The drawing below illustrates a small portion of the molecules that make up a cell membrane. What is the purpose of the phosphol
1 answer:
You might be interested in
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
Estimated distances are measured in centimorgans (cM).