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ziro4ka [17]
3 years ago
12

Why is a frameshift missense mutation more likely to have a severe effect on phenotype than a nucleotide-pair substitution misse

nse mutation in the same protein? Why is a frameshift missense mutation more likely to have a severe effect on phenotype than a nucleotide-pair substitution missense mutation in the same protein? A frameshift missense will cause the codons to be out of order, but a substitution missense does not change the order of the codons. A substitution missense mutation causes the protein to be shorter and thus non-functional. A substitution missense affects only one codon, but a frameshift missense affects all codons downstream of the frameshift. A frameshift missense mutation will cause an early Stop codon, but a substitution missense might be silent.
Biology
1 answer:
Serjik [45]3 years ago
4 0

A substitution missense affects only one codon, but a frameshift missense affects all codons downstream of the frameshift.

A frameshift mutation occurs during deletion of one of two nucleotides. The arrangement of codons downstream of the mutation will definitely change, hence it being called a frameshift mutation. On the other hand, a substitution mutation will only cause a change in one nucleotide of one codon.

Explanation:

A frameshift mutation causes new codons downstream of the mutation that will code for different amino acids hence changing the properties of the translated proteins. This will have great ramifications for the phenotype of the organism.

On the other hand, substitution mutation will only cause a change in the amino acid at that point of mutation during translation. This will not have a drastic change in the protein hence will not be as lethal as a frameshift mutation.

Learn More:

For more on mutations check out;

brainly.com/question/11278415

#LearnWithBrainly

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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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The right answer is hyphae for the first gap, mycelium for the second gap.

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