To maintain order within their cells and organs, all living things must

The following F2 results occur from a dihybrid cross (AaBb x AaBb): purple: A_B_ 9/16 white: aaB_ 3/16 white: A_bb 3/16 white: a

e-lub [12.9K]

Answer:

b. 3 (whilte) : 1(purple)

Explanation:

The first dihybrid crossing is between AaBb x AaBb

Now, if the double heterozygous traits self crossed, we have the following gametes shown below for the F₂ crossing.

AaBb = (AB, Ab, aB, ab)

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

We were being told that the results are;

purple: A_B_ 9/16

white: aaB_ 3/16

white: A_bb 3/16

white: aabb 1/16

From above, we can see that the same is true;

For Purple color; we Have (AABB and AaBb) since A is dominant to a and B is dominant to b.

∴ From the above punnet square; we have 9 purple colors which are:

(AABB, AABb, AaBB, AaBb, AABb, AaBb, AaBB, AaBb, AaBb) = 9/16

white: aaB_ ( since B is dominant to b, it can be either BB or Bb)

= (aaBB, aaBb, aaBb) = 3/16

white: A_bb ( since A is dominant to a, it can be either AA or Aa)

= (AAbb, Aabb, Aabb) = 3/16

white: aabb i.e homozygous recessive = (aabb) = 1/16

Furthermore, the question goes further by saying:

If a double heterozygote (AaBb) is crossed with a fully recessive organism (aabb), what phenotypic ratio is expected in the offspring?

If AaBb self crossed, we have: (AB, Ab, aB, ab)

If aabb self crossed, we have: (ab, ab, ab, ab)

AB Ab aB ab

ab AaBb Aabb aaBb aabb

AaBb ( purple)

= 4/16

= 1/4

= 0.25

= 25%

Aabb (white)

= 4/16

= 1/4

= 0.25

= 25%

aaBb (white)

= 4/16

= 1/4

= 0.25

= 25%

aabb (white)

= 4/16

= 1/4

= 0.25

= 25%

Now the proportion of white to purple = (25%+25%+25%): 25%

= 75%:25%

= 3(white):1(purple)

We can therefore conclude that the expected phenotypic ratio in the cross between a double heterozygote (AaBb) with a fully recessive organism (aabb) yeids;

3 (whilte) : 1(purple)

8 0

3 years ago

Which is the basic functional unit of life?

lord [1]

Answer:

It’s the cell

6 0

2 years ago

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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

What is a benefit to breeding purebreds?

Taya2010 [7]

Answer:

you get things like this

Explanation:

3 0

2 years ago

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A simplified model of protein synthesis is shown here: How might this model be revised to show a more accurate depiction of prot

Nady [450]

Answer:

The correct options are:

Prior to process 1, DNA should copy itself to allow the cell to divide.

Between processes 1 and 2, the mRNA must be modified before leaving the nucleus.

Explanation:

For eukaryotes, the mRNA is made in the nucleus from the DNA. This process in known as transcription. The process of making proteins from the mRNA takes place in the ribososmes of the cytoplasm. This process is known as translation. The mRNA has to be modified before it is transported for translation. Otherwise, the mRNA will be degraded by the enzymes of the nucleus.

7 0

2 years ago

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