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I was asked this question on a test and got it wrong, but I'd like to know how to do it. The answers are shown in the blanks below:

You are studying two recessive mutations in the fruit fly D. melanogaster. The red_ mutation causes flies to have red bristles (wild-type flies have black bristles) and the shiny- mutation causes flies to have shiny eyes (wild-type flies have eyes that are pebbly). You mate females from a true-breeding strain with red bristles and pebbly (wild-type) eyes to males from a true-breeding strain with black (wild-type) bristles and shiny eyes. F1 females are then mated to males that have red bristles and shiny eyes to produce F2 progeny.

If you analyzed 1000 male progeny in the F2 generation, how many flies of each possible phenotypic class would you expect, if:

The two traits are determined by two autosomal genes that are 15 cM apart:

Red bristles, pebbly eyes:_____ 425_______
Red bristles, shiny eyes: ______75____ 
Black bristles, pebbly eyes:____ 75_____ 
Black bristles, shiny eyes:____ 425______

My attempt:

So I know the 15cm=recombinants/parentals * 100 but I don't see how that formula can help.

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1 Answer

First, the recombination equation is:

cM = recombinants/(recombinants AND parentals) * 100

Let's assign the following genotypes for clarity's sake:

RR = black bristles
Rr = black bristles
rr = red bristles
SS = pebbly eyes
Ss = pebbly eyes
ss = shiny eyes

F1 cross: rrSS x RRss = RrSs (all progeny)
F2 cross: RrSs (F1) x rrss = ?

You would have the following gametes:

Parent 1, RrSs = RS, Rs, rS, rs
Parent 2, rrss = rs, rs, rs, rs

Draw your punnet square and you get the following genotypes of the offspring.

So your GENOTYPES of your F2 cross will be RrSs, Rrss, rrSs, rrss

Let's translate those to phenotypes.

RrSs: black pebbly (parental)
Rrss: black shiny (recombinant)
rrSs: red pebbly (recombinant)
rrss: red shiny (parental)

Now let's look at recombination frequency.

15cM = (recombinants/(recombinants+parentals)*100
recombinants/(parentals+recombinants) = 15/100

This tells you for every 15 recombinants, you have 85 parentals. If you scale this up to 1000, that's 150 recombinants for every 850 parentals.

Your 150 recombinants comes from 75 black/shiny and 75 red/pebbly.

Your 850 parentals ceoms from 425 black/pebbly and 425 red/shiny.

Anyways that was my attempt which appears to be directly opposite to what your answer key says. Did you copy it down correctly?

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