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## What is the probability of producing the genotype AaBbCc in a cross of individuals who both possess this genotype AaBbCc group of answer choices?

In a cross AaBbCc times AaBbCc, what is the probability of producing the genotype AABBCC? – 1/32.

## What is the probability of producing the genotype AaBbCc?

Last part of the given genotype-aaBbcc is ‘cc’. Now, let us look into a cross between parental parts cc and Cc. So, the probability of getting a homozygote (cc) is 2/4 i.e. 1/2. So, the correct answer is ‘1/8.

## What is the probability that the cross of AaBbCc with AaBbCc will produce an offspring that is pure breeding?

the probability of getting this unique offspring is going to be 1/64.

## Which is a possible genotype for a gamete produced by an organism with the 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 fraction of the offspring of AaBbCc and AaBbCc parents will have the same genotype as the second parent?

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

## Which of the following is true about a plant with the genotype AaBbCc?

Which of the following is true about a plant with the genotype AABbcc? It is homozygous at two loci.

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

Therefore, the ratio of hybrid for the three genes should be 1/8.

## How many gametes are formed by 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}) |

## What is genotypic ratio of Trihybrid cross?

a trihybrid cross yields a phenotypic ratio of 27:9:9:9:3:3:3:1. This reflects the phenotypes generated by the 64 genotypic combinations resulting from 8 different male gametes fertilizing 8 different female gametes.

## How many offsprings will be formed in the 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 many phenotypes are produced in a test cross of AaBbCc?

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 ${2^3} = 8$. Hence, the correct option is C. 8.

## Which of the following individuals can produce 16 different gametes?

Question : Which one of the following individuals can produce 16 different gametes?

- A. Aa Bb cc Dd.
- B. Aa Bb cc DD Ee Ff.
- C. Aa Bb Cc dd EE FF.
- D. Aa Bb Cc DD Ee Ff.
- Answer. B.
- It is hexahybrid and, therefore, number f gametes are (2)6 but cc and DD are homozygous and hence number of gametes will be (2)6-2=(2)4=16.

## How many genotypes are possible with 4 alleles?

4 alleles there are 1 + 2 + 3 + 4 = 10 genotypes.