Suppose that we are dealing with four genes, each gene consisting of a dominant allele (capital letter) and a recessive allele (small case letter). if the cross ccmmllpp x ccmmllpp is made, what is the probability of obtaining an individual who is ccmmllpp?

1/2 or 1/4 or 1/8 or 1/16 or 1/32

The probability of obtaining an individual who is CcmmLLPp 1/32.

Explanation: This can be achieved by crossing similar genes to obtain individual probability as shown in the attached image.

When you cross Cc in the first genotype with CC in the second genotype, the following probabilities will be obtained;

P (CC) = ¾, P(Cc) = ¼.

Similarly, crossing Mm with mm, we get;

P (Mm) = ½, P (mm) = ½

Crossing Ll with Ll, we get

P (LL) = ¼, P (Ll) = ½, P (ll) = ¼

Crossing PP with pp, we get

P (Pp) = 1

Therefore, the probability of individual with genotype CcmmLLPp will be;

P (Cc) x P (mm) x P (LL) x P (Pp)

= ¼ x ½ x ¼ x 1

= 1/32

thats not even an aswer choice