True or False.

What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent as

ASHA 777 [7]

Answer:

(a) AABBCC × aabbcc->AaBbCc → 1

(b) AABbCc × AaBbCc->AAbbCC → 1/32

(c) AaBbCc × AaBbCc->AaBbCc → 1/8

(d) aaBbCC × AABbcc->AaBbCc → 1/2

Explanation:

In such cases we calculate the probability of each allelic pair separately.

(a) AABBCC × aabbcc -> AaBbCc

Aa Aa Aa Aa Bb Bb Bb Bb Cc Cc Cc Cc

↓ ↓ ↓

4/4 = 1 4/4 = 1 4/4 = 1

In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.

In case of gene B, out of the 4 probable allelic combinations, 4 will be Bb so the probability is 4/4 = 1.

In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.

So, the total probability of getting AaBbCc = 1 x 1 x 1 = 1.

(b) AABbCc × AaBbCc -> AA bb CC

AA AA Aa Aa BB Bb Bb bb CC Cc Cc cc

↓ ↓ ↓

2/4 1/4 1/4

In case of gene A, out of the 4 probable allelic combinations, 2 will be AA so the probability is 2/4.

In case of gene B, out of the 4 probable allelic combinations, 1 will be bb so the probability is 1/4.

In case of gene C, out of the 4 probable allelic combinations, 1 will be CC so the probability is 1/4.

So, the total probability of getting AA bb CC = 2/4 x 1/4 x 1/4 = 1/32.

(c) AaBbCc × AaBbCc -> AaBbCc

AA Aa Aa aa BB Bb Bb bb CC Cc Cc cc

↓ ↓ ↓

2/4 2/4 2/4

In case of gene A, out of the 4 probable allelic combinations, 2 will be Aa so the probability is 2/4.

In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.

In case of gene C, out of the 4 probable allelic combinations, 2 will be Cc so the probability is 2/4.

So, the total probability of getting AaBbCc = 2/4 x 2/4 x 2/4 = 1/8.

(d) aaBbCC × AABbcc->AaBbCc

Aa Aa Aa Aa BB Bb Bb bb Cc Cc Cc Cc

↓ ↓ ↓

4/4 = 1 2/4 4/4 = 1

In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.

In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.

In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.

So, the total probability of getting AaBbCc = 1 x 2/4 x 1 = 1/2.

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jeyben [28]

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Acute exposure is radiation exposure that occurs in a short time period. It can be an exposure that occurs once in your lifetime. This might be that one foot x ray you had because you may have broken your foot. Acute exposure can also be exposures that occur more than once, but very infrequently and for short periods of time. This might be dental x rays you get every few years.</span>

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Following a bone fracture, the body must repair bone tissue by producing new bone cells. What is the source of these new cells?

LiRa [457]

Answer:

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Signed by, Virtuoso Sargedog

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