apparent in late pachytene or diplotene of meiosis, this cross-shaped connection is the physical manifestation of crossing-over between nonsister chromatids. 5. ________ The sex in which the two sex chromosomes are different; the male sex in mammals and most insects, the female sex in birds, moths, and butterflies. 6. ________ Position along the chromosome where the spindle fibers attach to be able to separate homologous chromosomes or chromatids during anaphase. 7. ________ Chromosome structure observed at diplotene that consists of a pair of sister chromatids of each of two homologous chromosomes, aligned along their length and connected by chiasmata. 8. ________ Term describing the result of a goodness-of-fit test in which the P value is smaller than 0.05 but greater than 0.01. 9. ________ For the type of chi-square test described in this chapter, this quantity equals the number of classes of data minus one. 10. ________ A numerical pyramid built from the values of the binomial coefficients n!/s!t!; each row corresponds to one value of n = 0, 1, 2, . . ., with the entries in each row corresponding to s = 0, 1, 2, . . ., n and t = n - s. Fill-ins Answers 1. S phase 2. anaphase I 3. nondisjunction 4. chiasma 5. heterogametic 6. centromere or kinetochore 7. bivalent 8. statistically significant 9. degrees of freedom 10. Pascals triangle Analysis and Applications ANS 4.1. Recall what distinguishes one stage of prophase I from the next. Leptotene literally means thin thread, when each chromosome is in an extended, thread-like condition prior to synapsis; this stage corresponds to diagram E. Zygotene means paired threads, and the pairing begins at the chromosome tips; this is configuration B. Pachytene means thick thread; it commences when pairing is completed and the homologous chromosomes still appear to be single, which corresponds to diagram D. Diplotene means double thread; at this time each homologous chromosome clearly consists of two sister chromatids, and chiasmata are apparent, which is shown in diagram A. Diakinesis means moving apart, and in this stage the synapsed homologous chromosomes begin to repel one another, being held together by the chiasmata, producing configuration C. Therefore, the identity of the stages is (A) diplotene, (B) zygotene, (C) diakinesis, (D) pachytene, and (E) leptotene. ANS 4.2. Illustration (A) is anaphase II of meiosis, because the chromosome number has been reduced by half. Illustration (B) is anaphase of mitosis, because the homologous chromosomes are not paired. Illustration (C) is anaphase I of meiosis, because the homologous chromosomes are paired. ANS 4.3. [4!/(3!1!)](3/4)3(1/4)1 = 27/64, or about 42 percent. ANS 4.4. Since the two genes are in different chromosomes (the mutation for phenylkentonuria is autosomal, that for hemophilia X linked), they will assort independently, and so we can consider them separately. 68 Because both parents are heterozygous for an autosomal recessive phenylketonuria mutation, there is a 1/4 chance that their child will be homozyous recessive. For X-linked hemophilia, only a boy could be affected; a girl would have inherited a normal X chromosome from her father, so she could not have hemophilia. So the probability of a girl being affected with both diseases is 0 and the probability of a boy being affected with both diseases is 1/4 - 1/2 = 1/8. Overall, the probability is (1/2)(0) + (1/2)(1/8) = 1/16. ANS 4.5. The mother of the woman III-3 is definitely a carrier because III-4 is affected, thus the probability that III-3 is a carrier is 1/2. ANS 4.6. Because the child has two Y chromosomes, the nondisjunction took place in the father, during the second meiotic division. ANS 4.7. a. 24 chromatids and 12 chromosomes. b. 24 chromatids and 12 chromosomes. c. 12 chromatids and 6 chromosomes. ANS 4.8. a. For each gene, half of the offspring will be heterozygous. Since the three genes undergo independent assortment, the proportions for each can be multiplied to get the expected frequency of offspring heterozygous for all three genes: (1/2)3 = 1/8. b. For each gene, half of the offspring will be homozygous, so you can multiply the proportions for each gene to get the expected frequency of offspring homozygous for all three genes: (1/2)3 = 1/8. (Note that the