1664 , or 25% of their offspring are expected to have the same phenotype as parent 2. Their genotypes are AabbCC and AabbCc .

## What is the probability that a cross between genotypes AaBbCc and AaBbCc would produce on offspring with genotype AaBbCc?

b) What is the probability that any of the offspring will have the genotype AabbCc? The probability off finding an offspring having the genotype AABBCC, will be **1/64 or 0.0156**, since the probability of finding AA,BB,CC from an individual monohybrid cross would all be 1/4.

## How many offspring will be formed in mating between AaBbCc and AaBbCc?

A total of **8 gametes** will form here. Now when we will perform a trihybrid cross using Punnett square. There are 8 different phenotypes for the given cross. And there are 27 different genotypes.

## How many gametes can be produced from AaBbCc?

Hints For Biology 101 Exam #4

No. of homologous chromosome pairs (heterozygous genes) | No. of different gametes from each parent |
---|---|

1 (Aa X Aa) | 2 (2^{1}) |

2 (AaBb X AaBb) | 4 (2^{2}) |

3 (AaBbCc X AaBbCc) | 8 (2^{3}) |

4 (AaBbCcDd X AaBbCcDd) | 16 (2^{4}) |

## How many genotypes are possible in a trihybrid cross?

With three unlinked genes, each parent can produce 8 different types of gametes, which generates **64 possible genotypic** combinations in the Punnett Square.

## How can you tell how many gametes a genotype can produce?

The number of gametes produced by a specific genotype is calculated by **the formula 2n**, where n= number of heterogeneous alleles present in the genotype. Here, the given genotype consists of two heterogeneous alleles Aa and Bb while CC is homozygous. So, it can produce 22 = 4 types of gametes.