Mendelian Genetics

Laws of inheritance, Punnett squares, and probability

🌱 Mendelian Genetics

Terminology

Gene: Unit of heredity, codes for trait Allele: Alternative version of gene Dominant: Allele that masks other (uppercase, e.g., A) Recessive: Allele that is masked (lowercase, e.g., a) Homozygous: Same alleles (AA or aa) Heterozygous: Different alleles (Aa) Genotype: Genetic makeup (e.g., Aa) Phenotype: Physical appearance (e.g., tall)

Mendel's Laws

Law of Segregation

Each parent has two alleles for each gene
Alleles separate during gamete formation
Each gamete gets one allele
Fertilization restores pairs

Law of Independent Assortment

Genes for different traits assort independently
Applies to genes on different chromosomes
Exception: linked genes on same chromosome

Monohybrid Cross

One trait considered

Example: Tall (T) × Short (t) pea plants

P generation: TT × tt
F₁ generation: All Tt (100% tall)
F₁ × F₁: Tt × Tt
F₂ generation: 1 TT : 2 Tt : 1 tt
- Genotypic ratio: 1:2:1
- Phenotypic ratio: 3:1 (3 tall : 1 short)

Dihybrid Cross

Two traits considered

Example: Round Yellow (RRYY) × Wrinkled Green (rryy)

F₁: All RrYy (round, yellow)
F₂ (RrYy × RrYy): 9:3:3:1 ratio
- 9 Round Yellow
- 3 Round Green
- 3 Wrinkled Yellow
- 1 Wrinkled Green

Testcross

Purpose: Determine unknown genotype

Method: Cross with homozygous recessive (tt)

If offspring all dominant phenotype → unknown is TT
If offspring 1:1 ratio → unknown is Tt

Probability Rules

Product Rule (AND)

Probability of independent events together
Multiply probabilities
Example: Probability of Tt AND Yy?
- P(Tt) = 1/2, P(Yy) = 1/2
- P(Tt AND Yy) = 1/2 × 1/2 = 1/4

Sum Rule (OR)

Probability of alternative events
Add probabilities
Example: Probability of TT OR Tt?
- P(TT) = 1/4, P(Tt) = 1/2
- P(TT OR Tt) = 1/4 + 1/2 = 3/4

Pedigree Analysis

Pedigree: Family tree showing trait inheritance

Symbols:

Circle = female
Square = male
Filled = affected
Half-filled = carrier
Horizontal line = mating
Vertical line = offspring

Determining inheritance pattern:

Dominant: appears in every generation
Recessive: skips generations, affected children from unaffected parents

Key Concepts

Dominant alleles mask recessive alleles
Segregation: alleles separate during gamete formation
Independent assortment: different genes assort independently
Monohybrid F₂: 3:1 phenotypic ratio
Dihybrid F₂: 9:3:3:1 phenotypic ratio
Testcross: reveals unknown genotype
Product rule: multiply probabilities (AND)
Sum rule: add probabilities (OR)

📚 Practice Problems

1Problem 1medium

❓ Question:

In pea plants, tall (T) is dominant over short (t), and yellow seeds (Y) are dominant over green seeds (y). Cross a heterozygous tall, heterozygous yellow plant (TtYy) with a short, green plant (ttyy). (a) Set up a Punnett square, (b) determine the phenotypic ratio, and (c) calculate the probability of getting a tall plant with green seeds.

💡 Show Solution

Given:

Parent 1: TtYy (tall, yellow - heterozygous for both)
Parent 2: ttyy (short, green - homozygous recessive)
T = tall (dominant), t = short
Y = yellow (dominant), y = green

(a) Punnett Square Setup:

Testcross (heterozygote × homozygous recessive)

Parent 1 gametes (TtYy): TY, Ty, tY, ty (4 types) Parent 2 gametes (ttyy): ty only (1 type)

Punnett Square:

         |  ty
    -----|--------
    TY   | TtYy (tall, yellow)
    -----|--------
    Ty   | Ttyy (tall, green)
    -----|--------
    tY   | ttYy (short, yellow)
    -----|--------
    ty   | ttyy (short, green)

Offspring genotypes:

TtYy: tall, yellow
Ttyy: tall, green
ttYy: short, yellow
ttyy: short, green

(b) Phenotypic Ratio:

Count each phenotype:

Tall, yellow: 1 (TtYy)
Tall, green: 1 (Ttyy)
Short, yellow: 1 (ttYy)
Short, green: 1 (ttyy)

$\boxed{\text{Phenotypic ratio: } 1:1:1:1}$

Or grouped:

