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madam [21]
3 years ago
12

The _____ is critical in the cell cycle because it indicates the cell is ready for division. G0 phase G1 phase G2 phase S phase.

Biology
2 answers:
prisoha [69]3 years ago
8 0
The S phase is critical in the cell cycle because it indicates the cell is ready for division.
zysi [14]3 years ago
8 0

The correct answer is S phase.

The critical phase of cell cycle is S phase, in this phase the DNA starts replicating.The cell cycle is called the synthesis phase because the synthesis of DNA starts in this phase. There is no chance that the cell can go back to the previous phase. The DNA within the cell starts replicating. Depending upon the level of nutrients, energy and external factors, cells must decide that they will enter into cell cycle or not.


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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. <u>Let's call the dominant allele associated with brown colour "B" and the recessive allele associated with blue colour "b"</u>. 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 <em>(square root of f(bb)).</em>

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

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3 0
3 years ago
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