Genetics Solving — Concepts, Formulas & Examples

Genetics numericals look scary but are predictable — there are only about six question types and each has a recipe. This topic is the recipe book. Master it and genetics becomes scoring rather than stressful.

Core Concepts

Monohybrid problem recipe

Identify parents’ genotypes. Write gametes. Combine in Punnett square. Read ratios. Double-check by counting — F2 should have 4 boxes summing to any integer multiple.

Worked example: In pea plants, tall (T) is dominant over dwarf (t). Cross Tt × Tt.

Gametes from each parent: T and t.

	T	t
T	TT	Tt
t	Tt	tt

Genotypic ratio: 1 TT : 2 Tt : 1 tt Phenotypic ratio: 3 tall : 1 dwarf

F1 (Tt × Tt): Phenotypic = 3:1, Genotypic = 1:2:1
Test cross (Tt × tt): Phenotypic = 1:1, Genotypic = 1:1
If F2 gives all dominant phenotype → parent was homozygous (TT)
If F2 gives 1:1 ratio → parent was heterozygous (Tt) — this IS a test cross

Dihybrid shortcut

Instead of a 16-box Punnett, use branch diagram — for each trait compute the 3:1 ratio separately, then multiply. $(3:1) \times (3:1) = 9:3:3:1$ . Much faster.

Let us break down why this works. If genes are on different chromosomes (independent assortment), the probability of each trait is independent. So:

Round and Yellow: $3/4 \times 3/4 = 9/16$
Round and Green: $3/4 \times 1/4 = 3/16$
Wrinkled and Yellow: $1/4 \times 3/4 = 3/16$
Wrinkled and Green: $1/4 \times 1/4 = 1/16$

Total = 9:3:3:1, exactly matching the 16-box Punnett square but computed in 30 seconds.

The branch method works for any number of genes. For a trihybrid: $(3:1)^3 = 27:9:9:9:3:3:3:1$ . Nobody draws a 64-box Punnett square in an exam — use the branch method.

Pedigree analysis

Look for the inheritance pattern first. Autosomal dominant — trait appears every generation, affects both sexes equally. Autosomal recessive — skips generations. X-linked recessive — more males affected, father to son never, grandfather to grandson through unaffected mother.

Step-by-step pedigree decoding:

If affected parents have unaffected children → trait is dominant. If unaffected parents have affected children → trait is recessive. Most genetic disorders are recessive.

If roughly equal numbers of males and females are affected → autosomal. If mostly males are affected and affected fathers never have affected sons → X-linked recessive.

Start with affected individuals (their genotype is known — aa for autosomal recessive, $X^a Y$ for X-linked). Work outward to parents and siblings using the inheritance rules.

Probability chain

Probability of a specific offspring genotype = product of probabilities of each parent giving that allele. For example, in Aa × Aa, probability of aa is $1/2 \times 1/2 = 1/4$ .

Product rule: Probability of A AND B = P(A) × P(B) (for independent events)

Sum rule: Probability of A OR B = P(A) + P(B) (for mutually exclusive events)

Example: In Aa × Aa, what is the probability of the first child being Aa? P(A from father) × P(a from mother) + P(a from father) × P(A from mother) = $1/2 \times 1/2 + 1/2 \times 1/2 = 1/2$

Incomplete dominance

When neither allele is fully dominant. The heterozygote shows an intermediate phenotype. Classic example: snapdragons — RR (red) × WW (white) → RW (pink).

The F2 ratio becomes 1:2:1 phenotypically (not 3:1), because the heterozygote is distinguishable from both homozygotes.

Codominance

Both alleles express fully in the heterozygote. ABO blood groups are the best example: $I^A I^B$ gives blood type AB — both A and B antigens are expressed on the red blood cell surface.

Genotype	Blood Type	Antigens	Antibodies
$I^A I^A$ or $I^A i$	A	A	Anti-B
$I^B I^B$ or $I^B i$	B	B	Anti-A
$I^A I^B$	AB	A and B	None
$ii$	O	None	Anti-A and Anti-B

Blood type AB is codominant. Blood type O is homozygous recessive.

Worked Examples

If a mother is type A and father is type B and a child is type O, both parents must be heterozygous ( $I^A i$ and $I^B i$ ). The child is $ii$ . Cross gives four equally likely outcomes — A, B, AB, O.

Queen Victoria was a carrier ( $X^H X^h$ ). Her sons had 50% chance of being hemophilic. Her daughters had 50% chance of being carriers. Classic X-linked recessive pattern.

Parents are both Aa. Probability of one child being aa = 1/4. Probability of both children being aa = $1/4 \times 1/4 = 1/16$ . We multiply because the two births are independent events.

