Answer:
0.4550
Explanation:
In this example, R allele is dominant, so individuals RR and Rr can roll their tongues. If in a population of 1000 gorillas, there are 575 gorillas who can roll their tongues, they will be RR and Rr.
In this case, there is Hardy Weinberg Equilibrium, so the following will be true:
Where:
frequency of RR
frequency of Rr
frequency of rr
The question is what is the frequency of heterozygotes, or, what is the value of 2pq.
We know that RR+Rr is 575 individuals in a population of 1000, or 0.575.
In other words:
So, it is possible to find 
:
Now, there are two alleles in the population, so the following will be true:
It is possible to find q (the frequency of allele r) and p (the frequency of allele p):
Therefore:
Now, the frequency of heterozygotes or 2pq is:
Answer:
B There will be blue colonies only
Explanation:
This screen in called the blue white screening. This is a rapid test that allows scientist to check for their insert of choice in the colonies at a glance. The concept is that where the insert is located the <em>lac Z </em>gene in the vector is interrupted and therefor Xgal, a reporter can be produced. Where there are blue colonies being formed the plasmid has self-ligated and the <em>lac Z</em> gene is not interrupted and the X-gal can be formed rendering a blue color.
Well, first off the spongy mesophyll does have some chloroplasts, however they are located quite far from the surface of the leaf where most of the chloroplasts are. Therefore they don't get much light and don't contribute a lot to photosynthesis in the leaf. So why should the leaf waste the energy in making chloroplasts if there is not enough light to make them all efficient enough at photosynthesis?