problem does not say that all three genes must be recessive or dominant. So AABBCC, aabbcc, AAbbcc, AAbbCC, etc. are all equivalent.) ANS 4.9. a. For each centromere, there are two possibilities (for example, A or a), and the centromeres of nonhomologous chromosomes segregate independently, so altogether there are 28 = 256 possible gametes. b. The probability of a gamete receiving any particular centromere designated by a capital letter is 1/2 for each chromosome, and since the chromosomes segregate independently, the probability of a gamete containing all eight centromeres designated by capital letters is (1/2)8 = 1/256. ANS 4.10. a. (1/2)6 = 1/64. b. 6(1/6)6 = 6/64. c. 15(1/6)6 = 15/64. d. 20(1/6)6 = 20/64. e. 1 " (1/64) " (6/64) " (15/64) = 42/64. 69 ANS 4.11. [(8!/0!8!) + (8!/2!6!) + (8!/4!4!) + (8!/6!2!) + (8!/8!0!)](1/2)8 = 128/28 = 0.500. ANS 4.12. The observed F2 ratio suggests that two different genes are involved in the determination of bulb color. Moreover, since the ratio departs from the classic Mendelian ratio of 9 : 3 : 3 : 1, it means that the genes interact with each other. (They show epistasis.) Let us use the symbol R to denote a dominant allele necessary to develop the red color, and r be a recessive allele of the same gene, so that onion bulbs with the genotype rr are white. Let us also use the symbol W to denote a dominant allele of another gene that suppresses the expression of R. On this hypothesis, bulbs with the genotypes R" W" are also white. Based on this hypothesis, the initial cross was between an RR ww red plant and an rr WW white plant. The F1 progeny of this cross are genotypically Rr Ww and white. The F2 progeny still have an underlying genotypic ratio of 9 R" W" : 3 R" ww : 3 rr W" : 1 rr ww. If onion bulbs with the genotype rr ww are yellow, those with the genotype R" ww are red, and those with the genotypes R" W" and rr W" are white, then we can explain the observations because 9 (R" W") + 3 (rr W") = 12. ANS 4.13. On her hypothesis, the expected numbers in each phenotypic class are: (9/16) - 160 = 90; (3/16) - 160 = 30; (3/16) - 160 = 30; (1/16) - 160 = 10. The number of degrees of freedom is one less than the number of classes of data. There are four classes of data, so the number of degrees of freedom equals 3. The P value for the calculated chi-square with 3 degrees of freedom turns out to be 0.22. Therefore, the geneticist need not reject the hypothesis. ANS 4.14. The chi-square value is 4.00. ANS 4.15. 1 " (1/2)7 " (1/2)7 = 63/64 = 0.984. ANS 4.16. There are 21 possible genotypes for an autosomal gene: 6 homozygous and 15 heterozygous. For an X-linked gene the number of possible genotypes is 27: 6 homozygous and 15 heterozygous genotypes in females, and 6 hemizygous genotypes in males. 2 2 91 902 2 90 21 30 30 37 = ' = ' + ' + ( exp) exp obs ( ) ( ) ( ' + ' = 30 30 11 10 10 4 44 )2 ( )2 . 70 ANS 4.17. The observed F2 ratio suggests that two independently assorting genes are involved in the determination of the growth habit in groundnuts. The departure of the F2 phenotypic ratio from the classic Mendelian ratio of 9 : 3 : 3 : 1 can be explained by epistatic gene interaction. The observed ratio of 9 : 7 and the observation that the cross of two bunch plants gives all runner progeny, indicates that at least one copy of the dominant allele of each gene is necessary to develop the runner habit. Let R be the dominant allele of one gene involved in the development of the growth habit, and r be the recessive allele of the same gene. Let B be the dominant allele of the other gene and b be the recessive allele of the same gene. Then the initial cross is between an RR bb bunch plant and an rr BB bunch plant. The F1 progeny of this cross is Rr Bb and has a runner habit. The F2 progeny have an underlying genotypic ratio of 9 R" B" : 3 R" bb : 3 rr B" : 1 rr bb. Among these, the 9/16 with the genotypes R" B" have the runner growth habit, whereas the 7/16 with the genotypes rr bb, R" bb, and rr B" have the bunch growth habit. Hence, the 9 : 7 ratio can be explained. ANS 4.18. a. The probability that their first child will have brown eyes is 3/4 because both parents must be heterozygous. b. 2/3. c. The probability that this couple will have three children with blue eyes is (1/4)3 = 1/64. The probability