Page 183 - Genetics_From_Genes_to_Genomes_6th_FULL_Part1

Problems 175

males. Assume 1000 flies are counted and that no a. Diagram the genotype of the female parent.
interference exists in this region. b. Map these loci.

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a b c + c. Do the data provide evidence of interference?
a b c Justify your answer with numbers.
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a b c 28. a. In Drosophila, crosses between F 1 heterozygotes of
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a b c + the form A b / a B always yield the same ratio of
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a b c phenotypes in the F 2 progeny regardless of the dis-
a b c + tance between the two genes (assuming complete
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a b c + dominance for both autosomal genes). What is this
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a b c ratio? Would this also be the case if the F 1 hetero-
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b. If the cross was reversed, such that a b c / a b c zygotes were A B / a b? (Hint: Remember that in
Drosophila, recombination does not take place
males are crossed to a b c / a b c females, how during spermatogenesis.)
many flies would you expect in the same pheno-
typic classes? b. If you intercrossed F 1 heterozygotes of the form
26. Drosophila females heterozygous for each of three re- A b / a B in mice, the phenotypic ratio among the F 2
progeny would vary with the map distance between
cessive autosomal mutations with independent pheno- the two genes. Is there a simple way to estimate the
typic effects [thread antennae (th), hairy body (h), and map distance based on the frequencies of the F 2
scarlet eyes (st)] were testcrossed to males showing phenotypes, assuming rates of recombination are
all three mutant phenotypes. The 1000 progeny of this equal in males and females? Could you estimate
testcross were
map distances in the same way if the mouse F 1
thread, hairy, scarlet 432 heterozygotes were A B / a b?
wild type 429 29. A true-breeding strain of Virginia tobacco has dominant
thread, hairy 37 alleles determining leaf morphology (M), leaf color
thread, scarlet 35 (C), and leaf size (S). A Carolina strain is homozygous
hairy 34 for the recessive alleles of these three genes. These
scarlet 33 genes are found on the same chromosome as follows:
a. Show the arrangement of alleles on the relevant M C S
chromosomes in the triply heterozygous females.
b. Draw the best genetic map that explains these data. 6 m.u. 17 m.u.
c. Calculate any relevant interference values. An F 1 hybrid between the two strains is now backcrossed
27. Male Drosophila expressing the autosomal recessive to the Carolina strain. Assuming no interference:
mutations sc (scute), ec (echinus), cv (crossveinless),
and b (black) were crossed to phenotypically wild- a. What proportion of the backcross progeny will
type females, and the 3288 progeny listed were resemble the Virginia strain for all three traits?
obtained. (Only mutant traits are noted.) b. What proportion of the backcross progeny will
resemble the Carolina strain for all three traits?
653 black, scute, echinus, crossveinless c. What proportion of the backcross progeny will have
670 scute, echinus, crossveinless the leaf morphology and leaf size of the Virginia
675 wild type strain but the leaf color of the Carolina strain?
655 black
71 black, scute d. What proportion of the backcross progeny will have
73 scute the leaf morphology and leaf color of the Virginia
73 black, echinus, crossveinless strain but the leaf size of the Carolina strain?
74 echinus, crossveinless 30. In humans, the correlation between recombination
87 black, scute, echinus frequency and length of DNA sequence is, on average,
84 scute, echinus 1 million bp per 1% RF. During the process of mapping
86 black, crossveinless the Huntington disease gene (HD), it was found that
83 crossveinless HD was linked to a DNA marker called G8 with an
1 black, scute, crossveinless RF of 5%. Surprisingly, when the HD gene was finally
1 scute, crossveinless identified, its physical distance from G8 was found to
1 black, echinus be about 500,000 bp, instead of the expected 5 million
1 echinus bp. How can this observation be explained?