Trisomic conditions is expressed as ____________.

2 n - 1

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2 n + 2

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2 n + 1

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2 n - 2

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Solution

The correct option is C $2 n + 1$
The different conditions of chromosomal variation due to a loss or a gain of one or more chromosomes resulting in the deviation from the normal or the usual number of chromosomes is called aneuploidy, Its types are nullisomy (2n-2), monosomy (2n-1), trisomy (2n+1), and tetrasomy (2n+2). Here n signifies the chromosomal number. Trisomy is characterized by the presence of one extra chromosome.
Thus the correct answer is option C.

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Simplify 16*2n+1-4*2n/16*2n/1*2n+2-2*2n+1

(^{2 n} C_{1})^{2} + 2. (^{2 n} C_{1})^{2} + 3. (^{2 n} C_{3})^{2} + . . . . + 2 n . (^{2 n} C_{2 n})^{2} = \frac{(4 n - 1)!}{{(2 n - 1)!}^{2}}

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 1}

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 2}

1 - 2 n + \frac{2 n (2 n - 1)}{2!} - \frac{2 n (2 n - 1) (2 n - 1)}{3!} + . . . + (- 1)^{n - 1} \frac{2 n (n - 1) . . . (n + 2)}{(n - 1)}

= (- 1)^{n + 1} \frac{(2 n)}{2 (n!)^{2}},

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