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Dominik [7]
2 years ago
10

AaBbCc x AaBbCc (use for all 3 questions) 1. What is the probability that this individual will: AABBcc 2. What is the probabilit

y that this individual will: AaBBcc 3. What is the probability that this individual will: AaBbCc
Mathematics
1 answer:
tankabanditka [31]2 years ago
4 0

Answer: 1. P = 1/64

             2. P = 1/32

             3. P = 1/8

Step-by-step explanation:

In genetics, an ofspring inherits one copy of an allele of each parent, which can be described in the tables below called Punnett Square:

For crossing Aa x Aa:

       A      a

A    AA   Aa

a    Aa    aa

For crossing Bb x Bb:

        B      b

B     BB    Bb

b     Bb    bb

For crossing Cc x Cc:

       C      c

C    CC    Cc

c     Cc     cc

We can separate them because they are assorted independently.

For offspring with <u>genotype</u> <u>AABBcc</u>, probability will be:

P(AA) = 1/4

P(BB) = 1/4

P(cc) = 1/4

As all three probabilities has to happen at the same time, it is a "E" rule:

P(AABBcc) = (\frac{1}{4}) (\frac{1}{4}) (\frac{1}{4})

P(AABBcc) = 1/64

Probability for the individual to be AABBcc is 1/64 or 1.56%.

<u>Genotype</u> <u>AaBBcc</u>:

P(Aa) = 2/4 = 1/2

P(BB) = 1/4

P(cc) = 1/4

P(AaBBcc) = (\frac{1}{2}) (\frac{1}{4}) (\frac{1}{4})

P(AaBBcc) = 1/32

Probability for the individual to be AaBBcc is 1/32 or 3.12%

<u>Genotype</u> <u>AaBbCc</u>:

P(Aa) = 1/2

P(Bb) = 1/2

P(Cc) = 1/2

P(AaBbCc) = (\frac{1}{2}) (\frac{1}{2}) (\frac{1}{2})

P(AaBbCc) = 1/8

Probability for the individual to be AaBbCc is 1/8 or 12.5%.

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