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p^(2)+2pq+q^(2)=1
In population genetics, the HardyWeinberg law, shown, describes the relative frequency of two alleles,
p and
q. The values of
p and
q are always between 0 and 1 . Which of the following correctly expresses
p in terms of
q ?
Choose 1 answer:
(A)
(p+q)^(2)=1
(B)
(p+q)^(2)=0
(C)
p=-q+1
(D)
p=-q

$p^{2}+2 p q+q^{2}=1$ $\newline$ In population genetics, the HardyWeinberg law, shown, describes the relative frequency of two alleles, $p$ and $q$ . The values of $p$ and $q$ are always between $0$ and $1$ . Which of the following correctly expresses $p$ in terms of $q$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $(p+q)^{2}=1$ $\newline$ (B) $(p+q)^{2}=0$ $\newline$ (C) $p=-q+1$ $\newline$ (D) $p=-q$

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Compare linear and exponential growth

Full solution

Q.

p^{2}+2 p q+q^{2}=1

\newline

In population genetics, the HardyWeinberg law, shown, describes the relative frequency of two alleles,

p

and

q

. The values of

p

and

q

are always between

0

and

1

. Which of the following correctly expresses

p

in terms of

q

?

\newline

Choose

1

answer:

\newline

(A)

(p+q)^{2}=1

\newline

(B)

(p+q)^{2}=0

\newline

(C)

p=-q+1

\newline

(D)

p=-q

Equation Verification: The Hardy-Weinberg law equation is given by $p^2 + 2pq + q^2 = 1$ . This equation resembles the expansion of $(p + q)^2$ . Let's verify if this is the case.
Expansion Comparison: We know that $(p + q)^2$ expands to $p^2 + 2pq + q^2$ . This matches the given Hardy-Weinberg law equation. Therefore, we can write the equation as $(p + q)^2 = 1$ .
Square Root Calculation: Since the values of $p$ and $q$ are always between $0$ and $1$ , we can take the square root of both sides of the equation to solve for $p + q$ . The square root of $1$ is $1$ , and since $p$ and $q$ are non-negative, we do not need to consider the negative square root. So, we have $p + q = 1$ .
Expressing $p$ in terms of $q$ : Now, we can express $p$ in terms of $q$ by rearranging the equation $p + q = 1$ to solve for $p$ . Subtracting $q$ from both sides gives us $p = 1 - q$ .

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