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## How many gametes are produced by genotype AaBbCc?

These types of genotypes that are ABC, ABc, AbC, and Abc will be found in the gametes. Thus the number of gametes produced by the plant having the genotype AABbCc is **four**.

## What are the possible gametes for 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.

## How many phenotypes are possible AaBbCc?

In the above crossAaBbCc×AaBbCc, there are three pairs of characters which are Aa, Bb, Cc. Therefore, the number of possible phenotypes are 23=**8**. Hence, the correct option is C. 8.

## When AaBbCc is crossed with AaBbCc then the ratio of hybrid for all the three?

The ratio of hybrid for all three genes for AaBbCc is **18**. Hence, the correct answer is option (A).

## How many gametes are produced in F1 generation of a Trihybrid?

The F1 hybrid produces **8 types** of gametes.

## What type of gamete is formed by genotype RrYy?

A plant with genotype RrYy, produce **four** types of gametes during fertilization. They are RY, Ry, rY, ry. So, the correct option is ‘four’.

## How many offsprings will be formed in mating between AaBbCc * AaBbCc?

Mating between the genotypes AAbbCCDd x AaBbCcdd produces **16 different types of offsprings**. Types of gametes formed are determined by 2ⁿ where n stands for degree of heterozygosity.

## How do you determine the number of phenotypes and 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. d.