Using multiplication, we can calculate (1) the number of possible allele combinations for a given cross, and (2) the probability of an offspring having a particular allele combination. 1. First, we’ll apply math to the sex chromosome example. Here’s the math for calculating the number of possible combinations: - The father can make sperm with 2 possible sex chromosomes: Z or the other Z.
- The mother can make eggs with 2 possible sex chromosomes: Z or W.
- Multiplying these numbers together gives us 4 possible offspring.
2. Here’s how to calculate the probability of the parents making a female offspring: - In the father, 2 of the 2 possible sex chromosomes (Z or Z) will contribute to making a female offspring.
- In the mother, 1 of the 2 possible sex chromosomes (just W) will contribute to making a female offspring.
Notice that the denominator also tells you the number of possible combinations. To see where these numbers came from and to check the calculation, you can look back at the first Punnet square near the top of the page. 3. We can also add crest to our calculations. We’ll jump straight to the probability calculation, since we know that will also tell us the number of possible combinations. a. To make a non-crested male, the father must contribute a Z chromosome and a ‘crest’ allele. 2 out of the 2 possible sex chromosomes will give us what we need. And 2 out of the 2 possible crest alleles will give us the desired offspring. Multiplying those numbers, we get 4 out of 4. Out of the 4 possible allele combinations in the sperm (denominator), 4 will be Z + ‘crest’ (numerator). b. The mother must contribute a Z chromosome and a ‘no crest’ allele. Just 1 out of the 2 possible sex chromosomes will give us what we need. Likewise, just 1 out of the 2 possible crest alleles will give us the desired offspring. Multiplying those numbers, we get 1 out of 4. Out of the 4 possible allele combinations in the egg, 1 will be Z + ‘no crest.’ c. To get our final number, we multiple the gamete fractions together. Out of 16 possible allele combinations in the offspring, 4 will be non-crested males (ZZ and ‘crest’ ‘no crest’). Our Punnett square above gave us the same values. Calculate these values for more-complex problems is simply a matter of adding another fraction to the multiplication problem. Tutorial
## Page 2In a dihybrid cross, AaBb x AaBb, what fraction of the offspring will be homozygous for both recessive traits?
The Biology Project University of Arizona Tuesday, August 13, 1996 Contact the Development Team http://www.biology.arizona.edu |