3. A fruit fly of genotype ara bb (parent 1) is crossed to another fruit fly of
ID: 215360 • Letter: 3
Question
3. A fruit fly of genotype ara bb (parent 1) is crossed to another fruit fly of genotype aa bb (parent 2). The progeny of this cross were: Genot a abb Number of individuals 38 37 12 13 a a bb aa b b wrte the genotype of the parents to ndicate the linkage phase of the genes on the chromosomes. (2 points) B. Calculate the distance between the two loci. (2 points) C. Give the expected genotype of the gametes produced by parent 1 and in what proportions? (8 points) D. Use a ?' analysis to test the hypothesis that the two genes assort independently. (10 points)Explanation / Answer
A.
Out of four offspring two genotypes with high freequenices are belongs to parental categoires two with low freequnces belongs to recombinant types. Here, Out of four genotypes, the categories with 38 and 37 belongs with parental categories and 12 and 13 belong to recombinant categories.
On the basis of the parents categories, the genes arrangement is cis arrangement.
a+ b+/a b.
B.
The Distance between genes = No. of. recomnants/ Total no. of progeny X 100.
= 25/100 x 100 = 25.
C.
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 = 25%
% of parental categories = 100 - 25 = 75.
As each category contains two genotypes, the % of individual genotypes are
% of individual parental categories = 75/ 2 = 37.5% or 37.5/100
% of individual recombinant categories. 25/ 2 = 12.5% or 12.5/100.
So, the % of genotypes from this cross =
Parenal categories =
a+ b+/ab = 37.5% or 37.5/100
ab/ab = 37.5% or 37.5/100
Recombinant categories =
a+ b/ab = 12.5% or 12.5/100.
a b+/ab = 12.5% or 12.5/100.
D. Null hypothesis: The obseved values are not deviating from the expected values.
Test static:
Category
a+ a b+ b
a a b b
a+ b a b
a b+ a b
Total
Observed values (O)
38
37
12
13
100
Exptected Ratio (ER)
1
1
1
1
4
Exprected Values (E)
25
25
25
25
Deviation (O-E)
13
12
-13
-12
D^2
169
144
169
144
D^2/E
6.76
5.76
6.76
5.76
25.04
X^2
25.04
Degrees of freedom
4
-
1
3
Inference: As the calculated chisquare value i.e. 25.04 is greater than the table value i.e. 7.82 at 3DF and 0.05 probability, the null hypothesis is rejected.
Category
a+ a b+ b
a a b b
a+ b a b
a b+ a b
Total
Observed values (O)
38
37
12
13
100
Exptected Ratio (ER)
1
1
1
1
4
Exprected Values (E)
25
25
25
25
Deviation (O-E)
13
12
-13
-12
D^2
169
144
169
144
D^2/E
6.76
5.76
6.76
5.76
25.04
X^2
25.04
Degrees of freedom
4
-
1
3
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