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4 years ago

Define cell conjunctions

1 answer:

8 0

It means that two or more cells connected together.
You can look at the word 'cell' and 'conjunctions' separately. 'Cell' is a tiny particle. 'Conjunctions' means 'two or more things connected', so just put the two meanings together.
Hope this helps. :)

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H2+N2 *arrow* 2NH3
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4 years ago

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a comet is cruising through the solar system at a speed of 50,000 kilometers per hour for 4 hours time what is the total distanc

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50,000 x 4 = 200,000

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3 years ago

What bird species was most noticeably impacted by West Nile virus when the virus was first detected? Was this the only bird spec

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Mostly crows, robins, and jays were the first of a few bird species impacted by the West Nile Virus when it was detected.

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3 years ago

Claire had a heavy breakfast early in the morning. Then she went to the gym to exercise. After the gym, Claire went to class. Wh

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3 years ago

The Aye-Aye exhibits incomplete dominance for eye color. In a population of Aye-Ayes there are red-eyed individuals exhibiting t

VMariaS [17]

Answer:

196.8

Explanation:

In a population of Aye-Ayes there are red-eyed individuals exhibiting the dominant phenotype -----i.e RR

yellow-eyed individuals exhibiting the recessive phenotype ----- i.e rr

in the next generation, we will have the following offspring

R R

r Rr Rr

All the offspring are heterozygous in nature.

Now, the question goes further by saying, if there are 66 yellow-eyed individuals in the current population of 459, i.e (q² =66)

How many would be expected to exhibit the heterozygous phenotype in the next generation which will consist of 410 total individuals.

In solving Hardy-Weinberg question, we use the following equation below.

p + q = 1

p² + 2pq + q² = 1

where;

p = the frequency of the dominant allele

q = the frequency of the recessive allele

p² = the frequency of individuals with homozygous dominant phenotype

2pq = the frequency of individuals with heterozygous phenotype

q² = frequency of individuals with the homozygous recessive phenotype

Now, If In the next generation, the total population = 410

and q² (yellow-eyed individuals that are recessive in nature) = 66

then; q²= $\frac{individuals with recessive phenotype}{total population}$

q² = $\frac{66}{410}$

q² = 0.16

q = $\sqrt{0.16}$

q = 0.4

Since q = 0.4, we can easily get p by using the formula above:

p + q = 1

p = 1 - q

p = 1 - 0.4

p = 0.6

∴ Since we've known our p and q , we can easily determine how many would be expected to exhibit the heterozygous phenotype in the next generation.

2pq = 2 ( 0.6 × 0.4)

= 2 ( 0.24)

= 0.48

2pq = $\frac{individuals with heterozygous phenotype}{total population}$

0.48 = $\frac{x}{410}$

x = 0.48 × 410

x = 196.8 individuals

4 0

3 years ago