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Laws of inheritance, Punnett squares, and probability
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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)
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.
Given:
(a) Punnett Square Setup:
Testcross (heterozygote × homozygous recessive)
Parent 1 gametes (TtYy): TY, Ty, tY, ty (4 types) ty only (1 type)
| Section | Format | Questions | Time | Weight | Calculator |
|---|---|---|---|---|---|
| Multiple Choice | MCQ | 60 | 90 min | 50% | 🚫 |
| Free Response (Long) | FRQ | 2 | 50 min | 30% | 🚫 |
| Free Response (Short) | FRQ | 4 | 40 min | 20% | 🚫 |
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One trait considered
Example: Tall (T) × Short (t) pea plants
Two traits considered
Example: Round Yellow (RRYY) × Wrinkled Green (rryy)
Purpose: Determine unknown genotype
Method: Cross with homozygous recessive (tt)
Pedigree: Family tree showing trait inheritance
Symbols:
Determining inheritance pattern:
Punnett Square:
| ty
-----|--------
TY | TtYy (tall, yellow)
-----|--------
Ty | Ttyy (tall, green)
-----|--------
tY | ttYy (short, yellow)
-----|--------
ty | ttyy (short, green)
Offspring genotypes:
(b) Phenotypic Ratio:
Count each phenotype:
Or grouped:
(c) Probability of tall with green seeds:
Phenotype: Tall, green Genotype: Ttyy
From Punnett square: 1 out of 4 offspring
Alternative calculation (using multiplication rule):
Since traits assort independently:
Key Concept - Testcross:
Mendel's Laws Applied:
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?
Given: AaBb × AaBb (dihybrid cross)
(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:
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:
Alternative method:
(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:
Step 2: Which are homozygous dominant for ≥1 trait?
Count: 1 + 2 + 2 = 5 out of 9
Verification:
Total A-B-: 9
Key Insights:
Genotypic ratio (9:3:3:1 expansion):
Probabilities for independent traits: