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lara [203]
3 years ago
5

Cells lining the small intestine are specialized for absorption of nutrients. their plasma membrane has _______________ which in

crease cell surface area.
Biology
1 answer:
Dimas [21]3 years ago
3 0
Cells lining the small intestine are specialized for absorption of nutrients. their plasma membrane <span>has </span><span>microvillus </span>(plural:<span> m</span>icrovilli)<span> which</span><span> increase cell surface area. These are cellular membranes that are microscopic in nature that increases the surface area of the cell for diffusion and minimize increase in volume.</span><span> </span>
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HPV has been recently been implicated as a risk factor for coronary artery disease
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The mouse and the ant are considered to be
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With a population in Hardy-Weinberg equilibrium, the A allele has a frequency of 0.60, and the frequency of the a allele is 0.40
valkas [14]

Following the Hardy-Weinberg equilibrium theory, the frequency of the heter0zyg0us genotype is 2pq. In the exposed example, 2pq = 0.48.

<h3>Hardy-Winberg equilibrium</h3>

The Hardy-Weinberg equilibrium theory states that the allelic frequencies in a locus are represented as p and q.

Assuming a diallelic gene,

→  The allelic frequencies are

  • p is the frequency of the dominant allele,
  • q is the frequency of the recessive allele.

→  The genotypic frequencies after one generation are

  • p² (H0m0zyg0us dominant genotypic frequency),
  • 2pq (Heter0zyg0us genotypic frequency),
  • q² (H0m0zyg0us recessive genotypic frequency).

If a population is in H-W equilibrium, it gets the same allelic and genotypic frequencies generation after generation.

The addition of the allelic frequencies equals 1  ⇒ p + q = 1.

The sum of genotypic frequencies equals 1 ⇒ p² + 2pq + q² = 1

If the allele A has a frequency of 0.6, and the allele B has a frequency of 0.4, then the frequency of the heter0zyg0us genotype is

2pq = 2 x 0.6 x 0.4 =<u> 0.48</u>

You can learn more about the Hardy-Weinberg equilibrium at

brainly.com/question/3406634

7 0
2 years ago
In an experiment involving an enzyme, the pH is raised from 2 to 11. The optimal pH for this enzyme is 2. What will be the most
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7 0
3 years ago
1. Which one lived in Antarctica earlier, the labyrinthodont, the lystrosaurus, or the duck-billed dinosaur? Explain
liubo4ka [24]

Answer:

The correct answer is - The labyrinthodont.

Explanation:

Fossils of These duck billed dinosaur were found in the Antarctica in sands that were 66 to 67 million years old, however these are also found in north America and north Europe.

About 110 million years ago, labyrinthodonts, the dinosaur that are amphibians. These are the animals that were lived in the river valleys of the Gondwana and the rift valley that was located at that time between Australia and Antarctica.

Thus, the correct answer is - The labyrinthodont.

7 0
3 years ago
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