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Gekata [30.6K]
1 year ago
13

Assortative mating changes ______ frequencies but does not change ______ frequencies.

Physics
1 answer:
andrew-mc [135]1 year ago
4 0

Allele frequencies are unaffected by assortative mating, but genotype frequencies .

<h3>Assortative mating: </h3>

Individuals with similar phenotypes and genotypes mate with others more frequently than is anticipated under a random mating pattern in assortative mating, which is a mating pattern and a type of sexual selection.

<h3>Frequencies of genotypes:</h3>

A population's genotype frequency is calculated by dividing the number of people having a particular genotype by the overall population size. The genotype frequency in population genetics is the frequency or ratio (i.e., 0 f 1) among genotypes inside a population.

<h3>The frequency for alleles in biology:</h3>

The term "allele frequency" describes the prevalence of an allele in a population. It is calculated by calculating the number of times the allele occurs in the population and dividing by the sum of all the gene copies.

To know more about Assortative mating visit:

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The answer is speed: 4.7 km/h, velocity: 3.3 km/h.

Distances and time are given:
d1 =  4 km
d2 = 3 km
d3 = 5 km
t = 1.5 h

The speed can be expressed as a distance (d) divided by time (t). The average speed (s) is total distance travelled divided by time:
s = (d1 + d2)/t = (4+3)/1.5 = 7/1.5 = 4.7 km/h

The average velocity (v) is total displacement (d₁) from the starting point divided by time. Since Mary's starting point was home, and she walked to the supermarket, which is 5.0 kilometers from her own home, her displacement is 5 km:
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3 years ago
In nuclear reaction 5 kg of reactants give 2kg of products
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3 years ago
State the law of conservation of momentum<br>​
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Answer:

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Explanation:

Hope it helps

8 0
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The electric field strength in the space between two closely spaced parallel disks is 1.0 10^5 N/C. This field is the result of
alex41 [277]

To solve this problem it is necessary to apply the concepts related to the capacitance in the disks, the difference of the potential and the load in the disc.

The capacitance can be expressed in terms of the Area, the permeability constant and the diameter:

C = \frac{\epsilon_0 A}{d}

Where,

\epsilon_0 = Permeability constant

A = Cross-sectional Area

d = Diameter

Potential difference between the two disks,

V = Ed

Where,

E = Electric field

d = diameter

Q = Charge on the disk equal to \rightarrow Q=ne=(3.9*10^9)(1.6*10^{-19})= 6.24*10^{-10}C

Through the value found and the expression given for capacitance and potential, we can define the electric charge as

Q = CV

Q = \frac{\epsilon A}{d}(Ed)

Q = \epsilon_0 AE

Q = \epsilon_0 \pi(\frac{d}{2})^2E

Q = \frac{\epsilon \pi d^2E}{4}

Re-arranging the equation to find the diameter of the disks, the equation will be:

d = \sqrt{\frac{4D}{\epsilon_0 \pi E}}

Replacing,

d = \sqrt{\frac{4(6.24*10^{-10})}{(8.85*10^{-12})\pi(1*10^{5})}}

d = 0.0299m

Therefore the diameter of the disks is 0.03m

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Evaporation (or another word to use is water vapor.)
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