Problem Set 3 Questions

Genome 371 Spring, 2005 

Chromosome Theory of Inheritance; Linkage & Recombination 

Problem Set #3 (not for credit): 

(thanks to Steve Jackson for some of these problems) 

1. A mating between yellow-bodied, vestigial (short) wing male Drosophila and wild type 

(tan bodied, long wings) females results in only wild type F1 progeny. F1 males and females 

were then crossed and found to produce 16 yellow-bodied males with vestigial wings, 48 

yellow-bodied males with normal wings, 15 males with tan bodies and vestigial wings, 49 

wild type males, 39 tan-bodied females with vestigial wings and 97 wild type females. 

Explain the inheritance of these genes based on these results. 

2. Each of the four pedigrees shown here represents a family within which a genetic disease 

is segregating. One of the diseases is transmitted as an autosomal recessive condition, one as 

an X-linked recessive, one as an autosomal dominant, and one as an X-linked dominant 

(assume all four diseases are rare in the population). 

Pedigree 1: 

Pedigree 2: 

1 2 

1 2 

1 2 3 4 

5 

1 2 3 4 

5 6 

1 

Pedigree 3: 

Pedigree 4: 

1 2 

1 2 

1 2 3 4 

5 6 

1 2 3 4 

5 6 

7 

1 

1 

(a). Indicate which pedigree represents which mode of inheritance and explain how you 

know. 

(b). For each pedigree, how would you advise the parents of the chance that their child 

(indicated by the hexagon shape) will have the condition? 

3. The pedigree shown here indicates the occurrence of albinism in a group of Hopi Indians, 

among whom the trait is unusually frequent. Assume that the trait is fully penetrant.


I 

1 2 

3 4 5 6 7 8 9 

II 

1 2 3 4 5 6 

7 

8 9 

III 

IV 

1 

2 3 4 5 6 7 

(a). Is albinism in this population caused by a recessive or dominant allele? 

(b). Is the gene sex-linked or autosomal? 

(c). What are the genotypes of the following individuals: 

Individual I-1 Individual I-8 Individual I-9 

Individual II-6 Individual II-8 Individual III-4 

4. The following is a pedigree of a family in which a rare form of colorblindness is found. 

Indicate as much as you can about the genotypes of all the individuals in the pedigree. 

1 

I 

1 

2 

II 

1 2 3 4 

III 

1 

2 3 4 

5. Several different antigens can be detected in blood tests. The following four traits were 

tested for each individual shown below. 

ABO type: [I A and I B codominant, i recessive] 

Rh type: [Rh + dominant to Rh - ] 

MN type: [M and N codominant] 

Xg(a) type: [X g(a+) dominant to X g(a-) ] 

All of these blood type genes are autosomal, except for Xg(a) which is X-linked. 

Mother: AB Rh - MN X g(a+) 

Daughter: A Rh + MN X g(a-) 

Alleged father 1: AB Rh + M X g(a+) 

Alleged father 2: A Rh - N X g(a-) 

Alleged father 3: B Rh + N X g(a-)


Alleged father 4: O Rh - MN X g(a-) 

(a). Which, if any, of the alleged fathers could be the real father? 

(b). Would your answer to part (a) change if the daughter had Turner syndrome (XO 

female)? If so, how? 

6. Hemophilia and red-green colorblindness are both X-linked recessive traits in humans. 

(X h = hemophilia allele, X H is normal; X g = colorblindness allele, X G = normal). A couple 

have had a hemophilic son with Klinefelter syndrome (XXY male). They would like to have 

a second child, but are worried about the risk of hemophilia. Their genetic counselor (who 

knows both their genotypes with respect to hemophilia) tells them that there is a 50% chance 

that their next child will have hemophilia regardless of whether it is a son or daughter 

(assuming that the child is normal). 

(a). What combinations of egg and sperm genotypes (with respect to the sex chromosome 

and hemophilia allele) could have given rise to the XXY son? 

(b). What are the genotypes of the parents with respect to hemophilia? Explain how you 

know this. 

(c). The woman is red-green colorblind, while the man and son have normal vision. 

