Exlain howthe palisade cellsadapted for photosynthesis(3).

1 answer:

3 0

The palisade cells are the main site of photosynthesis, as they have many more chloroplasts than spongy mesophyll, and also have several adaptations to maximizes photosynthesis efficiency. Large Vacuole restricts chloroplasts to a later near outside of the cell where they can be reached by light more easily.

You might be interested in

In a hypothetical population of 2500 people, 2275 people have brown eyes and 225 people have blue eyes (the homozygous-recessive

aev [14]

Answer:

In the next generation of 4000 children, 1680 of them will be heterozygous for the eye colour.

Explanation:

There's a population of 2500, 2275 of with have brown eyes and 225 blue eyes. Let's call the dominant allele associated with brown colour "B" and the recessive allele associated with blue colour "b". So the possible genotypes are BB, Bb and bb, being BB and Bb brown eyed individuals and bb blue eyed individuals.

If the population it's in Hardy-Weinberg equilibrium, it means genotypic and allelic frequencies don't change from one generation to the following.

From the information given, we can calculate both allelic and genotypic frequencies.

First, we know that the frequency of the genotype bb it's the amount of blue eyed individuals over the total population.

f(bb)=225/2500=0.09

Additionally we know the allelic frequencies can be related to the genotypic ones when the population it's in Hardy-Weinberg equilibrium. Particularly we can say:

f(bb)=[f(b)]^2 => f(b)=[f(bb)]^(1/2)= 0.3 (square root of f(bb)).

Also, we can calculate the frequency of the B allele, as the probability of all alleles of the gene sum 1. In other words:

f(b)+f(B)=1 => f(B)=1 - f(b) = 1 - 0.3 = 0.7

So far, we have calculated the allelic frequencies, f(b)=0.3 and f(B)=0.7.

Now we can calculate the genotypic frequencies, using the equations of the Hardy-Weinberg equilibrium.

f(bb)=[f(b)]^2 => f(bb)=0.3^2=0.09
f(Bb)=2*f(B)*f(b) => f(Bb)=2*0.7*03=0.42
f(BB)=[f(B)]^2 => f(BB)=0.7^2=0.49

Finally, knowing that there are 4000 children in the next generation, to know how many of them are expected to be heterozygous for the eye colour, we should multiply the number of children for the probability of being heterozygous for the eye colour (which is the genotypic frequency for the genotype Bb).

Nº of heterozygous individuals = f(Bb)*total population= 0.42*4000

=> Nº of heterozygous individuals =1680

3 0

3 years ago

Please help!!!! very confused pt 2

Schach [20]

Answer:

If the genotype of the parents are Iᵃi and Iᵇi, then four type of offspring will be produced.The genotype of the offspring are, IᵃIᵇ, Iᵃi,Iᵇi, and ii.

Explanation:

IᵃIᵇ = As the alleles are co-dominant in nature, so both type of alleles are expressed. The blood group will be AB. So, both A and B type of antigen will be found in plasma membrane of RBC.

Iᵃi= In this type of genotype only A type of antigen will be expressed in the membrane of RBC. The blood group will be A type.

Iᵇi= In this type of genotype only B type of antigen will be expressed in the membrane of RBC. The blood group will be B type.

ii= This is a recessive type of genotype. So, no antigen will be found on the membrane of the RBC. And the blood group will be O type.

The offspring's ratio will be 1:1:1:1.