that none of the three children will have blue eyes is (3/4)3 = 27/64. ANS 4.19. First, notice that there is no chance of the X-linked color blindness allele being passed to II-4 since he will inherit a Y chromosome from his father (I-2) and his X chromosome from his normal mother (I-1); thus, we do not have to consider the right-hand side of the pedigree at all. The mating between relatives (double line) is therefore irrelevant. On the left-hand side of the pedigree, II-2 must be a carrier for color blindness, since she will inherit her fathers (I-2) only X chromosome, which we know carries the allele for color blindness. Then there is a 1/2 chance that II-2 passes the allele to her daughter (III-2) and a 1/2 chance that III-2 would pass the allele to the male IV-1. So the total probability that IV-1 will be color blind is 1 - (1/2) - (1/2) = 1/4. ANS 4.20. For convenience we will use the symbols W and w to denote the wildtype and mutant alleles, respectively, of the X-linked gene. Let event A be that III-1 is heterozygous Ww. Let event B be that none of IV-1, 2, 3 are affected. The complement of A, the event A, is the event that III-1 is homozygous WW. Then, Pr{A} = 1/4 and Pr{A} = 3/4, since these do not depend on B. Furthermore, Pr{B| A} = 0 and Pr{B| A} = (1/2)2. The reason the 1/2 is squared and not cubed is that the female IV-2 is not informative 71 with regard to III-1; in particular, because the male III-2 has genotype W/Y, the female IV-2 will be unaffected no matter what the genotype of III-1. With these expressions in place we can use Bayes theorem as follows: ANS 4.21. The major difference between this Problem and Problem 4.20 is that in this case the individual IV-2 is informative with regard to the genotype of III-1, because the gel implies that IV-2 received the allele corresponding to the band S from III-1. Therefore Pr{A} = 1/4, Pr{A} = 3/4, and Pr{B | A} = 1 as in Problem 4.20, but in this case Pr{B | A} = (1/2)3. Bayes theorem therefore implies that ANS 4.22. For convenience we will use the symbols W and w to denote the wildtype and mutant alleles, respectively, of the X-linked gene. Let event A be that III-1 is heterozygous Ww and let event B be that the individuals IV-1, 2, 3 are all not affected. Since II-2 must be heterozygous Ww, then the probability that III-1 is heterozygous is Pr{A} = 1/2. The complement of A is the event A that III-1 is homozygous WW, and so Pr{A} = 1/2. Since the male III-2 is affected, and none of IV-1, IV-2, and IV-3 is affected, all of these offspring must have inherited the W allele from III-1. Hence Pr{B | A} = (1/2)3 and Pr{B | A} = 1. Applying Bayes theorem to this situation we have ANS 4.23. a. The cross is y/Y - y+/y+, in which Y represents Y chromosome; half of the progeny are y+/Y wildtype males and half of the progeny are y/y+ females (phenotypically wildtype, but heterozygous). Pr{ | } Pr{ | }Pr{ } Pr{ | }Pr{ } Pr{ | }Pr A B B A A B A A B A = + { } ( / ) ( / ) ( / ) ( / ) ( )( / ) = + = = A 1 2 1 2 1 2 1 2 1 1 2 1 9 0 3 3 .11 Pr{ | } Pr{ | }Pr{ } Pr{ | }Pr{ } Pr{ | }Pr A B B A A B A A B A = + { } ( / ) ( / ) ( / ) ( / ) ( )( / ) = + = = A 1 2 1 4 1 2 1 4 1 3 4 1 25 0 3 3 .04 Pr{ | } Pr{ | }Pr{ } Pr{ | }Pr{ } Pr{ | }Pr A B B A A B A A B A = + { } ( / ) ( / ) ( / ) ( / ) ( )( / ) = + = = A 1 2 1 4 1 2 1 4 1 3 4 1 13 2 2 0.077 72 73 b. The mating is y/y - y+/Y; in this case 1/2 of the progeny are y/Y yellow males and 1/2 of the progeny are y/y+ females (phenotypically wildtype, but heterozygous). c. The mating is y/y+ females - y+/Y males; here 1/2 of the progeny males are wildtype y+/Y and 1/2 of the progeny males are yellow y/Y. All of the female progeny are wildtype, and among these 1/2 are homozygous y+/y+ and 1/2 are heterozygous y/y+. d. The mating is y/y+ females - y/Y males; the result is that 1/2 of the female progeny are yellow (homozygous y) and 1/2 of the female progeny are heterozygous y/y+; 1/2 of the male progeny are yellow (y/Y) and 1/2 of the male progeny are wildtype (y+/Y). ANS 4.24. X-linked inheritance is suggested by the fact that each male has only one band, which is transmitted to his daughters but not his sons. The deduced genotypes are: I-1 (A1A1), I-2 (A2), II-1 (A1), II-2 (A1A2), II-3 (A1), II-4 (A1), II-5 (A1A2), II-6 (A2), III-1 (A2), III-2 (A1A2), III-3 (A1A2), and III-4 (A1). ANS 4.25. The 2 value equals (234 " 200)2/200 + (203 " 200)2/200 + (175 " 200)2/200 + (188 " 200)2/200 = 9.67. It has 3 degrees of freedom (four classes of data), and the P value is about 0.022. We therefore conclude that these data are not in satisfactory agreement with the hypothesis of 1 : 1 : 1 : 1 segregation. ANS 4.26. The accompanying Punnet square shows the outcome of the cross. The X- and Y-chromosomes from the male are denoted in boldface, the attached X-chromosomes are yoked by a ^ sign. The surviving progeny consists of yellow males and wildtype females in equal proportions. Attached-X inheritance differs from the typical situation in that the sons, rather than the daughters, receive their fathers X-chromosome. ANS 4.27. a. Since W is a rare allele, a woman is likely heterozygous for this gene (Ww) and thus produces two types of gametes with the equal probability: W I0 and w I 0. A man also produces two Sperm Eggs X^X X^X X X^X Y Y X Y YY X Y gamete classes: w IA and w IB in equal proportions. So the probability of a wooly-haired group B child is (1/2) - (1/2) = 1/4. b. The probability of a straight-haired group B child is also 1/4. c. The probability that after three straight-haired group A children, of a next child being wooly-haired group B is 1/4. This probability is not affected by previous births. ANS 4.28. The results of the crosses reveal how the three alleles control plumage pattern. If we let Pr be the allele controlling the restricted pattern, then this allele must be dominant to the Pm and Pd alleles responsible for the mallard and dusky patterns, respectively. Also the Pm allele is dominant to Pd. a. an F1 male from cross 1 has the genotype PrPm; an F1 female from cross 2 has the genotype PmPd. Half of their progeny would have restricted and the other half would have the mallard plumage patterns. b. In this case 3/4 of the progeny would have restricted and 1/4 would have mallard plumage pattern. ANS 4.29. a. The phenotype of the F1 progeny would be roan-colored and polled (hornless). b. Since the genes assort independently, we can consider them separately. There would be three phenotypic classes for the R, r allele pair: 1/4 red (RR) : 2/4 roan (Rr) : 1/4 white (rr). There would be two phenotypic classes for H, h allele pair: 3/4 polled (H") : 1/4 horned (hh). Therefore, there would be six phenotypic classes in following proportions: 3/16 red polled : 6/16 roan polled 3/16 white polled : 1/16 red horned : 2/16 roan horned : 1/16 white horned. c. A cross of the Rr Hh F1 to the white horned rr hh stock would yield 1/4 roan polled : 1/4 roan horned : 1/4 white polled : 1/4 white horned. ANS 4.30. a. All affected individuals, except one (individual 9), are homozygous for both FMF and VNTR allele number 1. This implies that the FMF mutation and VNTR1 are close together in the same chromosome. b. Affected individual 9 is heterozygous for the VNTR locus, which can be explained by recombination between the VNTR locus and the FMF locus in one of the parents. Challenge Problems ANS CP 1. X-linked inheritance for the morphological trait, because the affected male I-2 has normal sons and carrier daughters. Y-linkage for the molecular trait, because females have no DNA for the RFLP and because males receive the allele present in the father. 74 ANS CP 2. The surviving offspring are expected to be 2/3 tailless and 1/3 tailed. In a litter of five pups, the expected distribution is (1/3)5 = 1/243 with all 5 tailed, 5(1/3)4(2/3)1 = 10/243 with 4 tailed, 10(1/3)3(2/3)2 = 40/243 with 3 tailed, 10(1/3)2(2/3)3 = 80/243 with 2 tailed, 5(1/3)1(2/3)4 = 80/243 with 1 tailed, and 1(2/3)5 = 32/243 with 0 tailed. The two most probable outcomes are equally probable because, even though any particular litter with 2 tailed and 3 tailless is half as likely as any particular litter with 1 tailed and 4 tailless (8/243 versus 16/243), there are twice as many possible arrangements of the former (10 versus 5). ANS CP 3. For convenience we will use the symbols H and h to denote the wildtype and mutant alleles, respectively, of the X-linked hemophilia gene. Let event A be that III-5 is a heterozygous carrier Hh. Let event B be that none of IV-1, 2, 3, and IV-4 are affected. The complement of A, the