Six recessive mutant lines (A-F) of mouse have been identififed showedabmormal t
ID: 63581 • Letter: S
Question
Six recessive mutant lines (A-F) of mouse have been identififed showedabmormal tail. These lines were intercrossed, and the phenotypes of the F1offspring are shown below. + indicates normal tail development, - indicatesabnormal tail development.
A
B
C
D
E
F
A
-
+
+
-
+
+
B
-
-
+
-
+
C
-
+
-
+
D
-
+
+
E
-
+
F
-
a. What is the name of this genetic test?
b. How many genes were mutated in creating these 6 mutant lines of mice?
c. In which lines of mice has the same gene been mutated?
d. By intercrossing F1 individuals that are double-heterozygous for mutationsin two separate genes (for example, AaBb x AaBb mice), what proportions ofwild type and mutant progeny would you expect to find if the two genes areunlinked? 20 cM apart? Draw Punnet squares to help you
A
B
C
D
E
F
A
-
+
+
-
+
+
B
-
-
+
-
+
C
-
+
-
+
D
-
+
+
E
-
+
F
-
Explanation / Answer
a. The genetic test is called Complementation testing. This test allows to know the number of different mutations, and if the mutations lie in the same gene or different genes.
b. Three genes were mutated.
c. (A,D) have the same gene mutated
(B,C, and E) have the same gene mutated
(F) has a different gene mutated.
d. Since the two genes are 20 cM apart, there is 20% recombination between the gamete. If the parents are double heterozygotes, they are expected to produce the followign gametes:
NOn-recombinants: AB, ab
Recombinants : Ab, aB
In the progeny, 70% will be wildtype and 30% will be recombinants.
The cross between the gametes AB and ab gives wild:mutants in the ratio of 3:4.
Similarly, a cross between the gametes Ab and aB gives the ratio of 1:1.
Since the gametes are 20 cM apart, non recombinant gametes will produce 3/4 X 80 = 60 wild type offspring and 20 mutant offspring.
The recombinant gametes will produce 10 wild type and 10 mutant offspring.
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