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3 years ago

When a germ cell undergoes meiosis, which event would MOST likely result in four aneuploid daughter

Biology

1 answer:

sveta [45]3 years ago

6 0

Answer:

B. misalignment of the chromosomes during anaphase I

Explanation:

Aneuploidy refers to an error in the chromosomal number of a cell. This occurs when the chromosomes fail to separate equally into individual cells, a process called MEIOTIC NONDISJUNCTION. Meiotic disjunction can occur in either the anaphase stage of meiosis I or meiosis II, however, different outcomes are produced.

When chromosomes fail to separate at meiosis I, it results in two gametes that lack that particular chromosome (n-1) and two gametes with two copies of the chromosome (n+1). This means that the four daughter cells will have an incorrect number of chromosomes, hence, the misalignment of the chromosomes during anaphase I will result in four aneuploid daughter cells.

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KIM [24]

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3 years ago

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Darina [25.2K]

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3 years ago

Assume that the left ventricle of a child's heart has an EDV=90mL, and ESV=60mL, and a cardiac output of 2,400 mL/min. His SV is

Sliva [168]

Answer:

SV = 30 ml/ beat.

HR = 80.

Explanation:

The formula to calculate SV is as follows:

SV = EDV- ESV.

Here EDV = 90ml (end diastolic volume), ESV = 60ml (end systolic volume) ans SV is stroke volume.

SV = 90-60

SV = 30ml/beat.

The formula to calculate HR is as follows:

HR = CO/SV

Here, HR is heart rate, CO is cardiac output = 2,400 ml/min and SV is stroke volume = 30ml/beat

HR = 2,400/30

HR = 80.

Thus, the SV is 30ml/best and HR is 80.

4 0

3 years ago

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Gwar [14]

Answer:

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Explanation:

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Answer:

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