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SpyIntel [72]
3 years ago
11

9. Which describes a catastrophe that causes stress?

Health
2 answers:
anyanavicka [17]3 years ago
6 0
The threat of an unsafe environment
Contact [7]3 years ago
3 0

Answer:

The answer is <u><em>A. the threat of an everyday problem.</em></u>

Explanation:

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Which option is not a healthy practice? calorie restriction eating low-fat foods omitting carbohydrates small, frequent meals
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Arnold is 56 years old and considers himself to be in good health. He has a family history of heart disease, stroke, and diabete
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B. Increasing Vitamin D intake.

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In some regions where malaria is endemic, the frequency of the sickle-cell allele is as high as 15% due to heterozygous advantag
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There will be an increase in frequency, if there is continuous reproduction within humans with the heterozygous genotype but a decrease if there is a continuous reproduction of heterozygous humans and non-carriers of the allele.

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The heterozygous genotype gives an advantage for malaria resistance however, if malaria was completely eradicated and the heterozygous genotypes continue to reproduce within one another, there will be an increased frequency of sickle cell allele. However, if heterozygous alleles were to reproduce with individuals without the allele, there will be a decrease in the frequency of the allele in the human population. If this trend continues, there will be ao less a frequency of the allele that it's presence could be completely lost.

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3 years ago
Which of the following is a common consequence of teenage pregnancy?
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Not enough money, getting kicked out of home, drop out of school,  not going to college, baby has no father most of the time.

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Two parents who are heterozygous for type A blood and have sickle cell trait have children.
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!

3 0
3 years ago
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