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 1 year ago

In a particular trait in Drosophila a cross between a mutant male and wild type female...

In a particular trait in Drosophila a cross between a mutant male and wild type female results in all wild type offspring, while a cross between a wild type male and a mutant female produces all mutant offspring. Describe the genetics of the trait.

The allele b gives Drosophila flies a black body, and b+ gives brown, the wild-type phenotype....

The allele b gives Drosophila flies a black body, and b+ gives brown, the wild-type phenotype. The allele wx of a separate gene gives waxy wings, and wx+ gives nonwaxy, the wild-type phenotype. The allele cn of a third gene gives cinnabar eyes, and cn+ gives red, the wild-type phenotype. A female heterozygous for these three genes is testcrossed, and 2,424 progeny (you could also use n=2,423) are classified as follows: cn+ · wx+ · b 101 cn · wx+...

A female fruit fly with singed bristles was mated with a male from a true-breeding wild-type...

A female fruit fly with singed bristles was mated with a male from a true-breeding wild-type stock with long bristles. All of the F1 females had wild-type long bristles and all of the F1 males had singed bristles. If the F1 flies are intercrossed, the expected ratio of long to singed bristles in the F2 flies is Multiple Choice 3:1 in both sexes. 3:1 in females, while all the males will have singed bristles. 1:1 in both sexes. 1:1 in...

32) Normal (wild-type) Drosophila eyes are red. In a cross in Drosophila involving the X-linked recessive...

32) Normal (wild-type) Drosophila eyes are red. In a cross in Drosophila involving the X-linked recessive eye mutation white (resulting in a white eye) and the autosomal recessive eye mutation sepia (resulting in a dark eye), predict the sex and color of the eyes of the offspring from crossing true-breeding parents for parts A) and B) below. White is epistatic to the expression of sepia (i.e., if an eye is white, it cannot also be sepia; i.e., white masks the...

In the fruit fly, Drosophila melanogaster, a wingless female fly is mated to a male that...

In the fruit fly, Drosophila melanogaster, a wingless female fly is mated to a male that is white-bodied and legless. Assume that all alleles causing these phenotypes are recessive. Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males, and the following progeny were observed: Phenotypes Number Observed wingless 324 wild-type 34 white, wingless 135 white 8 white, legless 319 hairless, white, legless 32 hairless 140 hairless, legless 9 (a) With respect to the three genes mentioned in the problem, what are the genotypes...

50 units of protein A are required per cell to establish a wild-type phenotype. A single...

50 units of protein A are required per cell to establish a wild-type phenotype. A single wild-type allele of gene A produces 40 units of enzyme per cell. Which of the following statements is true regarding the wild type allele. Select ALL that apply. They are haplosufficient They are haploinsufficient They are recessive to null alleles They are dominant to hypomorphs making less than 10 units of enzyme per cell. They are recessive They are recessive to hypomorphs making less...

Parent phenotype: female: apterous and red eyes male: male winged and sepia eyes (note: sepia begins...

Parent phenotype: female: apterous and red eyes male: male winged and sepia eyes (note: sepia begins as orange and transitions to gray/black as flies mature) a) Considering both parents are homozygous for each trait, what are the parents' genotypes? b) Using a Punnett square for a dihybrid cross, calculate the genotype and phenotype for the F1 flies. c) With the information obtained from the F1 genotypr, use a Punnett square to calculate the expected genotype and phenotype ratios for the...

You cross a Drosophila female with kidney-shaped and brown eyes with a male that has wildtype...

You cross a Drosophila female with kidney-shaped and brown eyes with a male that has wildtype eye shape and color. Kidney-shaped eyes are found in females that are heterozygous for the bar eyed mutation. Bar eyed females are homozygous for the bar eyed mutant allele, whereas bar eyed males are hemizygous for the mutation. Brown eyes are an autosomal recessive mutation. Attention: Half of the resulting F1 has brown eyes! Would you expect to see bar eyed females in the...

A fly was found that had two mutations, Q&Z. Reciprocal crosses between wild type flies and...

A fly was found that had two mutations, Q&Z. Reciprocal crosses between wild type flies and Q&Z mutants gave all Q&Z mutant F1 progeny in all cases. These results suggest that the Q&Z mutations have an autosomal dominant inheritance pattern. The genes X, Q, and Z appear to be linked. To map the genes the following crosses were done. A homozygous Q, Z mutant female was crossed to a homozygous X mutant males shown in the diagram below. X, Q,...

Cinnabar eyes (cn) and reduced bristles (rd) are autosomal recessive characters in Drosophila. A homozygous wild-type...

Cinnabar eyes (cn) and reduced bristles (rd) are autosomal recessive characters in Drosophila. A homozygous wild-type female was crossed to a reduced, cinnabar male, and the F1 males were then crossed to the F1 females to obtain the F2. Of the 400 F2 offspring obtained, 292 were wild type, 9 were cinnabar, 7 were reduced, and 92 were reduced and cinnabar. Briefly explain (2 sentences) these results and estimate the distance between the cn and rd loci.