When a cell undergoing cellular division identifies that a chromosome misalignment had taken place the cell cycle stops until the error it's corrected.
Option two of the question states that the cell cycle will <u>proceed uncontrollably, this is not the case for the error at hand</u>. Although this can be a problem for cells and leads to the appearance of cancer, it is not caused by chromosome misalignment. As is the case with options 3 and 4.
The stop or delay in the <u>cell cycle is what normally takes place in these situations.</u> This stoppage is done by the <em><u>spindle checkpoint</u></em>, which prevents <u>duplicate chromosomes from separating.</u>
During this time, the error it's corrected. If a cell is not able to correct the error at this time, many situations can follow. <u>However, the most likely is that the cell will undergo a <em>programmed cellular death.</em> </u>
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Answer:
Your answer is Stigma.
Explanation:
The stigma is the sticky knob at the top of the pistil. It is attached to the long, tubelike structure called the style. Hope this helped :)
Answer:
The whole body works together, as a “team.”If one of the parts of the body won’t work together as a team, everything will get messed up. The stomach muscles churn and mix the food with digestive juices that have acids and enzymes, breaking it into much smaller, digestible pieces. An acidic environment is needed for the digestion that takes place in the stomach.
Explanation:
The frequency <em>p</em> of the yellow (A) allele is <em>p</em>= 0.3
The frequency <em>q</em> of the blue (a) allele is <em>q= </em><em>0.7</em>
Hardy–Weinberg equilibrium, states that allele and genotype frequencies in a population will remain constant from generation to generation. Equilibrium is reached in the absence of selection, mutation, genetic drift and other forces and allele frequencies p and q are constant between generations. In the simplest case of a single locus with two alleles denoted A and a with frequencies f(A) = p and f(a) = q, the expected genotype frequencies under random mating are f(AA) = p² for the AA homozygotes, f(aa) = q² for the aa homozygotes, and f(Aa) = 2pq for the heterozygotes.
p²+2*p*q+q²= 1 p+q= 1 q= 1-p
yellow (p²)= 9%= 0.09 p= √0.09= 0.3
green (2*p*q)= 42%= 0.42
blue (q²)=49%= 0.49 q=1-0.3= 0.7 <em>or</em> q= √0.49= 0.7
