Please answer this question in full, with explaination. Thanks Do only three gen

Question

Please answer this question in full, with explaination. Thanks

Do only three genes. Make up a three-point mapping problem using the following map as a basis. Pick three letters not adjacent (e.g. a, c and g, not a, b, c). Write this as a question with background on the parental generation and the phenotypes (freckled, not ""). Someone should be able to take what you have written and solve the problem. Solve the problem on the back of the paper. You may work person has to have theiro combinations of intervals so I shouldn't see repeats. b cd 9 86 111 27 12 4

Explanation / Answer

For framing question we will work with 3 genes- c, f & g

Q. In D. melanogaster cherub wings (c), black body (g) & cinnabar eyes (f) results from recessive alleles. A homozygous wild type fly was mated with this cherub wings (c), black body (g) & cinnabar eyes fly. The resulting F1 females are test-crossed cherub wings (c), black body (g) & cinnabar eyes male fly. The following progeny were produced-

a. Determine the linear order of gene on the chromosome.

b. Calculate recombinant distances between the three loci.

Answer:

a) At this point we don't know the gene order. So, we arbitrarily written it as cgf.

Now, non-recombinants= most frequent phenotypes. So, c+g+f+ & cgf must be non-recombinant.

Now, double cross-over= least frequent phenotypes. So, c+g+f & cgf+ must be double cross-over.

We can find the gene order by comparing alleles present in non-recombinant & double cross-over. It should be same in two locus & will differ in one locus. Comparing double cross-over progeny c+g+f+ & c+g+f we find that both have c+ & g+ but differ in cinnabar eyes (f) allele. So, f allele must be present in the middle. So, correct gene order is c f g.

b) Now, we write the phenotypes with correct gene order. We already find the non-recombinant & double cross-over progeny. So, the other 4 progeny must be single cross over type. To find the place of single cross-over we again compare the non-recombinant & single cross-over phenotypes. Crossing-over should take place in allele where they switch from their non-recombinant counterpart. Place of crossing over is shown by '/'.

So, c-f recombination frequency= [(52+50+6+4)/800] x 100% = (112/800) x 100% = 14%

So, f-g recombination frequency= [(22+24+6+4)/800] x 100% = (56/800) x 100% = 7%

As, 1% recombination frequency= 1 map unit (m.u)

So, from our solution we can check the given map (in the question) & find that it's matching with our answers. As distance between c & f is (1+11+2) or 14 m.u & distance between f & g is 7 m.u.

c+ g+ f+ 324 c+ g f 52 c g f 318 c g+ f+ 50 c+ g+ f 6 c g+ f 24 c g f+ 4 c+ g f+ 22 Total 800

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