event A is the event that III-1 is homozygous HH. Then, Pr{A} = 1/2 and Pr{A} = 1/2, since these do not depend on B. Furthermore, Pr{B | A} = 0 and Pr{B | A} = (1/2)2. The reason the 1/2 is squared and not cubed is that the females IV-1 and IV-4 are not informative with regard to the genotype of III-5; in particular, because the male III-6 has genotype H/Y, the female offspring will be unaffected no matter what the genotype of III-5. With these expressions in place we can use Bayes theorem to obtain: This is the probability that III-5 has the genotype Hh, given the pedigree. The probability that her next son will be affected is (1/5) - (1/2) = 1/10. (For any particular woman, of course, the answer is either 0 if she has genotype HH or 1/2 if she has genotype Hh, but among a large number of women with pedigrees like the one shown here, only 1/5 of the women corresponding to III-5 will have the genotype Hh.) Supplemental Problems with Answers S4.1. 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. Pr{ | } ( / ) ( / ) ( / ) A B = / + = 1 2 1 2 1 2 1 1 5 2 2 75 c. Chromosome condensation takes place in prophase; chromosome distribution to daughter cells takes place in anaphase. d. Chromosomes pair in metaphase. ANS S4.1. a. False. b. False. c. True. d. False. S4.2. 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. ANS S4.2. a. False. b. True. c. False. d. True. e. False. S4.3. 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. ANS S4.3. a. True. b. False. c. False. d. False. S4.4. 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. 76 d. There is a 3 : 1 ratio of F2 genotypes. e. All of the above. ANS S4.4. a is true and all the others are false. Statement d is false because, while it is true sometimes, it is not always true; segregation implies a 3 : 1 ratio in the F2 generation only in a cross between heterozygous genotypes when one of the alleles is dominant. S4.5. What is a chiasma? In what stage of meiosis are chiasmata observed? ANS S4.5. A chiasma is a cross-shaped connection between chromatids that results from a physical exchange between two chromatids of homologous chromosomes; chiasmata are observed late in pachytene and subsequent stages of meiosis until anaphase I. S4.6. When in meiosis does the physical manifestation of independent assortment take place? ANS S4.6. At metaphase I when nonhomologous chromosomes align independently on the metaphase plate. S4.7. In the life cycle of corn, Zea mays, how many haploid chromosome sets are there in cells of the following tissues? a. Megasporocyte b. Megaspore c. Root tip d. Endosperm e. Embryo f. Microspore ANS S4.7. a. 2. b. 1. c. 2. d. 3. e. 2. f. 1. 77 S4.8. A cytogeneticist examining cells in Tradescantia stamen hairs is trying to determine the length of the various stages in mitosis and the cell cycle. She examines 2000 cells and finds 320 cells in prophase, 150 cells in metaphase, 80 cells in anaphase, and 120 cells in telophase. What conclusion can be made regarding the relative length of each stage of the cell cycle including the time spent in interphase? Express each answer as a percentage of the total cell-cycle time. ANS S4.8. Assuming that she happens upon cells in each stage according to the length of time that an average cell spends in that stage, prophase takes up 320/2000 = 16.0% of the total time, metaphase 150/2000 = 7.5%, anaphase 4.0%, and telophase 120/2000 = 6.0%. The remaining time (66.5%) must be spent in interphase. S4.9. 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 ANS S4.9. a. 76. b. 76. c. 38. d. 19. S4.10. A woman who is homozygous for normal color vision mates with a color-blind male. They produce a female child who has only one X chromosome and who is also color-blind. In which parent did the nondisjunction take place? Can you specify which meiotic division? ANS S4.10. The mother, but one cannot specify which meiotic division. S4.11. One of the characteristics of rare X-linked recessive conditions is that the great majority of affected persons are males. Explain why this is expected. ANS S4.11. It requires two rare recessive conditions together to yield an affected, homozygous female, but only one rare recessive is needed to yield an affected male. 78 S4.12. The most common form of color blindness in human beings results from X-linked recessive alleles. One type of allele, call it cbr, results in defective red perception, whereas another type of allele, call it cbg, results in defective green perception. A woman who is heterozygous cbr/cbg mates with a normal male, and they produce a son whose chromosome constitution is XXY. What are the possible genotypes of this boy if: a. The nondisjunction took place in meiosis I? b. The nondisjunction took place in the cbr-bearing chromosome in meiosis II? c. The nondisjunction took place in the cbg-bearing chromosome in meiosis II? ANS S4.12. a. cbr cbg. b. cbr cbr c. cbg cbg. S4.13. 