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tiny-mole [99]
2 years ago
12

By what process do streams and rivers move material?

Biology
1 answer:
FromTheMoon [43]2 years ago
5 0

Answer:

infiltration

Explanation:

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Give an example of a basic scientific experiment, identify the control variable, independent and dependent variable in the exper
schepotkina [342]
The picture gives an example .

5 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
100 Points And Brainly If you answer correctly with detail.
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Answer:

I would first put jaws at number one because most of the other options are for mammals while some marine life has jaws too.

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Why is a second blood test necessary three months after a person thinks he or she has been exposed to hiv?
Elza [17]
The antibodies to the HIV virus may take up to three months to appear in the blood of the infected person. 
Therefore, the second blood test is necessary three months after a person thinks that he/she has been exposed to HIV to check the presence of these antibodies and confirm whether the person is infected or not.
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Which is Not a feature associated with valley glaciers
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I think its arroyo.
i hope this answer is correct and helps you :)
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