Lingkages Problems

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ISBN 10: 1-292-03922-1 ISBN 10: 1-269-37450-8 ISBN 13: 978-1-292-03922-0 ISBN 13: 978-1-269-37450-7

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America

Li n ka g e a nd C hr omos ome ma pp ing in eu k a ry ot es

InsIghts and solutIons 1. In rabbits, black color (B) is dominant to brown (b), while full color (C) is dominant to chinchilla (c ch). The genes controlling these traits are linked. Rabbits that are heterozygous for both traits and express black, full color are crossed to rabbits that express brown, chinchilla with the following results: 31

brown, chinchilla

34

black, full

16

brown, full

19

black, chinchilla

Solution: First, determine the arrangement of the alleles on the homologs of the heterozygous crossover parent (the female in this case). To do this, locate the most frequent reciprocal phenotypes, which arise from the noncrossover gametes—these are phenotypes (1) and (2). Each phenotype represents the arrangement of alleles on one of the homologs. Therefore, the arrangement is

Determine the arrangement of alleles in the heterozygous parents and the map distance between the two genes. Solution: This is a two-point map problem, where the two most prevalent reciprocal phenotypes are the noncrossovers. The less frequent reciprocal phenotypes arise from a single crossover. The arrangement of alleles is derived from the noncrossover phenotypes because they enter gametes intact. The single crossovers give rise to 35>100 offspring (35 percent). Therefore, the distance between the two genes is 35 mu. B

c ch

b

C

c ch

B

C

b

c ch

black, full

B

c ch

Noncrossovers

b

c ch

b

c ch

c ch

b brown, chinchilla Single crossovers

b c ch black, chinchilla

b

b

br







Ly

Sb

br







will yield Ly + br and + Sb + as phenotypes. Inspection shows that these categories (5 and 6) are actually single crossovers, not double crossovers. Therefore, the sequence is incorrect as written. Only two other sequences are possible: The br gene is either to the left of Ly (Case A), or it is between Ly and Sb (Case B).

C

brown, full

c ch

2. In Drosophila, Lyra (Ly) and Stubble (Sb) are dominant mutations located at locus 40 and 58, respectively, on chromosome III. A recessive mutation with bright red eyes is discovered and shown also to be located on chromosome III. A map is obtained by crossing a female who is heterozygous for all three mutations to a male that is homozygous for the bright-red mutation (which we will call br), and the data in the table are generated. Determine the location of the br mutation on chromosome III. Phenotype   

Sb

Second, find the correct sequence of the three loci along the chromosome. This is done by determining which sequence yields the observed double-crossover phenotypes, which are the least frequent reciprocal phenotypes (7 and 8). If the sequence is correct as written, then the double crossover depicted here,

 b

Ly

Number

(1) Ly

Sb

br

404

(2) +

+

+

422

(3) Ly

+

+

18

(4) +

Sb

br

16

(5) Ly

+

br

75

(6) +

Sb

+

59

(7) Ly

Sb

+

4

(8) +

+

br

2

Total

 

 

= 1000

br

Case A Ly

Sb

Ly

Case B br

Sb













Double crossovers  Sb

Ly

br

Double crossovers  Sb and

and 





Ly

br



Comparison with the actual data shows that Case B is correct. The double-crossover gametes yield flies that express Ly and Sb but not br, or express br but not Ly and Sb. Therefore, the correct arrangement and sequence are as follows. Ly

br

Sb







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L i n k a g e a n d C h r o mo s om e ma pping in eu ka r yo te s

Once this sequence is found, determine the location of br relative to Ly and Sb. A single crossover between Ly and br, as shown here, Ly

br

Sb







The final map shows that br is located at locus 44, since Lyra and Stubble are known. 40 Ly

yields flies that are Ly + + and + br Sb (phenotypes 3 and 4). Therefore, the distance between the Ly and br loci is equal to (18 + 16 + 4 + 2)>1000 = 40>1000 = 0.04 = 4 mu Remember to add the double crossovers because they represent two single crossovers occurring simultaneously. You need to know the frequency of all crossovers between Ly and br, so they must be included. Similarly, the distance between the br and Sb loci is derived mainly from single crossovers between them. Ly

br

Sb







This event yields Ly br+ and + + Sb phenotypes (phenotypes 5 and 6). Therefore, the distance equals

