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3 years ago

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In some instances, meiosis of a primary spermatocyte with six chromosomes results in to spermatozoa that contains for chromosome

s in to spermatozoa that contain two chromosomes. Predict the most likely cause

Advanced Placement (AP)

1 answer:

TiliK225 [7]3 years ago

4 0

Answer:

Nondisjunction

Explanation:

Nondisjunction occurs when chromosomes do not align and separate properly prior to the formation of daughter cells, during meiosis I or II. This means that some gametes are missing some chromosomes, and other gametes have extra chromosomes.

This is usually lethal, however, some genetic disorders in humans are caused by chromosome nondisjunction. For example, Down's syndrome is caused when the original gamete has an extra copy of chromosome 21, meaning the individual contains 3 copies of chromosome 21.

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Explain in detail, how you solved the following problem: The first two terms of a sequence are 10 and 20. If each term after the

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Answer:

$T_{2020} = 15$

Explanation:

Given

$T_1 = 10$

$T_2 = 20$

Each term after the second term is the average of all of the preceding terms

Required:

Explain how to solve the 2020th term

Solve the 2020th term

Solving the 2020th term of a sequence using conventional method may be a little bit difficult but in questions like this, it's not.

The very first thing to do is to solve for the third term;

The value of the third term is the value of every other term after the second term of the sequence; So, what I'll do is that I'll assign the value of the third term to the 2020th term

<em>This is proved as follows;</em>

From the question, we have that "..... each term after the second term is the average of all of the preceding terms", in other words the MEAN

$T_{n} = \frac{\sum T{k}}{n-1} ; where: k = 1 .... n -1$

<em>Assume n = 3</em>

$T_{3} = \frac{T_1 + T_2}{2}$

<em>Multiply both sides by 2</em>

$2 * T_{3} = \frac{T_1 + T_2}{2} * 2$

$2T_{3} = T_1 + T_2$

<em>Assume n = 4</em>

$T_{4} = \frac{T_1 + T_2 + T_3}{3}$

$T_{4} = \frac{(T_1 + T_2) + T_3}{3}$

Substitute $2T_{3} = T_1 + T_2$

$T_{4} = \frac{2T_3 + T_3}{3}$

$T_{4} = \frac{3T_3}{3}$

$T_{4} = T_3$

Assume n = 5

$T_{5} = \frac{T_1 + T_2 + T_3 +T_4}{4}$

$T_{5} = \frac{(T_1 + T_2) + T_3 +(T_4)}{4}$

Substitute $2T_{3} = T_1 + T_2$ and $T_{4} = T_3$

$T_{5} = \frac{2T_3 + T_3 +T_3}{4}$

$T_{5} = \frac{4T_3}{4}$

$T_{5} = \frac{(5-1)T_3}{5-1}$

<em>Replace 5 with n</em>

$T_{n} = \frac{(n-1)T_3}{n-1}$

<em>(n-1) will definitely cancel out (n-1); So, we're left with</em>

$T_{n} = T_3$

Hence,

$T_{2020} = T_3$

Calculating $T_3$

$T_{3} = \frac{10 + 20}{2}$

$T_{3} = \frac{30}{2}$

$T_{3} = 15$

Recall that $T_{2020} = T_3$

$T_{2020} = 15$

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