Diagram the meiosis where the nondisjunction gave rise to the aberrant (defective) gamete. 

Mark the genotypes on the chromosomes. Show the chromosomes and chromatids at the 

start of meiosis, at the end of meiosis I, and at the end of meiosis II. Use long lines for the X 

chromosome and short lines for the Y chromosome. 

7. Consider the following pedigree of a rare form of testicular cancer. Assume the allele 

responsible for the disease is completely penetrant. 

(a). Is the gene responsible for the disease autosomal or sex-linked? Why? 

(b). Would your answer change if the mutation responsible for the cancer is common? 

Explain. 

8. Choose the phrase from the right column that best fits the term in the left column. 

(a). recombination 

(b). linkage 

1. A statistical method for testing the fit between observed 

and expected results. 

2. An ascus containing spores of four different genotypes.


(c). chi square test 

(d). chiasma 

(e). tetratype 

(f). locus 

(g). coefficient of 

coincidence 

(h). interference 

(i). Parental ditype 

(j). ascospores 

(k). First division 

segregation 

3. One crossover along a chromosome makes a second 

nearby crossover less likely. 

4. When two loci recombine in less than 50% of gametes. 

5. The relative chromosomal location of a gene. 

6. The ratio of observed double crossovers to expected 

double crossovers. 

7. Fungal spores contained in a sac. 

8. Formation of new genetic combinations by exchange of 

parts between homologs. 

9. When the two alleles of a gene are segregated into 

different cells at the first meiotic division. 

10. An asci containing only two nonrecombinant kinds of 

spores. 

11. Structure formed at the spot where crossing over 

occurs between homologs. 

9. Two genes on chromosome 7 of corn are identified by the recessive alleles gl (glossy), 

which determines glossy leaves, and ra (ramosa), which determines branching of ears. When 

a plant heterozygous for each of these alleles was crossed with a homozygous recessive plant, 

the progeny consisted of the following genotypes: 

Gl ra 88 gl Ra 103 

Gl Ra 6 gl ra 3 

What is the frequency of recombination between these genes? 

10. In the yellow fever mosquito Aedes aegypti, a dominant gene DDT, for DDT resistance and 

a dominant gene Dl for Dieldrin resistance (Dieldrin is another long-lasting insecticide) are 

known to be on the same chromosome. A cross was carried out between a DDT-resistant 

strain and a Dieldrin-resistant strain, and dihybrids resistant to both insecticides were 

testcrossed with wildtype males. The progeny observed were: 88 resistant to DDT only, 89 

resistant to Dieldrin only, 98 sensitive to both insecticides, and 107 resistant to both 

insecticides. 

(a). Are these genes linked or unlinked? 

(b). A χ 2 test for linkage is to test whether the parental and recombinant classes are present 

in a 1:1 ratio. Test your hypothesis about whether or not the genes are linked with χ 2 test. 

(c). What can you say about the relative location of these genes on the chromosome? 

11. In the tomato, tall is dominant to dwarf, smooth-skinned fruit is dominant to fuzzyskinned 

fruit, and round fruit is dominant to oblate (pear) shaped. All of the dominant alleles


represent the wild type condition. A trihybrid of unknown origin is testcrossed to a dwarf, 

fuzzy, oblate tomato and the following progeny are observed: 

tall, smooth, oblate, 151 

tall, smooth, round 33 

tall, fuzzy, round 11 

tall, fuzzy, oblate 2 

dwarf, fuzzy, round 155 

dwarf, fuzzy, oblate 29 

dwarf, smooth, oblate 12 

dwarf, smooth, round 0 

(a). What is the allele configuration of the parent? 

(b). Draw a genetic map depicting these loci with map distances labeled. 

(c). Was interference observed in this cross? Show your work. 

12. That farmer in the Skagit Valley just won't shut up about his tulips. He tells you of other 

varieties of tulips he's found. One is short-stemmed, and examination of monohybrids 

suggests it is recessive to normal-stemmed. Another trait is red; to him, this is a mystery 

because when red tulips are crossed to white, all the progeny are pink, but when the pinks 

are selfed, they produce red, white and pink progeny. He wishes to map these traits with 

respect to each other and to his fringed petaled-plants (remember, recessive to normalshaped), 

so he can plant efficiently for marketing hybrid varieties. He has a short, white, 

fringed true-breeding stock, and a red true-breeding stock of normal height and petal shape. 