(4)

44

(14)

58

br

Sb

3. In reference to Problem 2, a student predicted that the mutation was actually the known mutation scarlet located at locus 44.0. Suggest an experimental cross that would confirm this prediction. Solution: Since the scarlet locus is identical to the experimental assignment, it is reasonable to hypothesize that the bright-red eye mutation is an allele at the scarlet locus. To test this hypothesis, you could perform complementation analysis by crossing females expressing the bright-red mutation with known scarlet males. If the two mutations are alleles, no complementation will occur and all progeny will reveal a bright-red mutant eye phenotype. If complementation occurs, all progeny will express normal brick-red (wild-type) eyes, since the bright-red mutation and scarlet are at different loci (they are probably very close together). In such a case, all progeny will be heterozygous at both the bright-red eye and the scarlet loci and will not express either mutation because they are both recessive. This type of complementation analysis is called an allelism test.

(75 + 59 + 4 + 2)>1000 = 140>1000 = 0.14 = 14 mu

probLEMs anD Discussion QuEstions How Do we Know

?

1. In this chapter, we focused on linkage, chromosomal mapping, and many associated phenomena. In the process, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. What answers would you propose to the following fundamental questions? (a) How was it established experimentally that the frequency of recombination (crossing over) between two genes is related to the distance between them along the chromosome? (b) How do we know that specific genes are linked on a single chromosome, in contrast to being located on separate chromosomes? (c) How do we know that crossing over results from a physical exchange between chromatids? (d) How do we know that sister chromatids undergo recombination during mitosis? 2. What is the significance of crossing over (which leads to genetic recombination) to the process of evolution? 3. Describe the cytological observation that suggests that crossing over occurs during the first meiotic prophase. 4. Why does more crossing over occur between two distantly linked genes than between two genes that are very close together on the same chromosome? 5. Why is a 50 percent recovery of single-crossover products the upper limit, even when crossing over always occurs between two linked genes?

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6. Why are double-crossover events expected less frequently than single-crossover events? 7. What is the proposed basis for positive interference? 8. What three essential criteria must be met in order to execute a successful mapping cross? 9. The genes dumpy wings (dp), clot eyes (cl), and apterous wings (ap) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the genetic distances shown below were determined. What is the sequence of the three genes? d p–a p

42

d p–c l

3

a p–c l

39

10. Colored aleurone in the kernels of corn is due to the dominant allele R. The recessive allele r, when homozygous, produces colorless aleurone. The plant color (not kernel color) is controlled by another gene with two alleles, Y and y. The dominant Y allele results in green color, whereas the homozygous presence of the recessive y allele causes the plant to appear yellow. In a testcross between a plant of unknown genotype and phenotype and a plant that is homozygous recessive for both traits, the following progeny were obtained: colored, green

88

colored, yellow

12

colorless, green

8

colorless, yellow

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Li n ka g e a nd C hr omos ome ma pp ing in eu k a ry ot es

Explain how these results were obtained by determining the exact genotype and phenotype of the unknown plant, including the precise association of the two genes on the homologs (i.e., the arrangement). 11. In the cross shown here, involving two linked genes, ebony (e) and claret (ca), in Drosophila, where crossing over does not occur in males, offspring were produced in a (2 + :1 ca :1 e) phenotypic ratio:

+

e

{

ca+

e

determination of sex was made in the F 2 data. (a) Using proper nomenclature, determine the genotypes of the P 1 and F 1 parents. (b) Determine the sequence of the three genes and the map distance between them. (c) Are there more or fewer double crossovers than expected? (d) Calculate the coefficient of coincidence; does this represent positive or negative interference? Phenotype

sc + + sc sc + sc +

ca +

* e + ca e + ca These genes are 30 mu apart on chromosome III. What did crossing over in the female contribute to these phenotypes? 12. In a series of two-point map crosses involving five genes located on chromosome II in Drosophila, the following recombinant (single-crossover) frequencies were observed: pr-adp

29

pr-vg

13

pr-c

21

pr-b

6

adp-b

35

adp-c

8

adp-vg

16

vg-b

19

vg-c

8

c-b

27

Female A

Female B

d b + + + c

d + + + b c

(1) d b c

v + v + v + + v

314 280 150 156 46 30 10 14

15. A cross in Drosophila involved the recessive, X-linked genes yellow body (y), white eyes (w), and cut wings (ct). A yellowbodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal, but whose wings were cut. The F 1 females were wild type for all three traits, while the F 1 males expressed the yellow-body, white-eye traits. The cross was carried to F 2 progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed. (a) Diagram the genotypes of the F 1 parents. (b) Construct a map, assuming that w is at locus 1.5 on the X chromosome. (c) Were any double-crossover offspring expected? (d) Could the F 2 female offspring be used to construct the map? Why or why not?

(a) If the adp gene is present near the end of chromosome II (locus 83), construct a map of these genes. (b) In another set of experiments, a sixth gene (d) was tested against b and pr, and the results were d - b = 17% and d - pr = 23%. Predict the results of two-point maps between d and c, d and vg, and d and adp. 13. Two different female Drosophila were isolated, each heterozygous for the autosomally linked genes black body (b), dachs tarsus (d), and curved wings (c). These genes are in the order d–b–c, with b closer to d than to c. Shown in the following table is the genotypic arrangement for each female, along with the various gametes formed by both. Identify which categories are noncrossovers (NCO), single crossovers (SCO), and double crossovers (DCO) in each case. Then, indicate the relative frequency with which each will be produced.

T

Offspring

s + s + + s s +

Gamete formation

T

(5) d + +

(1) d b +

(5) d b c

(2) + + +

(6) + b c

(2) + + c

(6) + + +

(3) + + c

(7) d + c

(3) d + c

(7) d + +

(4) d b +

(8) + b +

(4) + b +

(8) + b c

14. In Drosophila, a cross was made between females expressing the three X-linked recessive traits, scute bristles (sc), sable body (s), and vermilion eyes (v), and wild-type males. All females were wild type in the F 1, while all males expressed all three mutant traits. The cross was carried to the F 2 generation and 1000 offspring were counted, with the results shown in the following table. No

Phenotype

y + y + + y y +

+ w w + + w + w

Male Offspring

ct + ct + ct + + ct

9 6 90 95 424 376 0 0

16. In Drosophila, Dichaete (D) is a mutation on chromosome III with a dominant effect on wing shape. It is lethal when homozygous. The genes ebony body (e) and pink eye (p) are recessive mutations on chromosome III. Flies from a Dichaete stock were crossed to homozygous ebony, pink flies, and the F 1 progeny with a Dichaete phenotype were backcrossed to the ebony, pink homozygotes. Using the results of this backcross shown in the following table, (a) diagram the cross, showing the genotypes of the parents and offspring of both crosses. (b) What is the sequence and interlocus distance between these three genes? Phenotype

Dichaete ebony, pink Dichaete, ebony pink Dichaete, pink ebony Dichaete, ebony, pink wild type

Number

401 389 84 96 2 3 12 13

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L i n k a g e a n d C h r o mo s om e ma pping in eu ka r yo te s

17. Drosophila females homozygous for the third chromosomal genes pink eye (p) and ebony body (e) were crossed with males homozygous for the second chromosomal gene dumpy wings (dp). Because these genes are recessive, all offspring were wild type (normal). F 1 females were testcrossed to triply recessive males. If we assume that the two linked genes (p and e) are 20 mu apart, predict the results of this cross. If the reciprocal cross were made (F 1 males—where no crossing over occurs—with triply recessive females), how would the results vary, if at all? 18. In Drosophila, the two mutations Stubble bristles (Sb) and curled wings (cu) are linked on chromosome III. Sb is a dominant gene that is lethal in a homozygous state, and cu is a recessive gene. If a female of the genotype Sb cu + + is to be mated to detect recombinants among her offspring, what male genotype would you choose as her mate? 19. A female of genotype a b c + + + produces 100 meiotic tetrads. Of these, 68 show no crossover events. Of the remaining 32, 20 show a crossover between a and b, 10 show a crossover between b and c, and 2 show a double crossover between a and b and between b and c. Of the 400 gametes produced, how many of each of the eight different genotypes will be produced? Assuming the order a–b–c and the allele arrangement shown above, what is the map distance between these loci? 20. In a plant, fruit color is either red or yellow, and fruit shape is either oval or long. Red and oval are the dominant traits. Two plants, both heterozygous for these traits, were testcrossed, with the results shown in the following table. Determine the location of the genes relative to one another and the genotypes of the two parental plants. Progeny Phenotype

