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atroni [7]
3 years ago
13

. The mean grain yield of an open pollinated rye breeding population is determined to be 72 bushels/acre. Your goal is to improv

e grain yield in this population. The mean grain yield of selected individuals from this population is 93 bushels/acre. Selected individuals are intermated to form a new population. The new population mean is determined to be 84 bushels/acre. With this information, calculate the selection intensity, response to selection and heritability of grain yield. (3) Report your answers for A and B as whole numbers and your answer for C to 2 decimal places.
Biology
1 answer:
Free_Kalibri [48]3 years ago
4 0

Answer:

a) the selection intensity is 21bushels/acre

b) the response to selection is 12bushels/acre

c) The heritability is 0.57 (two decimal places)

Explanation:

Given:

mean grain yield population=72bushels/acre

mean grain yield selected individuals=93bushels/acre

mean grain yield new population=84busgels/acre

a) The selection intensity is given by:

SI=mean grain yield selected individuals-mean grain yield population=93-72=21bushels/acre

b) The response to selection is:

RS=mean grain yield new population-mean grain yield population=84-72=12bushles/acre

c) The heritability of grain yield is:

H=RS/SI=12/21=0.57

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8 0
4 years ago
I WILL GIVE BRAINLIEST!!!
Nesterboy [21]

Answer:

25%

Explanation:

When looking at a pedigree remember that:

  • squares are males
  • circles are females
  • the solid colored figure represents an individual affected by a disease
  • the empty figure represents a healthy individual

Let us assign the symbol X⁺ to represent the dominant allele linked to the X-chromosome and expressing healthiness, and X⁻ to represent the recessive allele expressing the dissease.

According to this pedigree

  • I1 is a man affected by the disease, YX⁻
  • I2 is a healthy woman X⁺X⁻
  • we can see that among the progeny (generation II) there are two individuals affected (a boy and a girl) and one healthy girl. This means that the mother I2 is heterozygous for the trait.

So, having their genotypes we can know what are the probabilities of getting a son with DMD

Parentals)    YX⁻     x     X⁺X⁻  

Gametes)   Y     X⁻      X⁺     X⁻

Punnett square)

                        X⁺             X⁻

            X⁻      X⁺X⁻         X⁻X⁻    

            Y        X⁺Y           X⁻Y

F1)

  • The probabilities of getting a healthy daughter X⁺X⁻ are 25%
  • The probabilities of getting a healthy son X⁺Y are 25%
  • The probabilities of getting a daughter with DMD X⁻X⁻ are 25%
  • The probabilities of getting a son with DMD X⁻Y are 25%
7 0
3 years ago
Read 2 more answers
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