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Vlad1618 [11]
3 years ago
6

If you were to take physical education classes daily, how do you think your grades would be impacted? Explain why you think this

using your own ideas and specific information from the infographic.
Health
2 answers:
aleksklad [387]3 years ago
5 0

Answer:

So, some studies show that working out (PE Classes included) help relieve stress. So they can actually benefit you with your other classes homework and projects that are due in the future.

krok68 [10]3 years ago
3 0
Not so much think PE class is just another class and I use it as a study just like hang with my boys and play basketball for a hour I make my brain get loose and not think about anything
Your grades should not be impacted at all just stay on task in your real classes though
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GuDViN [60]

This question is incomplete, but here is the complete question below.

In humans , blood type is a result of multiple alleles: I^{A}, I^{B}, i^{o}.

A few simple rules of blood type genetics are that:

I^{A} is dominant over i^{o}

I^{B} is dominant over i^{o}, and

I^{A}I^{B} are codominant

Two plants who are heterozygous for type A blood and have sickle cell trait have children. Answer the following questions:

a. What is the genotype of the parents?

b. What are the genetic make ups of all the possible gametes they can produce.

c. Complete the dihybrid Punnet square to determine the frequency of the different phenotypes in the offspring.

(NOTE: Consider blood type and normal versus mutant hemoglobin in the various phenotypes.)

Answer:

a) I^{A}i^{o}AS

b) I^{A}A,I^{A}S,i^{o}A,i^{o}S

c) The dihybrid table is shown in the explanation below and the frequency of the different phenotypes of the offspring are in the ratio of 3:6:3:1:2:1

Explanation:

a)

From the question the genotype of the parents can be determined:

Given that Two parents are heterozygous for  type  A and have sickle cell trait.

∴ heterozygous for  type  A will be  = I^{A}i^{o} &

heterozygous for having sickle cell traits = AS

Therefore, the traits can be determined the genotype of the parents can be determined as: I^{A}i^{o}AS

b)

the genetic make ups of all the possible gametes they can produce can be determined if the parent are self crossed.

If I^{A}i^{o}AS self crossed, we have:

                I^{A}                 i^{o}

A              I^{A}A               i^{o}A

S              I^{A}S                i^{o}S

∴ The genetic makeups of all the possible gametes they can produce are:

I^{A}A, I^{A}S, i^{o}A, i^{o}S

c)

Below shows the dihybrid Punnet square and  the frequency of the different phenotypes in the offspring

               I^{A}A               I^{A}S                    i^{o}A                 i^{o}S

I^{A}A     I^{A}I^{A}AA     I^{A}I^{A}AS        I^{A}i^{o}AA        I^{A}i^{o}AS

I^{A}S     I^{A}I^{A}AS      I^{A}I^{A}SS         I^{A}i^{o}AS        I^{A}i^{o}SS

i^{o}A     I^{A}i^{o}AA      I^{A}i^{o}AS         i^{o}i^{o}AA        i^{o}i^{o}AS

i^{o}S     I^{A}i^{o}AS       I^{A}i^{o}SS         i^{o}i^{o}AS         i^{o}i^{o}SS

\frac{3}{16}= 18.75% : Blood type A, normal hemoglobin (normal RBCs) = I^{A}I^{A}AA, I^{A}i^{o}AA, I^{A}i^{o}AA

\frac{6}{16}= 37.5% : Blood type A, normal and mutant hemoglobin (sickle cell trait) = I^{A}I^{A}AS, I^{A}i^{o}AS, I^{A}I^{A}AS, I^{A}i^{o}AS, I^{A}i^{o}AS, I^{A}i^{o}AS

\frac{3}{16}= 18.75% : Blood type A, mutant hemoglobin (sickle cell anemia) = I^{A}I^{A}SS, I^{A}i^{o}SS, I^{A}i^{o}SS

\frac{1}{16}= 6.25% : Blood type O, normal hemoglobin (normal RBCs) = i^{o}i^{o}AA

\frac{2}{16}= 12.5% : Blood type O, Blood type A, normal and mutant hemoglobin (sickle cell trait) = i^{o}i^{o}AA, i^{o}i^{o}AS

\frac{1}{16}= 6.25% : Blood type O, mutant hemoglobin (sickle cell anemia) = i^{o}i^{o}SS

I hope that helps alot!

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