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just olya [345]
2 years ago
10

I still dont get the stages of Mitosis. Please give me a descripition for each one

Biology
2 answers:
Inessa05 [86]2 years ago
8 0
Prophase- <span>chromosomes become visible as paired chromatids and the nuclear envelope disappears.

metaphase- </span><span>chromosomes become attached to the spindle fibers.

anaphase- </span><span>chromosomes move away from one another to opposite poles of the spindle.

telophase- </span><span>the final phase of cell division in which  chromatids, or chromosomes, move to opposite ends of the cell and two nuclei are formed.</span>
Tems11 [23]2 years ago
6 0
Prophase:chromatin are packaged to chromosomes
metaphase:chromosomes are lined up in lines that the centremore is in the half of the line
anaphase:the chromosomes are held from the opposite sides now we have 1 chromatid on a line but vertically we have a full set of chromosomes
telopphase:a nuclear membrane is formed around the chromosomes
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The question is incomplete. Here is the complete question.

You are studying a population of ground squirrels. You trap 100 individuals and find 9 that have a light colored coat and the rest have a dark coat. You know that light coat color is recessive in ground squirrels. What is the allele frequency for the light coat allele  and the dark coat allele? How many of the dark coated individuals do you expect to be heterozygotes at the coat color gene? Assume the population is at H-W equilibrium.

Answer: Allele frequency for light coat: q = 0.3;

              Allele frequency for dark coat: p = 0.7;

             There will be 42 heterozygote;

Explanation: A population in H-W equilibrium is a population which doesn't change from one generation to the next if there are no disturbing factors, as migration or mutation.

It follows the equation: p^{2}+2pq+q^{2}=1.

p represents the frequency of the dominant allele;

q represents the frequency of the recessive allele;

2pq represents heterozygote individuals;

The sum of frequencies of both alleles must be 1, i.e.: p+q=1

For the population of squirrels:

  • Frequency of recessive allele

q^{2} = \frac{9}{100}

q = \sqrt{\frac{9}{100}}

q = 0.3

  • Frequency of dominant allele

p+q=1

p=1-q

p=1-0.3

p = 0.7

  • Frequency of heterozygote

2pq = 2*0.7*0.3 = 0.42

In a population of 100 individuals: 0.42*100 = 42

The allele frequency for the light coat is 0.3; for the dark coat is 0.7.

It is expected to have 42 heterozygote individuals in this population.

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