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Mnenie [13.5K]
4 years ago
10

How does the average number of drops required to break open a whelk depend on platform height

Biology
1 answer:
Murljashka [212]4 years ago
3 0
The normal number of drops required to break whelks from different stage statures was recorded and arrived at the midpoint of over numerous trials with the device.Combining the information for every stage tallness, add up to "flight" stature was ascertained by duplicating the drop tallness by the normal number of drops required.
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Which viral replication cycle is represented in part B of this diagram
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. In this example, red flower coloring is dominant to white, and your population of 200 plants has 150 plants with red flowers.
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Answer:

a) The allele frequencies are - p= 0.8660\\q= 0.134

b) Genotype frequencies for this population- p^2=0.75, q^2= 0.0179, 2pq = 0.2321

Explanation:

Given ,

Red color of flower is dominant to white color flower

Let "R" represents allele for red color and "r" represents allele for white color

out of 200 flower, 150 are red flowers.

So the frequency of red flowers with genotype "RR" is equal to

p^{2} = \frac{150}{200} \\p^{2} =0.75

As per hardy Weinberg's I equilibrium equation -

p+q=1\\p=\sqrt{p^2} \\p=\sqrt{0.75} \\p= 0.8660\\q= 1-p\\q= 1-0.8660\\q = 0.134

a) The allele frequencies are - p= 0.8660\\q= 0.134

Frequency of genotype "rr" i.e homozygous recessive white flower is

q^{2} \\0.134^2\\q^2 = 0.0179\\

Substituting the values of p^2 and q^2, in below equation, we get -

p^2+2pq+q^2=1\\2pq= 1-p^2-q^2\\2pq= 1-0.75-0.0179\\2pq=0.2321

b) Genotype frequencies for this population- p^2=0.75, q^2= 0.0179, 2pq = 0.2321

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