red, long yellow, oval red, oval yellow, long Total

Plant A

Plant B

46 44 5 5

4 6 43 47

100

100

21. Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three autosomes (chromosomes II, III, and IV). A genetics student discovered a male fly with very short (sh) legs. Using this male, the student was able to establish a pure-breeding stock of this mutant and found that it was recessive. She then incorporated the mutant into a stock containing the recessive gene black (b, body color, located on chromosome II) and the recessive gene pink (p, eye color, located on chromosome III). A female from the homozygous black, pink, short stock was then mated to a wild-type male. The F 1 males of this cross were all wild type and were then backcrossed to the homozygous b, p, sh females. The F 2 results appeared as shown in the following table, and no other phenotypes were observed. (a) Based on these results, the student was able to assign sh to a linkage group (a chromosome). Determine which chromosome, and include step-bystep reasoning. (b) The student repeated the experiment, making the reciprocal cross: F 1 females backcrossed to homozygous b, p, sh males. She observed that 85 percent of the offspring fell into the given classes, but that 15 percent of the offspring were equally divided among b+ p, b+ +, +shp, and + sh + phenotypic

182

males and females. How can these results be explained, and what information can be derived from these data? Phenotype

wild pink* black, short black, pink, short

Female

Male

63 58 55 69

59 65 51 60

*

Pink indicates that the other two traits are wild type (normal). Similarly, black, short offspring are wild type for eye color. 22. In Drosophila, a female fly is heterozygous for three mutations: Bar eyes (B), miniature wings (m), and ebony body (e). (Note that Bar is a dominant mutation.) The fly is crossed to a male with normal eyes, miniature wings, and ebony body. The results of the cross are shown in the following list. Interpret the results of this cross. If you conclude that linkage is involved between any of the genes, determine the map distance(s) between them. miniature

111

wild type

29

Bar, ebony

101

Bar, miniature, ebony

31

Bar

117

Bar, miniature

26

ebony

35

miniature, ebony

115

23. An organism of the genotype AaBbCc was testcrossed to a triply recessive organism (aabbcc). The genotypes of the progeny are in the following table. AaBbCc

20

AaBbcc

20

aabbCc

20

aabbcc

20

AabbCc

5

Aabbcc

5

aaBbCc

5

aaBbcc

5

(a) Assuming simple dominance and recessiveness in each gene pair, if these three genes were all assorting independently, how many genotypic and phenotypic classes would result in the offspring, and in what proportion? (b) Answer part (a) again, assuming the three genes are so tightly linked on a single chromosome that no crossover gametes were recovered in the sample of offspring. (c) What can you conclude from the actual data about the location of the three genes in relation to one another? 24. Based on our discussion of the potential inaccuracy of mapping (see Figure 12), would you revise your answer to Problem 23(c)? If so, how? 25. In Creighton and McClintock’s experiment demonstrating that crossing over involves physical exchange between chromosomes (see Section 6), explain the importance of the cytological markers (the translocated segment and the chromosome knob) in the experimental rationale. 26. Traditional gene mapping has been applied successfully to a variety of organisms including yeast, fungi, maize, and Drosophila. However, human gene mapping has only recently shared a similar spotlight. What factors have delayed the application of traditional gene-mapping techniques in humans? 27. DNA markers have greatly enhanced the mapping of genes in humans. What are DNA markers, and what advantage do they confer? 28. Are sister chromatid exchanges effective in producing genetic variability in an individual? in the offspring of individuals?

