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3 years ago

Which of the following statements is true according to Mendel's law of segregation?

Biology

2 answers:

antoniya [11.8K]3 years ago

8 0

Answer:

Explanation:

musickatia [10]3 years ago

3 0

Answer:

D. Offspring only receive one version of their parents' traits.

Explanation:

To be specific, they receive on version of each parent. Traits are segregated during gamete formation, where one version of the trait comes from one parent, and the other version of that same trait comes from the other parent. When they join during fertilization, a new organism is made through the combination of the traits. However, the trait that will be expressed, according to Mendel, would be the one that is dominant.

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2 years ago

In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant

Ivenika [448]

Answer and Explanation:

Available data:

Comb shape is determined by genes at two loci (R, r and P, p).

The walnut comb genotype is R_P_.

The rose comb genotype is R_pp.

The pea comb genotype is rrP_.

The single genotype is rrpp.

a. Walnut crossed with single produces 1 walnut, 1 rose, 1 pea, and 1 single offspring:

Parental) RrPp x rrpp

Gametes) RP Rp rP rp rp rp rp rp

Punnet Square) RP Rp rP rp

rp RrPp Rrpp rrPp rrpp

rp RrPp Rrpp rrPp rrpp

F1 phenotype: 25% walnut, 25% rose, 25% pea, and 25% single.

F1 genotype: 4/16 RrPp, 1/16 Rrpp, 4/16 rrPp, 4/16 rrpp.

b. Rose crossed with pea produces 20 walnut offspring.

Parental) RRpp x rrPP

Gametes) Rp Rp Rp Rp rP rP rP rP

Punnet Square) Rp Rp Rp Rp

rP RrPp RrPp RrPp RrPp

F1 phenotype: 100% walnut.

F1 genotype: 16/16 RrPp.

c. Pea crossed with single produces 1 single offspring.

This is not possible, because the pea genotype involves at least one dominant allele P. There are two possible crosses: rrPp x rrpp, which must produce half of the progeny pea and the other half single, or rrPP x rrpp which produce a whole pea progeny with no single offspring.

Parental) rrPp x rrpp

Gametes) rP rp rP rp rp rp rp rp

Punnet Square) rP rp rP rp

rp rrPp rrpp rrPp rrpp

rp rrPp rrpp rrPp rrpp

F1 phenotype: 50% pea, and 50% single.

F1 genotype: 8/16 rrPp, 8/16 rrpp.

d. Rose crossed with pea produces 2 walnut, 1 single, and 1 pea offspring.

This is not possible, because having one of the parents with a rose phenotype involves at least one R allele, which means that there must be rose phenotype in the progeny.

Parental) Rrpp x rrPp

Gametes) Rp Rp rp rp rP rP rp rp

Punnet Square) Rp Rp rp rp

rP RrPp RrPp rrPp rrPp

rP RrPp RrPp rrPp rrPp

rp Rrpp Rrpp rrpp rrpp

rp Rrpp Rrpp rrpp rrpp

F1 phenotype: 25% walnut, 25% rose, 25% pea, and 25% single.

F1 genotype: 4/16 RrPp, 1/16 Rrpp, 4/16 rrPp, 4/16 rrpp.

e. Rose crossed with single produces 31 rose offspring.

Parental) RRpp x rrpp

Gametes) Rp Rp Rp Rp rp rp rp rp

Punnet Square) Rp Rp Rp Rp

rp Rrpp Rrpp Rrpp Rrpp

F1 phenotype: 100% rose (31 individuals equal 100% of the progeny).

F1 genotype: 16/16 Rrpp.

f. Rose crossed with single produces 10 rose and 11 single offspring.

Parental) Rrpp x rrpp

Gametes) Rp Rp rp rp rP rP rp rp

Punnet Square) Rp Rp rp rp

rp Rrpp Rrpp rrpp rrpp

F1 phenotype: 50% rose, 50% single.

F1 genotype: 8/16 Rrpp, 8/16 rrpp.

3 0

3 years ago