Question

In: Biology

A student crosses a phenotypically wild type Drosophila female to a male with unknown phenotype (that...

A student crosses a phenotypically wild type Drosophila female to a male with unknown phenotype (that unreliable lab partner confused the fly stocks again!). There are two genes segregating in the cross. The “vg” phenotype has small wings and the “cn” phenotype has light eye color, and both genes are on an autosome. Remember that Drosophila males do not exhibit meiotic recombination. (hint: you may need a page of blank paper to work possible crosses before you answer the questions. show the work

The phenotype of the progeny are: wt 250

vg 15

cn 15

vg cn 250

Are the vg and cn mutations dominant or recessive?

What is the genotype of the P1 female? Be sure to assign specific alleles to specific chromosomes.

What are the genotypes of the four classes of offspring (you do not need to designate which allele is on which chromosome)?

wt:

vg:

cn:

vg cn:

What is the genotype and the phenotype of the P1 male? Be sure to show which alleles are on which chromosomes.

What is the general formula for calculating a map distance in Drosophila (and many other organisms)? Explain how it relates to the definition of genetic map distance.

What is the map distance between cn and vg?

Expert Solution

a) most of the progenies are wildtype so the vg and cn alleles are recessive to wild-type alleles. cn and vg are

b)genotype of the wildtype female is vg+vgcn+cn ( that is the wildtype female is heterozygous).

the number vg and cn progenies are less than the other two phenotypes, so vg and cn are recombinant progenies.

recombination frequency is the percentage of the recombinant progenies, in the female parent both vg and cn genes are linked on a chromosome. if the genes are linked on a chromosome the number of recombinant gametes are less than the number of parental type gametes, vg and cn progenies are formed from recombinant gametes, so recombinant gametes are vg+cn ( cn progeny forms from this gamete) and vgcn+ (vg progeny forms from this gamete), parental type gametes are vgcn and vg+cn+, so in the wildtype female.

c) genotype of the P1 male is vgvgcncn, and its phenotype is vg, cn.

vg+cn+/vgcn

	vg+cn+	vg+cn	vgcn+	vgcn
vgcn	vg+cn+/vgcn ( parental type)	vg+cn/vgcn (recombinant)	vgcn+/vgcn ( recombinant)	vgvg/cncn ( parental type)

vgvgcncn male parent can produce only one type of gamete vgcn

d) map distance is the recombination frequency in percentage.

recombination frequency= (number of recombinant progenies/total number of progenies)100

e) if the recombination frequency between two genes is 1% the map distance between two genes is 1 cM.

f) vg and cn progenies are recombinants

recombination frequency= ( (number of vg + number of cn)/total number of progenies)100

= ( (15+15)/530)

= 5.66 %

distance between vg and cn= 5.66 cM.

gladiator answered 2 years ago

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