LECTURE 3: DISCUSSION ON AIPMT QUESTIONS (MENDEL'S LAW)

Why Mendel was Lucky?

Mendel was lucky in choosing Pea plant because all 7 characters that he chose in Pea plants were located on different chromosomes. Mendel's chosen characteristics did not blend, all characters showed independently. 

No linkage was found in the Mendelian experiment because genes were located on different chromosomes. Linkage occurs only when two or more genes are present on the same chromosome. 


What Mendel Couldn't Answer?

-- How linked pairs of alleles segregate dependently
-- How linkage affects recombination

=> Well, Mendel was not aware of genes and chromosomes because these things were not discovered that time. But what we should appreciate about Mendel's work is that he did all the experiments precisely and became the basis of all further genetic experiments and discoveries. And that is why we call Mendel is the Father of Genetics.

Most of the questions that have been asked in the medical entrance tests (from Mendel's law) include:

a. Find the number of gametes
b. What is the genotypic and phenotypic ratio of Dihybrid and Trihybrid cross
c. What are the different types of progenies formed in F2 generation
d. Test cross ratio


Q1. How many gametes would be formed from Aa Bb Cc?

There are two ways to solve this question:
1. Follow the steps mentioned in the picture
2. Use the formula 2^n

Note: It is always advised to use the first technique.









Q2. In a polygenic cross Aa Bb Cc x Aa Bb Cc, the phenotypic ratio of the offspring is
1 : 6 : X : 20 : X : 6 : 1. What is the value of X?

Q3. How many diferent types of phenotypes would be formed from a cross of RrYy x RrYy?

Q4. A plant of genotype AABbCC is selfed, what would be the phenotypic ratio of F2 generation?

Q5. Cross AABb x aaBb yields AaBB : AaBb : Aabb : aabb in the ratio of:

a. 1:2:1:0

b. 3:1:1:0

c. 1:2:1:1

d. 9:3:3:1

Solution: Cross AABb x aaBb

Form gametes from AABb = AB, Ab

Form gametes from aaBb = aB, ab

Progenies formed = AaBB, AaBb, AaBb, Aabb

So the ratio is 1 : 2 : 1 : 0

Q6. For a trihybrid cross A/a B/b C/c (considering that they are unlinked), what is the chance that a progeny would have a genotype of A/A and b/b and C/c?

A. Well, its a very interesting question, but quite simple as well. If you remember Monohybrid genotypic ratio (TT, Tt, tt) as 1:2:1, we can solve the above question easily.

The actual F2 ratio from the cross between a pure tall and a pure dwarf plant is:

TT: Tt: Tt: tt

Out of 4, TT (homozygous dominant) comes only 1/4 times

Out of 4, Tt (heterozygous) comes two times i.e., 1/2

Out of 4, tt (homozygous recessive) comes only 1/4 times

Now, look at the above question:

We have A/A b/b and C/c

(1/4). (1/4). (1/2) = 1/32