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Andrew [12]
3 years ago
10

Which of the following statements is NOT true of mitosis?

Biology
1 answer:
svlad2 [7]3 years ago
4 0
A is the correct answer
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What are the flowers in the picture​
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5 0
3 years ago
Read 2 more answers
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.

5 0
3 years ago
Compare the organization of DNA in prokaryotic and eukaryotic cells
Naddik [55]

Answer: prokaryotic cells have no nucleus, no organelles and a small amount of DNA. Eukaryotic cells do have a nucleus, along with many organelles, and more DNA.

Explanation:

5 0
3 years ago
Why is a lumbar puncture performed between the L4 and L5 vertebrae
ddd [48]
Because that how is belonge
4 0
3 years ago
Which of the following is not a step of Natural Selection?
kati45 [8]

Answer:

im pretty sure it is underpopulation.

6 0
3 years ago
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