If a plant showing both dominant traits (say Round Yellow) is test-crossed with a double recessive (wrinkled green), and the offspring show a 1:1:1:1 ratio, the unknown parent was RrYy (heterozygous for both). If the offspring are all Round Yellow, the parent was RRYY.

Six Question Types — Recognition Guide

Pattern in Question	Type	Method
Cross two organisms, predict offspring ratio	Monohybrid/dihybrid	Punnett square or branch
Given F2 ratio, determine parent genotypes	Reverse genetics	Match ratio to known cross
Pedigree chart given, find inheritance pattern	Pedigree analysis	Check dominant/recessive, autosomal/X-linked
What is probability of child having X?	Probability	Product and sum rules
Blood group of child/parent	Codominance	ABO genotype table
Colour of snapdragon flower	Incomplete dominance	1:2:1 ratio, intermediate phenotype

Common Mistakes

Applying dihybrid ratio blindly when genes are linked. Check if the problem states independent assortment.

Forgetting that pedigree squares are males and circles are females, filled shapes are affected.

Adding probabilities when you should multiply. For sequential events, multiply.

Confusing carrier and affected. A carrier ( $X^H X^h$ ) is phenotypically normal but carries one copy of the recessive allele. An affected individual ( $X^h X^h$ or $X^h Y$ ) shows the disease.

Saying a type O parent and type AB parent can have a type O child. Not possible. AB parent always gives either $I^A$ or $I^B$ — the child will be either type A or type B.

Exam Weightage and Revision

Genetics numericals appear in NEET every year — expect 2-3 questions from Mendelian genetics and inheritance patterns. CBSE Class 12 boards allocate 5-8 marks. Blood group inheritance and pedigree analysis are the two most predictable question types.

Question Type	NEET Frequency	Difficulty
Monohybrid cross	Every year	Easy
Dihybrid ratio	Most years	Medium
Blood group problem	Every 2 years	Medium
Pedigree chart	Every 2 years	Medium-Hard
Probability calculation	Occasional	Medium
Incomplete/codominance	Most years	Easy

If you can solve monohybrid crosses and blood group problems confidently, that covers 60% of NEET genetics questions. Add pedigree analysis and you cover 80%.

Practice Questions

Q1. In a cross between Tt (tall) and tt (dwarf), what fraction of offspring will be tall?

This is a test cross. Gametes: T and t from Tt parent; only t from tt parent. Offspring: Tt (tall) and tt (dwarf) in a 1:1 ratio. So 1/2 or 50% of offspring will be tall.

Q2. A couple has blood types A and B. Can they have a child with blood type O?

Only if both parents are heterozygous: mother is $I^A i$ and father is $I^B i$ . Then the cross can produce $ii$ (type O) with probability 1/4. If either parent is homozygous ( $I^A I^A$ or $I^B I^B$ ), a type O child is impossible.

Q3. In a pedigree, an affected father and carrier mother have children. What fraction of sons will be affected if the trait is X-linked recessive?

Father: $X^a Y$ (affected). Mother: $X^A X^a$ (carrier). Sons get Y from father and either $X^A$ or $X^a$ from mother. So sons: 1/2 are $X^A Y$ (normal), 1/2 are $X^a Y$ (affected). Answer: 1/2 or 50%.

Q4. A pink-flowered snapdragon (RW) is crossed with a white-flowered snapdragon (WW). What are the expected phenotypes?

This is incomplete dominance. Cross: RW × WW. Offspring: RW (pink) and WW (white) in 1:1 ratio. No red flowers appear because no RR genotype is possible from this cross.

FAQs

Why does the 3:1 ratio work only for large numbers of offspring?

The 3:1 ratio is a statistical probability. With 4 offspring, you might get 4 tall and 0 dwarf by chance. With 1000 offspring, the observed ratio will be very close to 3:1. This is why Mendel used thousands of pea plants — small samples give unreliable ratios.

How do linked genes affect the dihybrid ratio?

Linked genes (on the same chromosome) do not assort independently. Instead of 9:3:3:1, you get mostly parental combinations with a few recombinants (due to crossing over). The recombination frequency tells you how far apart the genes are on the chromosome.

Can two parents with blood type A have a child with blood type O?

Yes, if both parents are heterozygous ( $I^A i$ ). Cross: $I^A i \times I^A i$ gives $I^A I^A : I^A i : I^A i : ii$ = 3 type A : 1 type O.

Keep a recipe card — one line per question type. Before starting, identify which recipe applies.

Genetics solving is pattern recognition plus arithmetic. Spot the pattern, apply the formula, check by counting.