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? ANS S4.13. Because sister chromatids are genetically identical, the first meiotic division is reductional for A and equational for B whereas the second meiotic division is equational for A and reductional for B. S4.14. 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.) ANS S4.14. Males and females transmit autosomal genes in the same manner owing to segregation. In females, X-linked genes are transmitted to both sons and daughters, but in males, X-linked genes are transmitted only to daughters. Genes on the Y chromosome are transmitted from father to son to grandson and so forth. 79 S4.15. What observations about Drosophila provided proof of the chromosomal theory of heredity? ANS S4.15. Rare exceptions to the expected pattern of inheritance of X-linked genes were found by Calvin Bridges and correctly interpreted in terms of chromosome behavior. Bridges showed that the exceptions could be explained by occasional failure of the two X chromosomes in the mother to separate from each other in meiosis, which is called nondisjunction. S4.16. In Drosophila, sex is determined by the ratio of X chromosomes to autosomes. Below are several X/A combinations, in which A indicates one complete haploid set of autosomes. What is the expected sex of each of the following flies? a. X/AA b. XX/AA c. XXX/AA d. XX/AAA ANS S4.16. a. Ratio X/A = 0.5, phenotype male. b. Ratio X/A = 1.0, phenotype female. c. Ratio X/A 1.0, phenotype female. d. Ratio X/A = 0.66, phenotype intersex. S4.17. 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? ANS S4.17. The mother must be heterozygous, since her father was affected, hence her genotype is Hh, where h denotes the recessive hemophilia mutation. In meiosis, the h-bearing chromatids undergo nondisjunction and produce an XX-bearing egg with genotype hh. Fertilization by a normal Y-bearing sperm results in a 47, XXY zygote of genotype hh, hence the hemophilia. S4.18. A recessive mutation of an X-linked gene in human beings results in hemophilia, marked by a prolonged increase in the time needed for blood clotting. Suppose that a phenotypically normal couple produces two normal daughters and a son affected with hemophilia. 80 a. What is the probability that both of the daughters are heterozygous carriers? b. If one of the daughters mates with a normal man and produces a son, what is the probability that the son will be affected? ANS S4.18. a. The mother must be a heterozygous carrier of the mutation, and therefore the probability that any particular 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) - 1/2 (that is, the probability of an affected son if the daughter is a carrier) = 1/4. S4.19. In the mouse, an organism with a single X chromosome (XO) is a fertile female. What sex-chromosome constitutions are expected, and in what frequencies, from mating an XO female with a normal male? ANS S4.19. The expected zygotes are given by the accompanying Punnett square. Because the YO zygotes do not survive, the progeny are expected to be 1/3 normal female (XX), 1/3 normal male (XY), and 1/3 XO female. S4.20. 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 c. Y-linked d. X-linked recessive e. X-linked dominant 81 ANS S4.20. e, because all daughters, but none of the sons, receive the fathers X chromosome. S4.21. 