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iogann1982 [59]
3 years ago
14

One of your classmates submits the post below which is a very informative post about criteria for safe food production during th

e manufacturing process. What is missing from this post? "I recently read that industrial food processing factories are required to undergo periodic inspections by the United States Department of Agriculture as well as by the Food and Drug Administration. For the USDA, these inspections are conducted by staff in the Food Safety and Inspection Service, or FSIS."A. Did not refer to the source of the information in the body of the postB. Did not state what s/he specifically learned from the sourceC. Did not provide a citation for the source of the information
Biology
1 answer:
Oksi-84 [34.3K]3 years ago
5 0

Answer:

A. Did not refer to the source of the information in the body of the post

Explanation:

In the post written by his friend, we can see that he decided to write information about the periodic inspections carried out by the United States Department of Agriculture as well as by the Food and Drug Administration, because he read about these inspections somewhere. However, he forgot to show this place where he read this information, that is, he forgot to present the source where he got the information presented in the post. By presenting the source, it would make the post more trustworthy and relevant.

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

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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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