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2 years ago

Below is a segment of a normal DNA sequence:

1 answer:

4 0

This would be an example of ____________.
A) A deletion mutation.

Notice how the sequence continues after loss of the C in the middle. It will also shift the reading frame when translating, thus a frame-shift mutation.

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2 years ago

What type of offspring are produced by asexual reproduction

pickupchik [31]

Answer:

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Explanation:

hope the answer was helpful....

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2 years ago

Read 2 more answers

A genetically heterogeneous population of rice has a mean in the number of days to maturation of 30. Selection for decreased per

chubhunter [2.5K]

Answer:

h² = 0.6

Explanation:

Before answering the question, we need to know a few concepts.

Artificial selection is the selecting practice of a specific group of organisms in a population -that carry the traits of interest- to be the parents of the following generations.

Parental individuals carrying phenotypic values of interest are selected from the whole population. These parents interbreed, and a new generation is produced.

The selection differential, SD, is the difference between the mean value of the trait in the population (X₀) and the mean value of the parents, (Xs). So,

SD = X₀ - Xs

Heritability in the strict sense, h², is the genetic component measure to which additive genetic variance contributes. The heritability might be used to determine how the population will respond to the selection done, R.

h² = R/SD

The response to selection (R) refers to the metric value gained from the cross between the selected parents. R can be calculated by multiplying the heritability h², with the selection differential, SD.

R = h²SD

R also equals the difference between the new generation phenotypic value (X₁) and the original population phenotypic value (X₀),

R = X₀ - X₁

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Now that we know these concepts and how to calculate them, we can solve the proposed problem.

Available data:

You are selecting rice´s decreased period of maturation.
The population of rice has a mean maturation time of 30 days → X₀
Parental selected average period to maturation is 25 days → Xs
F1 plants mature on average in 27 days → X₁
N arrow sense Heritability → h²

According to what we sow previously, we need to find out the value of h².

We know that h² = R/SD, so we need to get R and SD first.

R = X₁ - X₀

R = 27 - 30

R = -3

SD = Xs - X₀

SD = 25 - 30

SD = -5

Knowing this, we can calculate h²

h² = R/SD

h² = -3/-5

h² = 0.6

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2 years ago