Question

1. if a homozygous recessive individual (rr) is crossed with a heterozygous individual (rr), what is the probability of their offspring having the dominant phenotype?

a. 75%
b. 25%
c. 50%
d. 0%

1. in a genetic cross, what does a recessive allele require to be expressed?

a. two recessive alleles
b. at least one dominant allele
c. a homozygous dominant genotype
d. a heterozygous genotype

1. in a monohybrid cross between two heterozygous parents (rr x rr), what is the probability of the offspring showing the recessive phenotype?

a. 75%
b. 25%
c. 50%
d. 0%

Explanation:

Question 13

Step1: Determine possible gametes

Homozygous recessive (rr) produces only r gametes. Heterozygous (Rr) produces R and r gametes.

Step2: Create Punnett square

	R	r
r	Rr	rr

Step3: Analyze phenotypes

Rr has dominant phenotype, rr has recessive. From the square, 2 out of 4 (50%) are Rr (dominant), 2 out of 4 (50%) are rr (recessive). So dominant phenotype probability is 50%.

A recessive allele is masked by a dominant allele. So, for a recessive allele to be expressed (show its trait), the individual must have two recessive alleles (homozygous recessive), as a dominant allele would hide it. Option a states this, while b (needs dominant, which hides recessive), c (homozygous dominant has no recessive alleles to express), and d (heterozygous has a dominant allele hiding the recessive) are incorrect.

Step1: Determine gametes for Rr parents

Each Rr parent produces R and r gametes.

Step2: Create Punnett square

	R	r
r	Rr	rr

Step3: Analyze recessive phenotype

Recessive phenotype is when genotype is rr (since R is dominant). From the square, 1 out of 4 (25%) is rr. So probability of recessive phenotype is 25%.

Answer:

c. 50%

Question 14