Sponges have no nerve tissue or sense organs.explain why this means that sponges have no cephalisation.

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

g Three autosomal genes are linked along the same chromosome. The distance between gene A and B is 7 mu, the distance between B

Alenkinab [10]

Answer:

a) F1 offspring will be AaBbCc

b) for part b and a explanation see the picture attached

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What lymphatic tissue in the submucosa of the small intestine prevents bacteria from entering the blood?

kolezko [41]

The answer to this question is the Pyers patches. The Pyers patches are lymphatic tissues found through the small intestines that have an important role in the immune system. Pyers patches are also known as the aggregated lymphoid nodules that monitors and prevents growth of bacteria in the intestine that is harmful in the intestines.

5 0

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Suppose you have a beaker of sugar and a beaker of salt. They are unlabeled and the white crystals look identical. Your assignme

DochEvi [55]

Answer:

I do not know

Explanation:

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John is completing an internship in biology at the Clearwater Marine Aquarium. While interning there, he monitors an outdoor phy

never [62]

Answer:A

Explanation:

It will decrease over time

7 0

3 years ago