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In mouse, genes X, Y and Z are on chromosome 2. The map of genes X, Y and Z is: X----------------Y-----------------Z 20 m.u. 20 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

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raduoprea160

Answer:

28%

i hope i'm right

Explanation:

In F1 we will get 100% XxYyZz

each of these organism has two chromosomes that look like this

X..........20..........Y...............30...............Z

x..........20..........y...............30...............z

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%

batolisis

The proportion of its gametes that would have xyz genotype = 28%

Given that F1 progeny =  XxYyZz i.e. 100%

From the map of genes there is a 20% chance of  x and y recombining and a 30% chance of y and z recombining and the aim of this question is to determine the chromosome where recombination does not occur and the possible gametes are :

XYZ, XYz, XyZ, Xyz, xYZ, xYz, xyZ, xyz

since; there is 20% of x and y recombining ; Non recombining % = 80%

there is also 30% chance of y and z recombining ; Non recombing% = 70%

Hence the percentage of the genes not recombining = 80% * 70% = 56%'

but since there are two(2) non- recombining gametes

∴ proportion of xyz genotype = 56 / 2 = 28%

learn more : https://brainly.com/question/15165836

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