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Mekhanik [1.2K]
3 years ago
12

Please give an perfect answer​

Biology
1 answer:
Anvisha [2.4K]3 years ago
3 0

Answer:

a neutral solution? i think

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as you start working out you notice that your heart rate and breathing rate starts to increase what division of your nervous sys
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Concerning your heart, that's the circulatory system
Your lungs are the respritaory system
4 0
3 years ago
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
If a parent with AB blood and a parent with
Vlad1618 [11]

Answer:

O blood is a recessive trait, so none of the children will have O blood. The genes for either type A or Type B will be determined by the egg/sperm being fertilized with the reproductive cell of the other parent. the probability of receiving either of the gene versions is 1/2, so half of the children will have A blood and the other half will have B blood

4 0
3 years ago
Placental 11β-hydroxysteroid dehydrogenase type 2 expression: correlations with birth weight and placental metal concentrations
Dafna11 [192]
The weight would be312
5 0
3 years ago
8)
elena-s [515]

Answer:

A

Explanation:

5 0
3 years ago
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