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b) If one parent seal is pure long-whiskered and the other is short-whiskered, what percent of offspring would have short whiskers? ANSWER: 0%. "Pure" is the same as homozygous, so "pure long-whiskered" would be "WW".

Predicting the genotype of offspring There are four possible combinations of gametes for the AaBb parent. Half of the gametes get a dominant A and a dominant B allele; the other half of the gametes get a recessive a and a recessive b allele. Both parents produce 25% each of AB, Ab, aB, and ab.

A dominant allele is denoted by a capital letter (A versus a). Since each parent provides one allele, the possible combinations are: AA, Aa, and aa. Offspring whose genotype is either AA or Aa will have the dominant trait expressed phenotypically, while aa individuals express the recessive trait.

To calculate the probability of getting a Bb genotype, we can draw a 4-square Punnett square using the parents' alleles for the coat color gene only, as shown above. Using the Punnett square, you can see that the probability of the Bb genotype is 1 / 2 1/2 1/2 .

II. For two heterozygote parents (Aa), 1/4 of all offspring would be expected to show the recessive trait of albinism.

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The P cross produces F1 offspring that are all heterozygous for both characteristics. The resulting 9:3:3:1 F2 phenotypic ratio is obtained using a Punnett square. In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y).

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The Punnett square below makes it clear that at each birth, there will be a 25% chance of you having a normal homozygous (AA) child, a 50% chance of a healthy heterozygous (Aa) carrier child like you and your mate, and a 25% chance of a homozygous recessive (aa) child who probably will eventually die from this ...