What term describes one organism eating another organism?

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.

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In both plant and animal cells, the cell membrane...?

prohojiy [21]

Both plant and animal cells have a nucleus and mitochondria

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Animals in phylum Arthropoda generally have these characteristics in common.

umka21 [38]

Answer:

Chitinous exoskeleton and segmented bodies.

Explanation:

Phylum Arthropoda are known as jointed legs species which include crabs, spiders, ants, etc. The important characteristic of Arthropoda is they have a hard chitinous exoskeleton that provides them protection and gives flexibility. These phylum species can survive in both terrestrial and marine habitats. Some characteristics of phylum Arthropoda are given below:

1) They have Chitinous exoskeleton which is hardened.

2) Their digestive tract is present at opposite ends of the body and complete with the mouth and anus.

3) They contain jointed limbs.

4) The body structure of these species is showing bilateral symmetry.

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Please HELP IM TIRED OF THOSE BOTS

elena-14-01-66 [18.8K]

Answer:

C

Explanation:

There are several cells, therefore it is multicellular. They also have nuclei, meaning that this organism is a multicellular eukaryote.

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Animals that chang greatly from the juvenile to the adult phase of life are said to undergo what?

Oduvanchick [21]

Puberty? I'm just taking a guess.

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