Li n ka g e a nd C hr omos ome ma pp ing in eu k a ry ot es

soLutions to sELEctED probLEMs anD Discussion QuEstions answers to now solve this

1. (a) 1/4 AaBb 1/4 Aabb 1/4 aaBb 1/4 aabb (b) 1/4 AaBb 1/4 Aabb 1/4 aaBb 1/4 aabb (c) If the arrangement is AB/ab * ab/ab then the two types of offspring will be as follows: 1/2 Ab/ab

1/2 aB/ab

If, however, A and B are not coupled, then the symbolism would be Ab/aB * aabb. The offspring would occur as follows: 1/2 Ab/ab

1/2 aB/ab

2. The most frequent classes are PZ and pz. These classes represent the parental (noncrossover) groups, which indicates that the original parental arrangement in the testcross was PZ/pz * pz/pz. Adding the crossover percentages together (6.9 + 7.1) gives 14 percent, which would be the map distance between the two genes. 3. Examine the progeny list to see which types are not present. In this case, the double crossover classes are the following: + + c and a b + (a, b) Gene b is in the middle and the arrangement is as follows. + b c/ a + + a - b = 7 map units

b - c = 2 map units

(c) The progeny phenotypes that are missing are + + c and a b +, of which, from 1000 offspring, 1.4 (0.07 * 0.02 * 1000) would be expected. Perhaps by chance or some other unknown selective factor, they were not observed. solutions to problems and Discussion Questions

4. With some qualification, especially around the centromeres and telomeres, one can say that crossing over is somewhat randomly distributed over the length of the chromosome. Two loci that are far apart are more likely to have a crossover between them than two loci that are close together. 6. If the probability of one event is 1/X, the probability of two events occurring at the same time will be 1/X2. 10. The heterozygous parent in the test cross is RY/ry * ry/ry with the two dominant alleles on one chromosome and the two recessives on the homolog. The map distance would be 10 map units between the R and Y loci. 12. The map for parts (a) and (b) is the following: d............b.........pr..........vg........c.........adp 31 48 54 67 75 83

(+++++)+++++* Map units The expected map units between d and c would be 44, between d and vg 36, and between d and adp 52. However, because

there is a theoretical maximum of 50 map units possible between two loci in any one cross, that distance would be below the 52 determined by simple subtraction.

14. (a) P1: sc s v/sc s v * + + +/Y F1: + + +/sc s v * sc s v/Y (b) The map distances are determined by first writing the proper arrangement and sequence of genes, and then computing the distances between each set of genes. sc v s + + +

sc - v = 33 percent (map units) v - s = 10 percent (map units) (c, d) The coefficient of coincidence = 0.727, which indicates that there were fewer double crossovers than expected; therefore, positive chromosomal interference is present. 16. (a) Represent the Dichaete gene as an uppercase letter because it is dominant. At this point, the gene sequence is not given. P1: D + +/ + + + * + e p/+ e p F1: D + +/+ e p * + e p/+ e p F2:

(b) F1:

D + +/+ e p + e p/+ e p D e +/+ e p + + p/+ e p D + p/+ e p + e +/+ e p D e p/+ e p + + +/+ e p

Dichaete ebony, pink Dichaete, ebony pink Dichaete, pink ebony Dichaete, ebony, pink wild type

D + +/+ p e * + p e/+ p e

D - p = 3.0 map units

p - e = 18.5 map units

18. One would use the typical testcross arrangement with the curled gene so the male parent arrangement would be + cu/ + cu. 20. Assign the following symbols, for example: R = Red O = Oval

r = yellow o = long

Progeny A: Ro/rO * rroo = 10 map units Progeny B: RO/ro * rroo = 10 map units 22. Begin with a set of symbols as indicated below: B+ = wild eye shape m+ = wild wings e+ = wild body color

B = Bar eye shape m = miniature wings e = ebony body color

Superficially, the cross would be as follows: B+B m+m e+e * B+? m? e (The ? is used at this point to indicate that we have no information allowing us to decide whether any of the alleles in the male are X-linked.) The arrangement is as indicated in the following: B m+/B+m; e+/e Notice that a semicolon is used to indicate that the ebony locus is on a different chromosome. At this point and without prior knowledge, we still don’t know whether any of the genes are X-linked; however, it is of no consequence to the solution of the problem. (In actuality, both B and m loci are X-linked.) To determine the map distances (again, ebony is out of the mapping picture at this point because it is not linked to either B or m): 111 117 26 29

+ + + +

115 101 31 35

= = = =

226 218 57 64

= = = =

parental parental crossover crossover

Mapping the distance between B and m would be as follows: (57 + 64)/(226 + 218 + 57 + 64) * 100 = 121/565 * 100 = 21.4 map units

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