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23. In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes (v+). All three recessive mutations are Xlinked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross. v.locus ct.locus s.locus num.offspring V ct S 442.00 + ct S 9.00 + + S 62.00 + + + 441.00 + ct + 129.00 V ct + 63.00 V + S 132.00 V + + 8.00 What the interference value shown by this cross?

          23. In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct+), sable
body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes
(v+). All three recessive mutations are Xlinked. A female fly with cut wings, sable body,
and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes.
The F1 females produced by this cross were mated with cut, sable, vermilion males in
a testcross. The following are the progeny resulting from the testcross.
v.locus ct.locus s.locus num.offspring
V ct S 442.00
+ ct S 9.00
+ + S 62.00
+ + + 441.00
+ ct + 129.00
V ct + 63.00
V + S 132.00
V + + 8.00
What the interference value shown by this cross?

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23. In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct+), sable
body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes
(v+). All three recessive mutations are Xlinked. A female fly with cut wings, sable body,
and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes.
The F1 females produced by this cross were mated with cut, sable, vermilion males in
a testcross. The following are the progeny resulting from the testcross.
v.locus ct.locus s.locus num.offspring
V ct S 442.00
+ ct S 9.00
+ + S 62.00
+ + + 441.00
+ ct + 129.00
V ct + 63.00
V + S 132.00
V + + 8.00
What the interference value shown by this cross?

Added by Salvador B.

Biology for AP Courses

Julianne Zedalis, John Eggebrecht

Instant Answer

Solved by Expert Sri K

04/19/2022

Step 1

Step 1: The interference value is calculated as 1 - (observed double crossovers / expected double crossovers). Show more…

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23. In *Drosophila melanogaster*, cut wings (ct) is recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes (v+). All three recessive mutations are Xlinked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross. v.locus ct.locus s.locus num.offspring V ct S 442.00 + ct S 9.00 + + S 62.00 + + + 441.00 + ct + 129.00 V ct + 63.00 V + S 132.00 V + + 8.00 What the interference value shown by this cross?

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Transcript

0:00 Hello everyone.

00:01 So for this type of question, there is one trick that is the genotypes that have the highest number will be the parents always.

00:12 So this is parent and this is also the parent.

00:17 And the ones with the lowest number will always be the recombinence.

00:22 This is also recombinant.

00:24 This is also recombinant.

00:26 So what we have, we have, we plus c t and s that is 510 and we have vc t plus s plus that is 497 so these are the parents and similarly we have v c t s equals to 2 and v plus c t plus and s plus that is equal to 3 as the three so how we determine the order of the gene is by comparing the genotype or the order of the parents with that of the recombinants.

01:07 So if you see here, cts, cts, cts, cts, it is normal and here ct plus s plus, this is normal and here also ct plus s plus is normal.

01:20 So only the v and v plus is the one that is different in both the cases.

01:25 So that gives us v as the middle gene and here we have ct and here we have s.

01:33 So we have v gene in the middle.

01:36 So this is the order of the genes.

01:38 For determining the distance we have to first draw the single cross over 1.

01:44 So how we do that? first we'll write the parent's genotype in the order that we have just founded.

01:52 That is v gene in the middle.

01:53 And similarly you will write the genotype of the second parent as well.

02:00 So for the single crossover, that means the crossover that is occurring between the c t and the v gene...

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