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## How do you determine the number of different genotypes?

The number of different genotypes is **3 ^{n} where n = number of genes**. For simple dominant–recessive relations, the number of different phenotypes is 2

^{n}, where n = number of genes.

## How many different gametes are in 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 unique gametes are produced?

The correct answer is (b) **8**. In this scenario, each gamete will consist of five genes.

## How many different types of gametes can be formed by individuals of the genotype?

An individual with the genotype Aa can make **two types** of gametes: A and a. Since this is a random process, the individual will, on average, make equal numbers of each gamete. (Frequency of 1/2 for each gamete). A second individual with genotype Aa can make two types of gametes: A and a.

## What is the number of phenotypes possible from AaBbCc AaBbCc?

There is a formula to know the number of possible phenotypes from two parents which is ${2^n}$ in which n describes the number of pairs of characters. In the above cross$AaBbCc times AaBbCc$, there are three pairs of characters which are Aa, Bb, Cc. Therefore, the number of possible phenotypes are **${2^3} = 8$**.