Question

Albino rabbits (lacking pigment) are homozygous for the recessive c allele (C allows pigment formation). Rabbits homozygous for the recessive b allele make brown pigment, while those with at least one copy of B make black pigment. True-breeding brown rabbits were crossed to albinos, which were BB. F1 rabbits, which were all black, were crossed to the double recessive (bb cc). The progeny obtained were 34 black, 66 brown, and 100 albino. a. What phenotypic proportions would have been expected if the b and c loci were unlinked? b. How far apart are the two loci?

Answer 1

Answer:

a. If the b and c loci were unliked the expected phenotypic proportions would be:

- 1/4 Brown

- 1/4 Black

- 1/2 Albino

b. The estimated distance between the two loci is equal to 17 centimorgans (cM).

Explanation:

a. If the loci b and c are unlinked or they are not in the same chromosome, it means that b and c are segregated independently (following Mendel's laws).  So, the first step is to state the genotypes of the rabbits.

Albino rabbits are homozygous for c (cc)

Brown rabbits are homozygous for b (bb).

Black rabbits have a copy of B (Bb or BB).

But, rabbits have the two genes (b and c), so they could be:

Albino: BBcc/Bbcc/bbcc.

Brown: bbCC/bbCc

Black: BBCC/BBCc/BbCC/BbCc

A cross was made with true-breeding brown rabbits( that means "pure" or homocygotes) bbCC and albino BBcc.

So, we have :

bbCC X BBcc

This is a dihybrid cross with only one possible genotype in F1: BbCc. -> Black rabbits.

After this, these rabbits (BbCc) were crossed with double recessive (albino) rabbits, like this:

BbCc X bbcc

To know the possible outcomes of F2, it is necessary to make a Punnett square 4x4. Following the independent assortment principle, we will have the following possible gametes:

Black Rabbits: Cb, CB, cB,cb.

Double recessive rabbits: cb, cb, cb, cb.

After doing the Punnett square, we will have 4 possible genotypes and 3 possible phenotypes in F2:

1. bbCc = Brown  

2 BbCc = Black

3. Bbcc= Albino

4. bbcc= Albino

That means:

1. 1/4 Brown

2. 1/4 Black

3. 1/2 Albino

b. To estimate the distance between the two loci is necessary to use the data provided in the question.  

In this case, we are assuming that b and c are in the same chromosome, so they don't follow Mendel's laws. The first step is to know which are the initial genotypes.

The initial cross was between a brown rabbit (bbCC ) and an albino rabbit (BBcc). If the two genes are linked, we can say that in the brown rabbit bC are always together and in the albino Bc will be together, because they're linked.

F1 will be black rabbits bCBc, (remember bC and Bc go always together).

So, after this, a new cross was made with a double recessive:

bCBc X bcbc

The possible outcomes are:

bCbc Brown

bCbc Brown

Bcbc Albino

Bcbc Albino

As you see, there are not black rabbits, so to find black rabbits, we need to find a BCbc genotype. This genotype can be produced by recombination or crossingover in the same chromosome in parentals. So, the distance between two loci is equal to the proportion of individuals with a recombinant genotype, in this case, BCbc or black rabbits.

So, we have:

34 black

66 brown

100 albino

Total: 200

D$Distance=\frac{Recombinants}{Total } *100= \frac{34}{200}*100= 17 cM$

Estimated distances are measured in centimorgans (cM).