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3 years ago

In the term 2pq of the Hardy-Weinberg equation. Why is the multiplication by two necessary?

Biology

1 answer:

Gnesinka [82]3 years ago

3 0

Answer:

The correct answer is : option B.

Explanation:

Hardy–Weinberg equilibrium is a mathematical model that states that genotype percentage or frequency in a population and allele in a population remains stable or constant to generations if there are no evolutionary influences present.

Hardy Weinberg equation is :

P²+2pq+q² = 1

Here, p² represents the frequency of genotypes dominant homozygotes (AA) in a population in equilibrium. The term 2pq represents the frequency of heterozygotes (Aa) as there are two ways of developing heterozygous individuals either dominant from the egg and recessive from sperm or vice versa. and the term q² represents the frequency of recessive homozygotes (aa).

Thus, the correct answer is : option B.

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Mountain ranges form along ___ plate boundaries via orogenic processes.

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2 years ago

A certain buffer is made by mixing a weak acid HA and its conjugate base A–. When HA is present at a concentration of 0.5 mM and

MAVERICK [17]

Answer:The ratio of the concentrations of $HA$ and $A^-$ when the buffer has a pH of 7.02 is 0.69

Explanation:

The dissociation constant for formic acid = $K_a=1.8\times 10^{-4}$

Concentration of HA= 0.5 mM

Concentration of $A^-$ = 0.1 mM

pH = 6.16

First we have to calculate the value of $K_a$ .

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[A^-}{[HA]}$

Now put all the given values in this expression, we get:

$6.16=pK_a+\log (\frac{[0.1]}{0.5})$

$pK_a=6.86$

To calculate the ratio of the concentrations of $HA$ and $A^-$ when the buffer has a pH of 7.02.

Using Henderson Hesselbach equation :

$7.02=6.86+\log \frac{[A^-]}{[HA]}$

$\frac{[A^-]}{[HA]}=1.44$

$\frac{[HA]}{[A^-]}=0.69$

Thus the ratio of the concentrations of $HA$ and $A^-$ when the buffer has a pH of 7.02 is 0.69

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3 years ago

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galina1969 [7]

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3 years ago