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Sindrei [870]
3 years ago
12

In rabbits, black fur (B) is dominant to white fur (b). A purebred black furred male is bred with a female that had the recessiv

e white fur
Biology
1 answer:
ValentinkaMS [17]3 years ago
8 0

Answer:

All the offsprings will be black-furred (Bb)

Explanation:

This question involves a single gene coding for fur length in rabbits. The allele for black fur (B) is dominant to the allele for white fur (b). This means that a rabbit heterozygous for this gene (Bb) will have a black fur.

According to this question, a purebred black furred male (BB) is bred with a female that had the recessive white fur (bb). The parents will produce gametes as follows:

BB - B only

bb - b only

Using these gametes in a punnet square (see attached image), the genotypic proportion of the produced offsprings is as follows:

Bb, Bb, Bb, Bb

All Bb (heterozygous) means that all of the offsprings will be black-furred.

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How does molecules in the air that come into the plants leaves affect the plants mass
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3 years ago
Can someone help me with a biology related question? I'm offering good points!!!!! Help me out please.
miss Akunina [59]
First, write out all combinations for the parents:
<u>SsBB</u>:
SB
sB
SB
sB

<u>ssBb</u>:
sB
sb
sB
sb

So now that you know what to put on as the x and y for the Punnett Square, set it up:

<u>     |    SB   |    sB    |   SB    |  sB     </u>
<u>sB | SsBB |  ssBB |  SsBB |  ssBB </u>
<u>sb | SsBb |  ssBb  |  SsBb |  ssBb </u>
<u>sB | SsBB |  ssBB |  SsBB |  ssBB </u>
<u>sb | SsBb |  ssBb  |  SsBb |  ssBb </u>

Now, listing each possible genotype and the number of times they show up in the square:
SSBB
0 times
SSBb/SSbB
0 times
SSbb
0 times
SsBB/sSBB
4 times
SsBb/SsbB/sSBb/sSbB/
4 times
Ssbb/sSbb
0 times
ssBB
4 times
ssBb/ssbB
4 times
ssbb
0 times

Since there are only 4 genotypes are possible, set up ratios of those four:

SsBB:
4 times/16 total = .25 = 25%

SsBb:
4 times/16 total = .25 = 25%

ssBB:
4 times/16 total = .25 = 25%

ssBb:
4 times/16 total = .25 = 25%

This means that the<em> genotypes</em> have a 25% : 25% : 25% : 25% likelihood.


Now, let's figure out the phenotypes of the 4 genotypes:
SsBB:
Ss = dominant
BB = dominant

SsBb:
Ss = dominant
BB = dominant

ssBB:
ss = recessive
BB = dominant

ssBb:
ss = recessive
Bb = dominant

So we have 2/4 that are S-dominant B-dominant and 2/4 that are S-recessive B-dominant.
This means that the phenotypes are:
2/4 = 50% Thin stripes with black eyes, and
2/4 = 50% Thin stripes with gray eyes.

I hope this helped you out :)
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Pie

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Explanation:

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