**Contents**show

## How many gametes can be formed from AABbCC and 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 f1 genotypes of AABbCC AABbCC are there?

AABbCC will produce two type of gametes ABC and AbC. Thus in F, generation **three genotypes** will be obtained. These are AABBCC, AABbCC and AAbbCC in the ratio of 1 : 2 : 1. Phenotypically AABBCC and AABbCC are same.

## What is the genotype of AABbCC?

N is the number of heterozygous gene pairs. For instance, if a genotype is AaBbCc, n will be 3. However, if the genotype is AABbCC, the **n is 1**. When analyzing two parents, it is easier to determine how many possible phenotypic and genotypic combinations of each gene pair, and then multiply them all together.

## What gametes can be produced by AABbCC?

The plant having genotype AABbCC is heterozygous for only one character B. Number of “gametes = 2n”, where n is the heterozygosity.” Since n = 1 so 2 gametes will be formed. Those are **ABC and AbC**. So the two types of gametes will be ABC and AbC.

## 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 **${2^3} = 8$**.

## How many types of gametes are found in F1 progeny of cross between AaBbCc and AaBbCc?

Therefore, types of gametes possible is 23=**8**.

## How many different types of gametes would be produced by an organism of genotype AaBbCcDdEe if all of the genes assort independently?

The number of different possible gametes produced by the diploid genotype (AaBbCcDdEe) is 2 x 2 x 2 x 2 x 2 = **32** (2 for each pair of heterozygous genes).

## How do you calculate genotypes?

Number of genotypes for a given number of alleles Given n alleles at a locus, the number genotypes possible is **the sum of the integers between 1 and n**: With 2 alleles, the number of genotypes is 1 + 2 = 3. 3 alleles there are 1 + 2 + 3 = 6 genotypes. 4 alleles there are 1 + 2 + 3 + 4 = 10 genotypes.

## How many offsprings will be formed in mating AABbcc?

Mating between the genotypes AAbbCCDd x AaBbCcdd produces **16 different types** of offsprings.