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BONUS (5 pts) The genes A and B are on one chromosome, 10 cM apart. The genes C

ID: 207222 • Letter: B

Question

BONUS (5 pts) The genes A and B are on one chromosome, 10 cM apart. The genes C and D are on another, different chromosome, 30 cM apart. You cross a homozygous A B C D individual with an a bed individual, and then cross the resultant F, tetrahybrid with a homozygous recessive (testcross). What are the chances of getting individuals with the following phenotypes in the progeny of the testcross? 1. Hint: you need to consider (1) how genetic linkage/distance influence the proportion of gametes and (2) which rule of probability applies to segregation of these events. A. ABCD B. ABCd

Explanation / Answer

The map distance = % of recombinantion.
The total offspring (100%) = % of parental categories + % of the recombinant categories.
So, the % of parental categories = The total offspring (100%) - % of the recombinant

categories.

For AB genes,

The map distance = % of recombinantion = 10%

% of parental categories = 100 - 10 = 90.

As each category contains two genotypes, the % of individual genotypes are

% of individual parental categories = 90/ 2 = 45% or 45/100
% of individual recombinant categories. 10/ 2 = 5% or 5/100.

So, the % of genotypes from this cross =

Parenal categories =
AB/ab = 45% or 45/100
ab/ab = 45% or 45/100
Ab/ab = 5% or 5/100
aB/ab = 5%. or 5/100

For CD genes,

The map distance = % of recombinantion = 30%

% of parental categories = 100 - 30 = 70.

As each category contains two genotypes, the % of individual genotypes are

% of individual parental categories = 70/ 2 = 35% or 35/100
% of individual recombinant categories. 30/ 2 = 15% or 15/100.

So, the % of genotypes from this cross =

Parenal categories =
CD/cd = 35% or 35/100
cd/cd = 35% or 35/100
Cd/cd = 15% or 15/100
cD/cd = 15% or 15/100

The rule of probabilty applies here is multiplcation and the probabilites is

A. ABCD = 45/100 (probability of AB) * 35/100 (probability of CD) = 1575/10000

B. ABCd = 45/100 (probability of AB) * 15/100 (probability of Cd) = 675/10000