Black skin color is dominant to pink skin color in pigs. Two heterozygous black pigs are crossed. a. What is the probability that their offspring will have pink skin? b. What is the probability that the first and second off- spring will have black skin? c. If these pigs produce a total of three piglets, what is the probability that two will be pink and one will be black?

Answer:

[tex]P(Pink) = \frac{1}{4}[/tex]

[tex]P(Both\ Black) = \frac{9}{16}[/tex]

[tex]Probability = \frac{3}{64}[/tex] --- Two pink and One black

Step-by-step explanation:

Represent the pink skin with P and the black skin with B.

Since black skin is dominant to pink skin, there will be more occurrence of B than P

The punnet square for the breeding of the two pigs is then represented as:

[tex]\begin{array}{cc} {B} & {B} \ \\ {B} & {P} \ \ \end{array}[/tex]

Solving (a): Probability of Pink

In the above square, there is only 1 occurrence of P out of a possible of 4.

So, the probability is:

[tex]P(Pink) = \frac{n(P)}{4}[/tex]

[tex]P(Pink) = \frac{1}{4}[/tex]

Solving (b): Probability that first and second are black

In the above square, there is only 3 occurrence of B out of a possible of 4.

[tex]P(Black) = \frac{n(B)}{4}[/tex]

[tex]P(Black) = \frac{3}{4}[/tex]

So, the probability that both are black is:

[tex]P(Both\ Black) = P(Black) * P(Black)[/tex]

[tex]P(Both\ Black) = \frac{3}{4} * \frac{3}{4}[/tex]

[tex]P(Both\ Black) = \frac{9}{16}[/tex]

Solving (c): Probability of two pink and 1 black

This is calculated as:

[tex]Probability = P(Pink) * P(Pink) * P(Black)[/tex]

[tex]Probability = \frac{1}{4} * \frac{1}{4} * \frac{3}{4}[/tex]

[tex]Probability = \frac{3}{64}[/tex]

A) The probability that their offspring will have pink skin is; ¹/₄

B) The probability that the first and second off- spring will have black skin is;

9/16  or 56.25%

C) The probability that two will be pink and one will be black  If the pigs produce a total of three piglets is; ³/₆₄ or 4.6875%

To answer this question, we need to first understand Mendel's law of dominance. This law states that in a heterozygote, one trait will conceal the presence of another trait for the same characteristic. This means that the dominant trait always conceal the recessive trait.

Now in this question, we are told that;

Let's denote the black color gene by ‘P’ and pink colour by ‘p’. Therefore, if cross is held between these two genes then according to mendel's law, the heterozygous pigs is specified as "Pp"

Now when the heterozygous pigs specified as "Pp" are crossed, we have;

Pp × Pp = PP, Pp, Pp, pp

where;

(a) From the 4 cross possibilities above;

Probability that their offspring will have pink skin = ¹/₄

(b) From the 4 cross possibilities gotten earlier;

Probability that their offspring will have black skin = ³/₄

Thus;

Probability that the first and second off- spring will have black skin is;

³/₄ × ³/₄ = 9/16  or 56.25%

(c) We are told that the pigs produced a total of three piglet. Thus, the probability that two will be pink and one black will be black is;

P(2 pink and 1 black) =   ¹/₄ × ¹/₄ × ³/₄ = ³/₆₄ or 4.6875%

Read more about Mendel's law of dominance at; https://brainly.com/question/14436268

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