In mice, the gene that encodes the enzyme inosine triphosphatase is 12 mu from t
ID: 57949 • Letter: I
Question
In mice, the gene that encodes the enzyme inosine triphosphatase
is 12 mu from the gene that encodes the enzyme ornithine
decarboxylase. Let's suppose you have identified a strain of mice
homo zygous for a defective inosine triphosphatase gene that
does not produce any of this enzyme and is also homozygous
for a defective ornithine decarboxylase gene. In other words,
this strain of mice cannot make either enzyme. You crossed this
homozygous recessive strain to a normal strain of mice to produce
heterozygotes. The heterozygotes were then backcrossed to the
strain that cannot produce either enzyme. What is the probability
of obtaining a mouse that cannot make either enzyme?
Explanation / Answer
Gene A: inosine triphosphatase
GeneB: ornithine decarboxylase
Homozygous recessive strain aabb X normal strain (AABB)
F1 progeny heterozygous AaBb
Back Cross
AaBb X aabb (strain doesn’t produce either enzyme)
ab
AB
AaBb
Ab
Aabb
aB
aaBb
ab
aabb
In the backcross, the two parental types would be the homozygotes (aabb) that cannot make either enzyme, and the heterozygotes (AaBb) that can make both enzymes. The recombinants would make one enzyme but not both. Because the two genes are 12 mu apart, 12% would be recombinants and 88% would be parental types. Because there are two parental types are produced in equal numbers.
We would expect 44% of the mice to be unable to make either enzyme.
Note: Distance between the genes= % of recombinants
ab
AB
AaBb
Ab
Aabb
aB
aaBb
ab
aabb
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