"consider the following cross:
parent 1: yytt
parent 2: yytt
using the rules of probability, determine the probability that the offspring will show yytt genotype?

a.1/8
b.1/16
c.1/2"

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

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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explanation:

answer: - b cells

explanation: -

so, the b cells mediate their action by identifying the antigens present on the surface of the pathogens and then producing antibodies in return to mediate its effect.