- The genes for ruby eyes (rb), tan body (t) and cut wings (ct) are all found on the X-chromosome of Drosophila melanogaster. All of these are recessive traits. They map in the order rb, ct, t with 12.5 map units between rb and ct and 7.5 map units between ct and t. Suppose you cross a cut wing male with a homozygous female that is both tan and has ruby eyes.
- What will the F1 females look like?
- Draw map of the section of the X chromosomes that has these 3 genes for the F1 females
- Assume you testcross your F1 females.
- What progeny classes would you expect? ii. Give approximate numbers for each class based on a total of 2000 progeny.
- Assuming the i=1 and there are no double crossovers.
- Assuming the i=0 and there are the expected number of double crossovers.
i need help finding the answer for this question
There are lots of things to consider when answering this question.Firstly all of the traits are recessive. Males = XY Females =XXHomozyous means both chromosomes have the same alleles (form of a gene)I would start by mapping out the traits like this:rb<---12.5----->ct<----7.5----->tWe start by crossing a ct Male (who therefore has the chromosomes XY) with the female (XX) who is homogenous.This means the male only has the ct trait on his X chromosome and the female has the recessive versions of rb and t on each X chromosome.To cross them make a grid like on the whiteboard. The F1 female will inherit one X chromosome from the male and one from the female. So on one X chromosome will have the ct and the other will have rb and t. However because all the traits are recessive and feature on only one of the two chromosomes none of these traits will actually display themselves and so the F1 female will normal. Or wont have cut wings, ruby eyes or tan body.
1b) ......ct.....rb<-----20...>t1c) cut wing 500, ruby eyed and tan body 500, normal 1000
To answer the last part you need to know the expected number of double crossovers which you normally calculate using observed data e.g. 498 flies had ruby eyes