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: Mendels seven genes are actually located in only four chromosomes, but only two of the genestall versus short plant and smooth versus constricted podare close enough together that independent assortment would not be observed; he apparently never examined these two traits for independent assortment.) ANS S4.21. No matter which chromosome the first locus is in, the chance that the second will be in a different chromosome is 6/7; the chance that the third locus will be in a different chromosome from either of the first two is 5/7. Continuing this argument, we finally obtain 6/7 - 5/7 - 4/7 - 3/7 - 2/7 - 1/7 = 0.00612, or less than 1 percent. The likelihood of having every gene on different chromosomes is quite small, and for this reason Mendel has been accused of deliberately choosing unlinked genes. In fact, it is now known that not all of Mendels genes are on different chromosomes. Three of the genes ( fa, determining axial versus terminal flowers; le, determining tall versus short plants; and v, determining smooth versus constricted pods) are all located in chromosome 4. The le and v genes undergo recombination at the rate of about 12 percent, but Mendel apparently did not study this particular combination for independent assortment. The fa gene is very distant from the other two and shows independent assortment with them. S4.22. 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? ANS S4.22. a. (2)6 = 64. b. (1/2)6 = 0.016. S4.23. How many different genotypes are possible for a. An autosomal gene with four alleles? b. An X-linked gene with four alleles? 82 ANS S4.23. a. 10, because there are 4 homozygous and 6 heterozygous genotypes. b. 14, because, there are 4 homozygous female genotypes, 4 hemizygous male genotypes, and 6 heterozygous female genotypes. S4.24. Among sibships of size 6 that include three or more girls, what is the ratio of girls : boys? Assume an overall sex distribution of 1 : 1. ANS S4.24. Among sibships of size 6, the sex distribution is 1/64 with 0 girls, 6/64 with 1, 15/64 with 2, 20/64 with 3, 15/64 with 4, 6/64 with 5, and 1/674 with 6. Among those with three of more girls, the sex distribution is 20/42 with 3, 15/42 with 4, 6/42 with 5, and 1/42 with 6. Among these sibships the average proportion of girls is (20/42)(3/6) + (15/42)(4/6) + (6/42)(5/6) + (1/42)(6/6) = 0.619, and so the ratio of girls : boys is 0.619 : 0.381 or 1.625 : 1. S4.25. Duchenne-type muscular dystrophy is an inherited disease of muscle due to a mutant form of a protein called dystrophin. The pattern of inheritance of the disease has these characteristics: (1) affected males have unaffected children, (2) the unaffected sisters of affected males often have affected sons, and (3) the unaffected brothers of affected males have unaffected children. What type of inheritance is suggested by these findings? Explain your reasoning. ANS S4.25. These are typical characteristics of X-linked inheritance. Affected males transmit the mutant X chromosome only to their daughters. A carrier female will have both affected and unaffected sons as well as both carrier and noncarrier daughters. The carrier daughters (sisters of the affected sons) will transmit the mutant gene, but the unaffected sons (brothers of the affected sons) will not. S4.26. For an autosomal gene with m alleles, how many different genotypes are there? ANS S4.26. There are m homozygous genotypes and m(m ' 1)/2 heterozygous genotypes, for a total of m(m + 1)/2. S4.27. Consider the mating Aa - 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? 83 ANS S4.27. a. An absence of aa means that all progeny are A", each of which has a probability of 3/4; hence, the probability of all A" offspring is (3/4)8 = 0.10. b. From the binomial formula, [8!/6!2!](3/4)6(1/4)2 = 0.31. S4.28. Tall, red-flowered hibiscus is mated with short, white-flowered hibiscus. Both varieties are true breeding. All the F1 plants are backcrossed with the short, white-flowered variety. This backcross yielded 188 tall red, 203 tall white, 175 short red, and 178 short white plants. Does the observed result fit the genetic hypothesis of 1 : 1 : 1 : 1 segregation as assessed by a 2 test? ANS S4.28. The 2 value equals (188 " 186)2/186 + (203 " 186)2/186 + (175 " 186)2/186 + (178 " 186)2/186 = 2.56. It has 3 degrees of freedom (four classes of data), and the P value is about 0.5. We therefore conclude that there is no reason, from these data, to reject the hypothesis of 1 : 1 : 1 : 1 segregation. S4.29. What are the values of chi-square that yield P values of 5% (statistically significant) when there are 1, 2, 3, 4, and 5 degrees of freedom? For the types of chi-square tests illustrated in this chapter, how many classes of data do these degrees of freedom represent? Because the significant chi-square values increase with the number of degrees of freedom, does this mean that it becomes increasingly harder (less likely) to obtain a statistically significant chi-square value when the genetic hypothesis is true? ANS S4.29. 