In a Cc x Cc crossing of peas, 5 offspring are of genotype CC, 12 are of genotype Cc, and 6 are of genotype cc. What is the probability of this event?
To determine the probability of this event, we need to use the principles of Mendelian genetics.
In a Cc x Cc crossing, where both parents are heterozygous (Cc), we can create a Punnett square to understand the possible outcomes. The capital "C" represents the dominant allele for a certain trait, while the lowercase "c" represents the recessive allele.
The Punnett square for a Cc x Cc cross will look like this:
| C | c
--------------
C | CC | Cc
--------------
C | Cc | cc
From the Punnett square, we can see that there are four possible combinations of alleles: CC, Cc, Cc, and cc.
Given that 5 offspring are CC, 12 are Cc, and 6 are cc, we can calculate the probability of each genotype. The probability can be determined by dividing the number of occurrences of each genotype by the total number of offspring.
Total number of offspring = 5 (CC) + 12 (Cc) + 6 (cc) = 23
Probability of CC genotype = 5 / 23 ā 0.217
Probability of Cc genotype = 12 / 23 ā 0.522
Probability of cc genotype = 6 / 23 ā 0.261
Therefore, the probability of obtaining 5 offspring with genotype CC, 12 offspring with genotype Cc, and 6 offspring with genotype cc in a Cc x Cc crossing is approximately 0.217, 0.522, and 0.261, respectively.