Question

Please answer these questions

1: You conduct an experiment to test inheritance of a trait in pea plants.  You conduct a monohybrid cross with two herterozygous individuals.  From this cross you would expect 25% homozygous dominant, 25% homozygous recessive and 50% heterozygous.  What is the probability that a randomly selected individual will be heterozygous?

2: In the experiment from question 1 what is the probability that 2 individuals out of 10 randomly selected will be heterozygous?

3: In the experiment from question 1 what is the probability that 2 of 10 randomly selected individuals will be homozygous recessive?

4: What is the expected value for the mean number of heterozygous individuals our of 10 total individuals?

5: What is the expected number of homozygous individuals of both types combined in 10 total individuals?

please show detail as well

Answer 1

Let Rr*Rr be the monohybrid cross then RR, Rr, Rr, rr will be the outcomes.

1.

P(A randomly selected individual will be heterozygous) = 2/4

= 0.5

2.

This can be modelled using binomial distribution where p = 0.5 and n = 10. Let x be the random variable then,

P(X=x) = (nCx)*p^x*(1-p)^n-x

So,

P(X = 2) = 0.04395

3.

P(Homozygous recessive) = 0.25 = p

n = 10

Using the above formula of binomial distribution,

P(X = 2) = 0.28157

4.

n = 10

p = 0.5

Expected value = np

= 5

5.

P(Homozygous individual) = 0.5 = p

n = 10

Expected value = np

= 5

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