What is the possibility to get AaBbCc when AaBbCc is crossed with AaBbCc?

1/8
So, the correct answer is ‘1/8. ‘

How many types of phenotypes will develop from AaBbCc X AaBbCc cross?

8
Hence, the correct option is C. 8.

What is the phenotypic ratio of AaBbCc X AaBbCc?

Solution : In a polygenic trihybrid cross `(“AaBbCc” xx “AaBbCc”)` the phenotypic ratio is 1:6:15:20:15:6:1.

How many gametes does AaBbCc produce?

two types
Explanation: A plant having genotype AABbCC will produce only two types of gametes, ABC and AbC.

How many offsprings will be formed in the mating of AaBbCc?

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

How many types of phenotypes would be produced in F1 progeny AaBbCc AaBbCc?

The frequency for production of ABC gamete is 1. Genotype AaBbCc can produce 8 types of gametes, i.e., ABC, Abc, ABc, AbC, abc, aBC, aBc and abC.

How many gametes are formed by AaBbCc?

Explanation: A plant having genotype AABbCC will produce only two types of gametes, ABC and AbC.

How many organisms with dominant trait are obtained when AaBbCc * AaBbCc are crossed?

So, the correct option is ‘8’.

What is the genotype of AaBbCc?

The plant having the genotype AaBbCc is a triploid plant. To calculate the total number of gametes that are produced by a particular genotype, a specific formula 2n is used, where n= number of heterogeneous alleles that are found in the genotype.

How many type of gametes can be produced by YyRr?

After fertilisation, the F1 heterozygote will be formed i.e YyRr (yellow and round). On selfing F1 we get two possible combinations i.e YR, Yr, yR, yr. Therefore four types of gametes are formed in the ratio of 1:1:1:1. Thus, the correct answer is option D.

How many different genotypes can be produced from a cross between parents with genotype AaBbCc?

So, the correct answer is ’27’.

How do you make a Trihybrid cross?

How to do a trihybrid cross Punnett square?

  1. A giant, 8×8 table of results;
  2. 64 possible sets of crosses in each try;
  3. 27 possible genotypes;
  4. 8 possible mother’s alleles combinations;
  5. 8 possible father’s alleles combinations; and.
  6. 729 possible Punnett square trihybrid cross examples!