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Dima020 [189]
3 years ago
8

(07.02 HC) A certain type of heart-muscle disease thickens the heart muscle so that the heart chambers shrink in size. How would

this disease affect a person over time, as the volume of the heart chambers continues to shrink? (2 points) The person would become more alert and anxious over time because the enlarged heart muscles would pump more blood to the brain than before the disease progressed. The person would develop low blood pressure and feel faint because the heart will be pumping less blood with each heartbeat. The person would become more tired and weak because the volume of oxygenated blood pumped with each heartbeat would decrease over time. The person would develop high blood pressure because the thicker heart muscles and smaller heart chamber volume would cause more pressure in the heart.
Biology
1 answer:
mario62 [17]3 years ago
8 0
The correct answer is The person would develop low blood pressure and feel faint because the heart will be pumping less blood with each heartbeat.   
<span>
The thickening of the heart muscle may block blood flow out of the ventricle which then must work hard to pump blood. As a result, the symptoms can include chest pain, dizziness or fainting.</span>
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Field mice in beach mice are closely related species. The Field mice are various shades of brown in the beach mice or later 10 a
wariber [46]

Answer:

The process of natural selection is acting on <u>Field mouse individuals</u> where as evolution is occurring on <u>Field Mouse population. </u>

Explanation:

As the problem shows, the field mice are the ones that natural selection is acting on, but only the ligher shade ones. Because the lighter shade individuals are more likely to survive in the beach because they blend with the environment.

Evolution occurred on the field mouse population because if you read the script in the end, they were able to successfully reproduce. This means that they evolved in such a way that they were able to survive in their new environment. (most likely carrying on the trait of light-colored fur)

Natural selection acts on traits, phenotypic traits, favorable to the environment. Evolution occurred because of the natural selection, because the favored trait enabled the organism to adapt to the environment.

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2 years ago
Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre
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Answer:

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Explanation:

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8 0
3 years ago
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2. Dominant trait: cleft chin (C) Mother’s gametes: Cc
andre [41]

.2. Offspring Genotypes will be Cc or cc.

     Offspring phenotypes : Cleft chin or no cleft chin.

    % chance child will have cleft chin: 50%

3.  % chance child will have arched feet: 25%

4.  % chance child will have blonde hair:  50%

5.  % chance child will have normal vision: 25%

 

Explanation:

CASE 1 :

 Dominant trait: cleft chin (C)

    Recessive trait: lacks cleft chin (c)

    Father’s gametes: cc

    Mother’s gametes: Cc

There are two possible combination of Gametes ,

C fom mother and  c from father= Cc

c from mother and c from father = cc

Gametes of Cc Parents=  \frac{1}{2}C + \frac{1}{2} c........(i)

Gametes of cc parents =<u> </u>\frac{1}{2}c + \frac{1}{2}c .........(ii)

Combining (i) and (ii) we get,

\frac{1}{2}  Cc + \frac{1}{2} cc                              

There fore offspring Genotypes will be Cc or cc

Offspring phenotypes :

Genotype Cc then phenotype= Cleft chin

Genotype cc then phenotype = Lacks cleft chin.

percentage chance child will have cleft chin  =\frac{0.5}{1} ×100

Therefore the chance is 50%.

CASE 2 :

Dominant trait: flat feet (A)

Recessive trait: arched feet (a)

Mother’s gametes: Heterozygous (Aa)

Father’s gametes: Heterozygous   (Aa)

There are four possible combination of genotypes are =AA , Aa, Aa and aa

i.e. A from mother, A from father= AA

     A from mother, a from father =Aa

     a from mother, A from Father = Aa

     a from mother, a from father = aa

Gametes of Aa parent =\frac{1}{2} A + \frac{1}{2} a

Gametes of other Aa parent = \frac{1}{2} A + \frac{1}{2} a

                                       <u>..................................................................................</u>

                                              \frac{1}{4} AA + \frac{1}{4} Aa

                                                                           +  \frac{1}{4} Aa +\frac{1}{4} aa

                                   <u>..........................................................................................</u>

                                <u>\frac{1}{4}AA + \frac{1}{2}Aa +\frac{1}{4} aa</u>

Offspring Genotypes will be: AA or Aa or aa

Offsprings phenotype will be:

Genotype AA then phenotype will be Flat feet

Genotype Aa then phenotype will be flat feet

Genotype aa then Phenotype will be arched feet.

Percentage chance child will have arched feet = \frac{0.25}{1} × 100 = 25%

CASE 3:

Dominant trait: Brown hair (B)

Recessive trait: Blonde hair (b)

Mother’s gametes: Homozygous recessive  (bb)

Father’s gametes: Heterozygous  (Bb)

This case is very similar to the case 1 as one parent is homozygous recessive and other parent is heterozygous.

Resulting in  half  Bb and halve bb combination.

Genotypes will be Bb or bb

Phenotypes will be :

Genotype Bb then phenotype Brown hair

Phenotype bb then Phenotype bb.

% chance child will have blonde hair: 50%

CASE 4:

Dominant trait: farsightedness (F)

Recessive trait: normal vision (f)

Mother’s gametes: Heterozygous  (Ff)

Father’s gametes: Heterozygous  (Ff)

This Case is similar to case 2

it will result in one-fourth FF , half Ff and one-fouth ff combination.

Therefore Genotypes will be: FF, Ff and ff

Phenotypes:

Genotype FF  then phenotype farsightedness

Genotype Ff then phenotype  farsightedness

Genotype ff then phenotype normal vision.

% chance child will have normal vision: 25%

 

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3 years ago
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Answer: i'm pretty sure its B!!!

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