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Hello everyone.
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In this question we have been given two crosses in which the individual having the dominant characters for all the genes is crossed with the recessive character and the c here is in heterozygous condition.
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Then we need to tell the probability that the offspring produce will have the heterozygous character for a and b and homozygous dominant character for capital c.
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And let's cross it individually.
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So when capital a crossed with small a small a, this will produce individuals having capital a small a, capital a small a, capital a small a, capital a small a.
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So this all the four individuals produce here have the heterozygous condition.
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So the probability of heterozygous condition for the a gene is 1.
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Now let's cross for the b gene here.
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So capital b, capital b crossed with small b, small b will produce all the heterozygous characters here.
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So we have capital b small b, capital b and small b.
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So here we have 4 divided by 4 that is 1.
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The probability one is for capital b and small b that is in the heterozygous condition...