In mouse, genes X, Y and Z are on chromosome 2. The map of genes X, Y and Z is: XYZ
20 m. u. 30 m. u.
You cross an individual with genotype XXYYZZ to an individual with genotype xxyyzz, and F1 progeny are collected. Assume interference between regions (X-Y and Y-Z) is 100%. When an F1 individual produces gametes, what proportion of its gametes would have xyz genotype?
a. 10%
b. 20%
c. 25%
d. 30%
e. 40%
f. 50%
g. 0%
h. None of these above

In F1 we will get 100% XxYyZz

each of these organism has two chromosomes that look like this

X20Y30Z

x20y30z

Those two chromosomes are homologous

we want a gamete that contains the second chromosome where crossing-over didn't occur. The distance between the genes of the chromosomes in map units is equal to the frequency of recombination in percentage. Therefore there is a 20% that x and y will recombine and a 30% chance that y and z will recombine. We want a chromosome that is not recombined.

The possible gametes are

XYZ XYz XyZ Xyz xYZ xYz xyZ xyz if the genes were on different chromosomes xyz would be 1/8 or 12,5%

Since there is a 20% that the genes x and y will recombine that means there is a 80% chance that they won't and since there is a 30% chance that the genes y and z will recombine that means there is a 70% chance that they won't.

80%*70%=56% chances that the genes won't recombine. But, there are two non-recombinant gametes:XYZ and xyz therefore the chances for xyz is 56/2=28%

b because they naturally adapt to their environment plz mark brainliest

explanation:

the answers are a and c