3.84 for 1 df, 5.99 for 2 df, 7.82 for 3 df, 9.49 for 4 df, 11.07 for 5 df (these values are from statistical tables, values read from the graph will not be as accurate). These values correspond to 2, 3, 4, 5, and 6 classes of data, respectively, for the type of chi-square tests in this chapter. Perhaps surprisingly, the increasing chi-square value needed for significance does not imply that one is less likely to obtain a statistically significant chi-square with increasing degrees of freedom. The probability of obtaining a statistically significant chi-square value, given that the genetic hypothesis is true, is 5%, no matter what the number of degrees of freedom. S4.30. 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? 84 ANS S4.30. Hypothesis a predicts a 3 : 1 ratio of normal : hyperactive, hence the expected numbers are 3/4 - 64 = 48 and 1/4 - 64 = 16. The 2 = (53 ' 48)2/48 + (11 ' 16)2/16 = 2.1. With one degree of freedom, the P value from the 2 chart is 0.15, so there is no reason in these data to reject the single-gene hypothesis. On the other hand, hypothesis b predicts a ratio of 15:1, hence the expected values are 60 and 4 and the 2 = (53 ' 60)2/60 + (11 ' 4)2/4 = 13.1. With one degree of freedom, P = 0.0003, and so the two-gene hypothesis can be rejected. S4.31. Are the observed progeny numbers of 11, 11, 22, and 22 consistent with a genetic hypothesis that predicts a 1 : 1 : 1 : 1 ratio? ANS S4.31. 2 = (16.5 - 11)2/16.5 + (16.5 ' 11)2/16.5 + (22 ' 16.5)2/16.5 + (22 ' 16.5)2/16.5 = 7.33 with 3 df, which yields a P value of about 0.08. The difference from expectation is not statistically significant, so the genetic hypothesis cannot be rejected even though the discrepancies from the expected numbers may seem to be large. S4.32. A dihybrid cross yielded four classes of phenotypes in the percentages 59%, 18%, 17%, and 6%, among a total of 1000 progeny. Are these data consistent with the independent segregation of two genes? ANS S4.33. Remember that chi-square values must be calculated based on the observed numbers, not the observed percentages. In this case, the chi-square = 3.38 with 3 df, for which the P value is approximately 0.30. The fit is satisfactory. S4.33. Are the observed progeny numbers of 230, 135, and 35 consistent with a model of epistasis between two unlinked genes yielding an expected ratio of phenotypes of 9 : 6 : 1? ANS S4.33. The 2 = (230 - 225)2/225 + (135 - 150)2/150 + (35 - 25)2/25 = 5.5 with 2 df, for which the P value is approximately 0.07. These data would not lead to the rejection of the hypothesis. S4.34. In a 2 test with 1 degree of freedom, the value of 2 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 85 a. The total number of progeny equals 50 b. The total number of progeny equals 100 ANS S4.34. a. Let x be the observed number in one class of progeny and (50 " x) be the number of progeny in the other. The expected numbers are 25 each. The 2 equals (x - 25)2/25 + (50 - x - 25)2/25 = 2(x - 25)2/25. Setting this equal to 4 yields (x - 25)2 = 50, or x - 25 = 7 is the deviation that yields statistical significance at the 5 percent level. In other words, a ratio of progeny of 18 : 32 (or of 32 : 18) or worse leads to rejection of the hypothesis. b. Solve as in part a but

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Textbook: Genetics: Analysis of Genes and Genomes

Created: 2010-01-31

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Textbook: Genetics: Analysis of Genes and Genomes

Created: 2010-01-31

Updated: 2010-01-31

File Size: 10 page(s)

Keywords: flash card flashcards digital flashcards note sharing notes textbook wiki college dorm class classroom exam homework test quiz university college education learn student teachers tutors share, study blue studyblue studyblu

Views: 324

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