(a). What is the first cross you tell him to do? 

(b). He does it, testcrosses the trihybrid to short, white, fringed plants, and obtains the 

following results 

Short, white, fringed 623 

Short, pink, fringed 2 

Short, white 62 

Short, pink 29 

White, fringed 25 

White 3 

Pink, fringed 63 

Pink 630 

Draw a genetic map of these loci, including map distances, gene order, and any interference. 

(c). The farmer wants to establish a true-breeding stock of his short, pink, fringed plants. 

Can he do it? How, or why not? 

13. The following is a genetic map of a region of a chromosome: 

gene a -10cM- gene b - 5cM- gene c -20cM- gene d


Start with a heterozygote with all the mutant alleles of these genes in cis, perform a testcross with 

this tetrahybrid as one of the parents, and examine 10,000 progeny: 

(a). How many progeny do you expect to have the phenotype a + + +? 

(b). How many will have the phenotype a b + +? 

(c). How many will have the phenotype a + c +? 

14. If the a and b loci are 20 map units apart in humans, and an AB/ab woman marries an ab/ab 

man, what is the probability that their first child will be Ab/ab? 

15. A tomato farmer wants to grow and sell fuzzy, yellow, beaked (pear-shaped) tomatoes. Each of 

these traits is determined by a single recessive gene (fz, y and bk, respectively). Although he has a 

true-breeding stock containing all these mutations, it is very unhealthy and not practical for 

growing lots of tomatoes. He therefore wants to construct a new stock. He knows that beaked and 

fuzzy are normally 38 map units apart in normal tomato plants, and he knows that yellow is normally 

unlinked to these genes. He plans to cross bk fz / bk fz plants to y / y plants, cross the trihybrid to his 

sick testcross stock, and isolate the resulting bk fz y recombinant plants. 

(a). When he does this, what percentage of the progeny will be fuzzy, yellow and beaked? Do not 

ignore recombination between bk and fz. 

(b). The farmer also knows that as a result of the Tomato Genome Project, variants have been 

isolated in which the yellow locus is now linked to both fuzzy and beaked. Furthermore, a map of 

these variants has been determined: beaked -38cM- fuzzy -18cM -yellow. His alternate plan is to 

obtain the + + yellow variants from the Project and cross them to the corresponding bk fz + variants. 

He will then testcross this trihybrid, and isolate the resulting bk fz y recombinant plants. Using this 

approach, what percentage of the progeny will be fuzzy, yellow and beaked? Assume all the 

chromosomes used for this have the same translocation, and again, do not ignore recombination 

between bk and fz. 

(c). What do you think is therefore the most economical way for the farmer to obtain the 

plants he wants? 

16. A cross was performed between a yeast strain that requires methionine and lysine for 

growth (met - , lys - ) and another yeast strain which is met + lys + . One hundred asci were 

dissected and colonies were grown for the four spores in each ascus. Cells from these 

colonies were tested for their ability to grow on Petri plates containing either minimal 

medium (min), min + lysine (lys), min + methionine (met), or min + lys + met. The asci could 

be divided into two groups based on this analysis: 

Group 1: In 89 asci, cells from two of the four spore colonies could grow on all four 

kinds of media, while the other two spore colonies could only grow on min + lys + met. 

Group 2: In 11 asci, cells from one of the four spore colonies could grow on all four 

kinds of petri plates. Cells from a second one of the four spore colonies could grow only on 

min + lys plates and on min + lys + met plates. Cells from a third of the four spore colonies 

could only grow on min + met plates and on min + lys + met plates. Cells from the 

remaining colony could only grow on min + lys + met. 

(a). What are the genotypes of the spores within each of the two types of asci? 

(b). Are the lys and met genes linked? If so, what is the map distance between them?


