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In D. melanogaster, a homozygous female with mutations for vestigial wings (vg), black body (b) and...

In D. melanogaster, a homozygous female with mutations for vestigial wings (vg), black body (b) and purple eye color (pr) was mated to a homozygous wild-type male. All the F₁ progeny, both male and female, had normal (wild-type) phenotypes for the three characteristics, indicating that the mutations are autosomal recessive. In a testcross of the F₁ females with males that have vestigial wings, black body and purple eyes, the progeny were of eight different phenotypes reflecting eight different genotypes. The order in which the genes are list below is completely arbitrary.

vg b pr 1779

vg⁺ b⁺ pr⁺ 1654

vg⁺ b pr 252

vg b⁺ pr⁺ 241

vg⁺ b pr⁺ 131

vg b⁺ pr 118

vg b pr⁺ 13

vg⁺ b⁺ pr 9

4197 Total

a. Determine the linear order of the genes on the chromosome (which gene is in the middle)?

b. Calculate the recombinant frequency between the three loci.

c. Draw a genetic map of the three loci, with the correct map distance between each.

Solutions

Expert Solution

P generation: vg b pr/ vg b pr X vg⁺ b⁺ pr⁺/ vg⁺ b⁺ pr⁺

F1 offsprings: vg⁺ b⁺ pr⁺ /vg b pr (phenotype: wild type)

Testcross: vg⁺ b⁺ pr⁺ /vg b pr X vg b pr/ vg b pr

In the test cross results, we can observe that majority of the offsprings have character combinations that is similar to one of the parents. This is because the genes are linked, within a chromosome, and therefore they are not assorting independently in their offspring. However, we can also observe progenies with a combination of characters not shown by any of the parents. This is due to recombination. Recombination is essentially an exchange of genetic information among homologous chromosomes by means of crossing over in the process of gamete production through meiosis. Progenies that inherited recombinant gametes are less in number signifying that a recombination is a rare event in case of linked genes. However, their frequency of particular genotypes also signifies how far or close they are for crossing over is hard when they are very close. Two genes with greater distance will have greater chance of recombination, and hence greater frequency of that type of recombinant progenies will be observed. In the table below the two progeny combinations with the lowest frequencies are double crossover types due to their drastically low chance of occurrence compared to single crossover between two gens at a time.

genotype	reorganized genotype after finding the middle gene	number of this type	crossover
vg b pr	vg pr b	1779	parental type1
vg⁺ b⁺ pr⁺	vg⁺ pr⁺ b⁺	1654	parental type2
vg⁺ b pr	vg⁺ pr b	252	single crossover (SCO)between vg and pr
vg b⁺ pr⁺	vg pr⁺ b⁺	241	single crossover (SCO)between vg and pr
vg⁺ b pr⁺	vg⁺ pr⁺ b	131	single crossover (SCO) between pr and b
vg b⁺ pr	vg pr b⁺	118	single crossover (SCO) between pr and b
vg b pr⁺	vg pr⁺ b	13	double cross over 1
vg⁺ b⁺ pr	vg⁺ pr b⁺	9	double cross over2

a. Finding the middle gene.

The key here is to identify the double cross over (DCO) and parental types. Its already discussed that the progenies with parental combination will have the highest frequencies ( owing to linkage) and progeny combination with the lowest frequency are the double crossover recombinants. Also, realize that double crossover places the middle gene between two parental genes. So by comparing the genotype of DCO , the middle gene will be found. You can see that if we switch the form of pr gene in DCO1,we get the genotype of the parental type 1. The same thing occurs when we compare DCO2 with parental type 2. So, by comparing, we can see that pr is the middle gene. Hence, the linear order of the gene will be vg pr b .

Following is a representation:

b. Calculating the recombination frequencies.

Noe that we know the middle gene we write the genotype by re-organizing it for every type of progeny. Comparing it with the parental genotype, we can then determine which genes are involved in the particular single crossover. It is worked above , in the table.

Recombination frequency between vg and pr = [(Total SCO for vg and pr) + (total DCO]/(total progeny)

= [(252+241) + (13+9)]/4197=0.123

Recombination frequency between pr and b = [(Total SCO for pr and b) + (total DCO]/(total progeny)

= [(131+118) + (13+9)]/4197=0.064

c. The map

We already know that the middle gene is pr. Now, the question of how far pr is from vg, and how far b is from pr. This can be estimated by recombination frequency. Recall that we first discussed that the distant two genes are more likely to recombine. In other words, recombination frequency is directly proportional to the distance between the gene. Indeed recombination frequency (RF) is the basis of map distance measured in centimorgan(cM). 1 cM= RF*100.

So, map distance between vg and pr = 0.123*100= 1.23 cM.

and map distance between pr and b -0.064*100=0.64 cM

This is represented in a map below:

gladiator answered 1 year ago

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