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EleoNora [17]
2 years ago
11

A mutation within a gene that will insert a premature stop codon in mrna would _______

Biology
1 answer:
AveGali [126]2 years ago
6 0

A mutation within a gene that will insert a untimely cease codon in mRNA would result in a shortened polypeptide chain.

<h3>What occurs if there is a untimely end codon?</h3>

Thus, nonsense mutations occur when a premature nonsense or end codon is added in the DNA sequence. When the mutated sequence is translated into a protein, the resulting protein is incomplete and shorter than normal. Consequently, most nonsense mutations result in nonfunctional proteins

<h3>What mutation motives untimely cease codon?</h3>

In genetics, a nonsense mutation is a factor mutation in a sequence of DNA that effects in a premature stop codon, or a nonsense codon in the transcribed mRNA, and in a truncated, incomplete, and normally nonfunctional protein product.

Learn more about mutation here:

<h3>brainly.com/question/17031191</h3><h3 /><h3>#SPJ4</h3>

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This is known as parasitism. 
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Explain why an individual that is heterozygous for this inversion would be partially sterile
Ipatiy [6.2K]
<span>Heterozygotes for inversion have a serious problem chromosome pairing at meiosis and recombination within the characteristic loop leads to chromosome duplication, deficiencies and in some cases two centro meters after recombination in meiosis.The abnormalities are usually not recovered in next generation because gametes or zygotes are receiving them are in-viable.</span>
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3 years ago
small group of 100 people decide to isolate themselves from the world and move to a small and remote deserted island. Out of thi
irina [24]

Answer:

Frequency of p = 0.684

Frequency of p = 0.316

Number of individuals with homozygous dominant (AA) = 47

Number of individuals with heterozygous (Aa)= 43

Number of individuals with homozygous recessive (aa) = 10

Explanation:

Out of 100 people, 10 have albino skin (aa)

So, the frequency of homozygous recessive individuals (q^{2}) is \frac{10}{100} = 0.1

Now, q will be

= \sqrt{q^{2} } = \sqrt{0.1} \\= 0.316

As per Hardy Weinberg's equation -

p + q = 1

Substituting the value of q in above equation, we get -

p + 0.316 = 1p = 1 -0.316\\p = 0.684

Now the frequency of homozygous dominant (AA) will be

p^{2} = 0.684^{2} \\= 0.467

Hence, out of 100 people 0.467 * 100 = 46.7 or 47 people are homozygous dominant (AA)

Like wise out of 100 people 0.1 * 100 = 10 people are homozygous recessive (aa)

As per As per Hardy Weinberg's equation-

p^{2} + q^{2} + 2pq = 1\\

Substituting the values in above equation, we get -

0.467 + 0.316 + 2pq = 1\\2pq = 1 -( 0.467+ 0.1)\\2pq = 0.433

So, out of 100 people 0.433 * 100 = 43.3 or 43 people are heterozygous (Aa)

7 0
4 years ago
Estimates of current rates of extinction
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Estimates of current rates of extinction indicate that rates may be greater than extinction rates seen in the fossil records.

<h3>What is extinction of species?</h3>

Extinction of species refers to the event in which species of organisms are entirely wiped from existence either by man-made causes, natural disasters or disease.

The recent industrial age and many technological advancements by man has resulted in rapid extinction of species .

Estimates of current rates of extinction indicate that rates may be greater than extinction rates seen in the fossil records.

Learn more about extinction at: brainly.com/question/1048615

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