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Lina20 [59]
3 years ago
13

Describe how the rabbit population changed over the course of 10 years.

Biology
1 answer:
valkas [14]3 years ago
6 0
So, the population went up with every spring season and fell slightly in the winter; it increased within the course of 10 years
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Industrial melanism refers to the dark pigmentation that evolved in some insects giving them protective coloration on vegetation
Natalka [10]

Answer:

  • The frequency of the dominant allele, p =  0.542
  • The proportion of black moths that are heterozygous 2pq = 0.496

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the allelic dominant or recessive forms. The genotypic frequencies after one generation are p² (Homozygous dominant), 2pq (Heterozygous), q² (Homozygous recessive). 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 same way, the sum of genotypic frequencies equals 1, this is

p² + 2pq + q² = 1

Being

  • p the dominant allelic frequency,
  • q the recessive allelic frequency,
  • p² the homozygous dominant genotypic frequency
  • q² the homozygous recessive genotypic frequency
  • 2pq the heterozygous genotypic frequency

In the exposed example, 79% of the moths of the species Biston betularia were black due to the presence of a dominant gene for melanism.

If the genotypic frequency of back moths is 0.79, then, by performing the following equation we can get the not-black moths genotypic frequency:

p² + 2pq + q² = 1

where p² is the homozygous dominant genotypic frequency, q² the homozygous recessive genotypic frequency, and 2pq is the heterozygous genotypic frequency.

As 0.79 is the phenotypic frequency of black moths, then this frequency equals p²+2pq.

Clearing the equation:

p² + 2pq + q² = 1

0.79 + q² = 1

q² = 1 - 0.79

q² = 0.21

The genotypic frequency of non-black moths is 0.21. So, from here we can calculate the allelic frequency:

q² = 0.21

q= v 0.21

q = 0.458

If 0.46 is the allelic frequency of non-black moths, then by clearing the equation p + q = 1, we can get the p allelic frequency:

p + q = 1

p + 0.458 = 1

p = 1 - 0.458

p = 0.542

  • The genotypic frequency p² = (0.542)² = 0.294
  • The heterozygote genotypic frequency

        2 x p x q = 2 x 0.542 x 0.458 = 0.496

Finally, we can check this answer by clearing the following equation:

p² + 2pq + q² = 1

0.294 + 0.496 + 0.21 = 1                  

4 0
3 years ago
What is tonsillitis​
serious [3.7K]

Answer: inflammation of the tonsils.

Explanation:

Pretty much a sore throat

8 0
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What is the condition, resulting from hypersecretion of cortisol, whose clinical manifestations include "moon" facies, obesity o
vampirchik [111]

Answer: Cushing Syndrome

Explanation:

Cushing Syndrome occurs when the body is exposed to high level of cortisol for a long time either externally (taken into the body from outside ) or internally (secretion in excess of the hormone called cortisol by the body) . It can also be called hypercortisolism .

The symptoms of Cushin Syndrome may vary depending on the severity of the excess cortisol secretion. Females may have irregular menstrual period due to this condition.

Asthma and arthritis patients who take corticosteroids and experiences the mentioned symptoms should seek medical attention for diagnosis.

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Glycolysis provides a cell with a net gain of?
Vinil7 [7]
Glycolysis provides a cell with a net gain of 2 ATP molecules
8 0
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Large multicellular organisms are made of a wide variety of cell types. Do
Murrr4er [49]

Answer:

Yes, All of the cells divide at approximately the same rate, although they may divide at different times

7 0
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