(c). If you could extend this analysis to many more asci, you would eventually find some asci 

with a different pattern. For these asci, describe the phenotypes of the four spores. List this 

as the ability of dissected spores to form colonies on the four kinds of petri plates. 

17. A single yeast cell placed on a solid agar will divide mitotically to produce a colony of 

about 10 7 cells. A haploid yeast cell that has a mutation in the ade2 gene will produce a red 

colony; an ade2 + colony will be white. Some of the colonies from a diploid yeast cell with a 

genotytpe of ade2 + /ade2 - will contain sectors of red within a white colony. How would you 

explain these sectors? 

18. One form of hemophilia, an X-linked disorder of blood clotting, is caused by mutation in 

clotting factor VIII. Many single-nucleotide mutations of this gene have been described, 

making detection of mutant genes by Southern blots inefficient. There is, however, an RFLP 

for the enzyme HindIII contained in an intron of the factor VIII gene that can often be used in 

screening, as shown below. 

hindIII hindIII hindIII 

hindIII hindIII 

region 

detected 1'-6" 

by probe 

region 

detected 1'-6" 

by probe 

4kb 

3kb 

A female whose brother has hemophilia is at 50% risk of being a carrier of this disorder. To 

test her status, DNA is obtained from white blood cells from family members, cut with 

HindIII, and the fragments probed and visualized by Southern blotting. The results are 

shown below (the mothers symbol represents a probable carrier):


? ? 

= carrier of sex-linked recessive mutation. 

Determine whether any of the females in generation II are carriers for hemophilia. 

19. A man heterozygous for a disease gene D is also heterozygous for an RFLP linked at a 

distance of 6 map units from the disease gene. His chromosomal arrangement is D RFLP-1/d 

RFLP-2 (where d represents the wild type allele of the disease gene). What percentage of his 

sperm will be D RFLP-2? 

20. A woman heterozygous for a dominant disease gene D is also heterozygous for an RFLP 

linked 10 map units away. The chromosomal arrangement is D RFLP-1/d RFLP-2. Her 

unaffected mate is homozygous for RFLP-2. Her fetus is tested and is shown to be 

homozygous for RFLP-2. What is the probability that the child will have the disease? 

21. A gene for a genetic disease in cats (D) is linked to an RFLP. A female from a pure-bred 

strain of cats (a pure-bred strain of cats means that the cats are homozygous at every locus, 

ideally) unaffected by the disease is crossed to a male from a pure bred strain affected by the 

disease. The F 1 were intercrossed to produce an F 2 that included 8 kittens, as shown in the 

pedigree below:


I 

1 2 

II 

3 4 

III 

7kb 

4kb 

5 6 7 8 9 10 11 12 

(a). Assuming the disease is fully penetrant, does the disease appear to be dominant or 

recessive? 

(b). What is the genotype of the II-3 individual? 

22. A pedigree is shown below involving seven different alleles at an RFLP locus. The 

autoradiogram of an electrophoretic gel shows the distinctive positions of eight fragments (1- 

8) from the individuals shown in the pedigrees: 

I 

1 2 

3 4 

II 

III 

1 

2 

3 

4 

5 

6 

7 

8 

1 2 

1 2 3 4 5 6 7 

(a). Which alleles should each of the parents (II-1 and II-2) have on the basis of the 

grandparents' and children's genotypes? 

(b). If individuals III-1 and III-3 have children together, what combinations of alleles would 

be observed in their offspring?


23. Filled symbols in the pedigree indicate people with a rare autosomal dominant genetic 

disease. DNA samples were prepared from each of the individuals in the pedigree. The 

samples were digested with a restriction endonuclease and analyzed by Southern blot 

analysis using a probe corresponding to a human sequence called DS12-88, with the results 

shown in the figure below. Do the data in these two figures support the hypothesis that the 

gene for the disease that is segregating in this family is linked to the region homologous to 

DS12-88? 

generation: 

I 

1 2 

II 

1 2 

3 4 5 6 

III 

1 2 3 4 5 6 

7 8 9 10 11 12 13 14 

IV 

1 2 3 4 5 6 

7 

I II III IV 

1 2 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 2 3 4 5 6 7

Problem Set 3 Questions

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