"Consider the following cross:
1 answer:
The answer is <span>a.1/8</span>
<span>This is
an example of Mendelian dihybrid cross (a cross between two different traits).
If:
Y - (dominant) allele for the first trait
r - (recessive) allele for the first trait
S - (dominant) allele for the second trait
s - (recessive) allele for for the second trait
</span>The cross will be like:
Parents: YyTt x YyTt
As shown in image, there will be in total 16 different genotypes. Two of them are YyTT genotypes. So the probability that the offspring will show YyTT genotype is 2 of 16:
= 2/16 = 1/8
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