Tall: 2 (50%)
Short: 2 (50%)
Yellow: 2 (50%)
Green: 2 (50%)

(c) Probability of tall with green seeds:

Phenotype: Tall, green Genotype: Ttyy

From Punnett square: 1 out of 4 offspring

$P(\text{tall, green}) = \frac{1}{4}$

$\boxed{\text{Probability} = 25\% \text{ or } 0.25}$

Alternative calculation (using multiplication rule):

$P(\text{tall}) = P(T-) = \frac{1}{2}$ $P(\text{green}) = P(yy) = \frac{1}{2}$

Since traits assort independently: $P(\text{tall AND green}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Key Concept - Testcross:

Crossing with homozygous recessive reveals genotype of unknown
Used to determine if organism is homozygous or heterozygous
Mendel used this to confirm his laws

Mendel's Laws Applied:

Law of Segregation: Each gamete gets one allele per gene
Law of Independent Assortment: T/t segregates independently of Y/y

2Problem 2hard

❓ Question:

In a dihybrid cross (AaBb × AaBb), where both traits show complete dominance: (a) determine the phenotypic ratio of offspring, (b) calculate the probability of offspring being homozygous for both traits, and (c) what fraction of the dominant phenotype offspring are homozygous dominant for at least one trait?

💡 Show Solution

Given: AaBb × AaBb (dihybrid cross)

Both traits show complete dominance
A and B are dominant alleles

(a) Phenotypic Ratio:

Classic 16-square Punnett square:

Gametes from each parent: AB, Ab, aB, ab (each ¼ probability)

        AB      Ab      aB      ab
   |-------|-------|-------|-------|
AB | AABB  | AABb  | AaBB  | AaBb  |
   | A-B-  | A-B-  | A-B-  | A-B-  |
   |-------|-------|-------|-------|
Ab | AABb  | AAbb  | AaBb  | Aabb  |
   | A-B-  | A-bb  | A-B-  | A-bb  |
   |-------|-------|-------|-------|
aB | AaBB  | AaBb  | aaBB  | aaBb  |
   | A-B-  | A-B-  | aaB-  | aaB-  |
   |-------|-------|-------|-------|
ab | AaBb  | Aabb  | aaBb  | aabb  |
   | A-B-  | A-bb  | aaB-  | aabb  |

Phenotype counts:

A-B- (both dominant): 9
A-bb (A dominant, b recessive): 3
aaB- (a recessive, B dominant): 3
aabb (both recessive): 1

$\boxed{\text{Phenotypic ratio: } 9:3:3:1}$

This is Mendel's classic dihybrid ratio!

(b) Probability of homozygous for both traits:

Homozygous for both means: AABB or AAbb or aaBB or aabb

From Punnett square:

AABB: 1/16
AAbb: 1/16
aaBB: 1/16
aabb: 1/16

$P(\text{homozygous both}) = \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16}$

$\boxed{P = \frac{4}{16} = \frac{1}{4} = 25\%}$

Alternative method:

$P(\text{homozygous for A}) = P(AA) + P(aa) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

$P(\text{homozygous for B}) = P(BB) + P(bb) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

$P(\text{both}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

(c) Fraction of A-B- that are homozygous dominant for ≥1 trait:

Step 1: Identify A-B- offspring (9 total)

From Punnett square, A-B- genotypes:

AABB (1) - homozygous for BOTH
AABb (2) - homozygous for A only
AaBB (2) - homozygous for B only
AaBb (4) - heterozygous for both

Step 2: Which are homozygous dominant for ≥1 trait?

AABB: homozygous dominant AA and BB ✓
AABb: homozygous dominant AA ✓
AaBB: homozygous dominant BB ✓
AaBb: neither ✗

Count: 1 + 2 + 2 = 5 out of 9

$\boxed{\text{Fraction} = \frac{5}{9} \approx 55.6\%}$

Verification:

Total A-B-: 9

Homozygous dominant (AA or BB): 5
Heterozygous for both (AaBb): 4
Check: 5 + 4 = 9 ✓

Key Insights:

Genotypic ratio (9:3:3:1 expansion):

AABB: 1
AABb: 2
AaBB: 2
AaBb: 4
AAbb: 1
Aabb: 2
aaBB: 1
aaBb: 2
aabb: 1 Total: 16 (ratio 1:2:1:2:4:2:1:2:1)

Probabilities for independent traits:

Each trait alone: 3:1 ratio (¾ dominant : ¼ recessive)
Combined: (¾)² : (¾)(¼) : (¼)(¾) : (¼)² = 9/16 : 3/16 : 3/16 : 1/16

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