How to calculate number of unique gametes

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.

 1. How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCdd?

Possible no. of gametes= 2n  ’n’ is the no. of heterozygous gene pairs

AaBbCCdd

No. of heterozygous gene pairs n=2

Possible no. of gametes= 2n

Answer: 2n  =22=2x2=4

You can watch this video for better understanding

2. How many different possible gametes are produced by the diploid genotype aaBbCC ?

aaBbCC

Possible no. of gametes= 2n  ’n’ is the no. of heterozygous gene pairs

No. of heterozygous gene pairs n=1

Answer: 21  =2

3. How many different possible gametes are produced by the diploid genotype (AaBbCcDdEe)?

AaBbCcDdEe

n=5

Answer: 2n  =25= 2 x 2 x 2 x 2 x 2 = 32

4.For the following genotypes, How many gametes will be produced?

a) AA

b)Bb

c)BBCc

d)cCDd

Answers:

Possible no. of gametes= 2n  ’n’ is the no. of heterozygous gene pairs

a) AA     2n =20 =1; the only gamete is A

b)Bb      2n =21 =2; the gametes are B and b

c)BBCc  2n =21 =2; the gametes are BC and Bc

d)cCDd 2n =22 =4; the gametes are cD,CD,Cd and Cd

Thank you so much:)