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Ulleksa [173]
3 years ago
5

In an experiment that tests how

Biology
1 answer:
katrin2010 [14]3 years ago
5 0

The growth of the plant

Explanation:

Dependant variable is the variable you measure

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A particular recessive genetic disorder is fatal before birth, so there are no homozygous recessive individuals. In a particular
Ksenya-84 [330]

Answer:

  • the allelic frequency for p is 0.967
  • the allelic frequency for q is 0.033

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the exposed example,

  • A recessive genetic disorder is fatal before birth, so there are no homozygous recessive individuals
  • In a particular population, one in 15 individuals is a carrier for this disorder.

What are the allele frequencies of the dominant (p) and recessive (q) alleles in this population?

If 1 of 15 individuals are carriers for this disorder, this means that 1/15 are heterozygous, 2pq. So, 2pq = 1/15 = 0.066

Now we must calculate the allelic frequencies.

We know that 1 in 15 individuals are heterozygous, and we also know that there are no recessive homozygous, q², because they can not survive, so of the 15 individuals only one is heterozygous and the rest 14 individuals must be dominant homozygous, p².

The dominant homozygous genotypic frequency is

p²= 14/15 = 0.933

And by clearing the next equation we can get the allelic frequency for p

p²= 0.933

p = √0.933

p = 0.967

So now we know that the allelic frequency for p is 0.967  

This means that the allelic frequency for q or p is 0.033, which we deduce by clearing the equation p + q = 1

                          0.967 + q = 1

                         q = 1 - 0.967

                          q = 0.033

  • the allelic frequency for p is 0.967
  • the allelic frequency for q is 0.033
5 0
3 years ago
Phosphorus is difficult for plants and animals to access in nature because___
Alika [10]

C should be your answer

8 0
3 years ago
Two rabbits that were homozygous for both coat and eye color were crossed. The following are the phenotypes of their F2 generati
laiz [17]

Answer:

D

Explanation:

According to the law of inheritance, the phenotypes of their F2 generation will have 95 brown coat brown eye, 12 brown coat blue eye, 16 white coat blue eye, 30 white coat blue eye.

4 0
4 years ago
The uptake of material through the plasma membrane by the formation of a vesicle is _______, whereas the fusion of a vesicle wit
lidiya [134]

Answer:

Endocytosis, exocytosis

Explanation:

The uptake of material through the plasma membrane by the formation of a vesicle is endocytosis, whereas the fusion of a vesicle with the plasma membrane and the release of its contents outside of the cell is called exocytosis.

Endocytosis and exocytosis are two active transport processes through which cells move macromolecules across the cell membrane.

<em>While the former involves the capturing of materials from the surrounding of the cell and engulfing such through the cell membrane by forming a phagocytic or pinocytic vesicle, the latter involves the fusion of vesicles to the plasma membrane and the release of the contents of the vesicles to the outside of the cell.</em>

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3 years ago
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What is an analogy to a farm for chromatin of an animal cell?
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Ok the hemophilia is a recessive disorder found in the X chromosome of males abd females. For an example Patricia is a healthy carrier of hemophilia and that is for another example Sam is completely healthy then Patricia hope that helped
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4 years ago
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