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White raven [17]
3 years ago
5

Describe the outcome of irregular mitosis, meiosis I and meiosis II. be sure use all relevant terminology and to note the impact

on chromosomal number in each circumstance.
Biology
1 answer:
Serhud [2]3 years ago
6 0

Answer:

Outcome of irregular mitosis:

Mitosis may be defined as the process of cell division in which a single diploid parent cell divides to form two diploid daughter cells. The irregular mitosis may result in aneuploidy condition of chromosome in which the the extra (trisomy) or missing (nullisomy) of one or more chromosome may occur. This condition also leads the genetic disease and cancer condition. Non disjunction may occur in which the sister chromatid are unable to separate.

Outcome of irregular meiosis I and meiosis II.

Meiosis is the cell division in which a single diploid parent cell divides into four haploid daughter cells. Nondisjunction at meiosis I prevent the segregation at least one homologous chromosomes. This causes the two cells with extra copy of chromosome and two cells with deleted chromosome. Nondisjunction at meiosis II. Two cells have normal chromosomes and two cells with the abnormal (missing and extra) chromosomes.

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The first organism to move into an area after primary disturbance are
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When pink sweet peas were self-pollinated and the seeds were collected and sown, the following flower colors were obtained: Red
Pavlova-9 [17]

Answer:

c. 1:2:1

The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.

Explanation:

If flower color were determined by a gene showing incomplete dominance, the possible genotypes and phenotypes are as follows:

  • RR- red
  • ww - white
  • Rw - pink

If pink sweet peas are self-pollinated, then a cross between two heterozygous individuals is done (Rw x Rw).

<u>From this cross the expected ratios are:</u>

  • 1/4 RR (red)
  • 2/4 Rw (pink)
  • 1/4 ww (white)

So the null hypothesis is that the observed results exhibit a 1:2:1 ratio.

<h3><u>Chi square test</u></h3>

X^{2} = \sum \frac{(Observed - Expected)^2}{Expected}

<u>The observed frequencies were:</u>

  • 34 Red
  • 76 Pink
  • 40 White

Total 150

<u>The expected frequencies for our null hypothesis are:</u>

  • 1/4 x 150 = 37.5 Red
  • 2/4 x 150 = 75 Pink
  • 1/4 x 150 = 37.5 white

X^{2} = \frac{(34- 37.5)^2}{37.5} + \frac{(76- 75)^2}{75} + \frac{(40- 37.5)^2}{37.5}

X^2=0.5067

The degrees of freedom (DF) are calculated as number of phenotypes - 1; in this case DF = 3-1 = 2.

If we look at the Chi square table, for 2 DF and a probability of p0.05, the critical value is 5.991

Our X^2 value of 0.5067 is less than the critical value, so we do not reject the null hypothesis. The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.

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andriy [413]

Answer:

Humans maintain a body temperature of 37oC; .

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