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In a dihybrid cross between two heterozygous parents ($AaBb \times AaBb$), what is the phenotypic ratio observed in the offspring, assuming complete dominance and independent assortment?
Options
1
$1:2:1$
2
$3:1$
3
$9:3:3:1$
4
$1:1:1:1$
Correct Answer
$9:3:3:1$
Solution
1

$AaBb \times AaBb$: Gametes from each parent: $AB, Ab, aB, ab$ (4 types each)

Punnett square gives 16 combinations. Phenotypic classes:

$A\_B\_: 9$ | $A\_bb: 3$ | $aaB\_: 3$ | $aabb: 1$

2

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

Mendel's 2nd Law: Independent Assortment (genes on different chromosomes)

Dihybrid: $AaBb \times AaBb \to 9:3:3:1$
9(A_B_) : 3(A_bb) : 3(aaB_) : 1(aabb)
Theory: Genetics
1. Mendel's Laws of Inheritance

Gregor Mendel (1866): studied 7 traits in garden pea (Pisum sativum). Selected pea for: short generation time, many offspring, ability to self-fertilise AND cross-pollinate, distinct contrasting characters. 7 traits studied: seed shape (round/wrinkled), seed colour (yellow/green), pod shape (inflated/constricted), pod colour (green/yellow), flower colour (purple/white), flower position (axial/terminal), stem height (tall/dwarf). Law 1 — Segregation: each individual carries two alleles for each character; alleles separate during gamete formation; each gamete carries only one allele. Ratio from monohybrid (Aa × Aa): phenotypic 3:1, genotypic 1:2:1. Law 2 — Independent Assortment: alleles for different characters assort independently of each other during gamete formation. Applies only to genes on different chromosomes (non-linked). Dihybrid (AaBb × AaBb): 9:3:3:1 phenotypic ratio.

2. Dihybrid Cross Analysis

$AaBb \times AaBb$: Each parent produces 4 types of gametes: $AB, Ab, aB, ab$ (each with frequency 1/4). Punnett square (4×4 = 16 boxes). Phenotypic classes: A\_B\_ (9/16): dominant for both traits. A\_bb (3/16): dominant A, recessive b. aaB\_ (3/16): recessive a, dominant B. aabb (1/16): recessive for both. Genotypic ratio (9 different genotypes): AABB(1) + AABb(2) + AaBB(2) + AaBb(4) + AAbb(1) + Aabb(2) + aaBB(1) + aaBb(2) + aabb(1) = 16 total. The factor 9 = (3/4)²: probability of A dominant × probability of B dominant = (3/4)(3/4) = 9/16. Factor 3 = (3/4)(1/4). Factor 1 = (1/4)². The multiplicative rule works because genes assort independently.

3. Chromosomal Basis of Inheritance

Sutton and Boveri (1902-1903): proposed chromosomes carry genes (chromosome theory of heredity). Evidence: chromosomes segregate during meiosis (matches allele segregation). Chromosomes assort independently during meiosis I (matches independent assortment). Morgan (Nobel 1933): confirmed with Drosophila. Discovered: sex-linked inheritance (white eye in Drosophila, X-linked). Genetic linkage (genes on same chromosome tend to be inherited together). Crossing over produces recombinant chromosomes → explains why linked genes sometimes separate. Chromosomal maps: frequency of crossing over between two genes ∝ distance between them. 1 map unit (centiMorgan, cM) = 1% recombination frequency. Morgan's group mapped genes on Drosophila chromosomes — first genetic maps.

4. Deviations from Mendelian Ratios

Incomplete dominance: Aa phenotype intermediate between AA and aa. Antirrhinum (snapdragon) flower: RR=red, rr=white, Rr=pink. F2 ratio: 1:2:1 (phenotypic = genotypic). Codominance: both alleles expressed equally in heterozygote. Blood groups: IA and IB are codominant → IAIB = AB blood group. Sickle cell trait: HbA and HbS both expressed in HbA/HbS. Epistasis: one gene masks expression of another gene (gene-gene interaction). Recessive epistasis (9:3:4): aaBB = aaBb = aabb phenotype same (a epistatic to B). Dominant epistasis (12:3:1): AABB = AaBb = Aabb same phenotype. Duplicate genes (15:1). Pleiotropy: one gene affects multiple phenotypes. Phenylketonuria: PKU gene affects multiple systems. Multiple alleles: more than two alleles for one gene (e.g., ABO blood groups: IA, IB, i).

5. Linkage and Crossing Over

Linked genes: genes on same chromosome → tend to be inherited together (parental combinations more frequent). Complete linkage: no crossing over → only parental types. Incomplete linkage: some crossing over → recombinants less frequent than 50%. Recombination frequency: number of recombinants / total offspring × 100. < 50% for linked genes. = 50% for unlinked genes (on different chromosomes OR very far apart on same chromosome). Chi-square test: statistical test to determine if observed ratios fit expected Mendelian ratios. $\chi^2 = \sum\frac{(O-E)^2}{E}$. If $\chi^2 < $ critical value: accept null hypothesis (no significant deviation). Crossing over: physical exchange between non-sister chromatids during prophase I. Chiasma (plural: chiasmata): the X-shaped structure visible at crossing over points. Produces recombinant gametes.

6. Sex Determination

XX-XY type (humans, Drosophila, most mammals): females = XX, males = XY. Y chromosome carries SRY gene (sex-determining region Y) → triggers testis development. All eggs carry X. Half sperm carry X (→ female offspring), half carry Y (→ male offspring). So sex determined by father. Ratio: 1:1 (males:females). ZW-ZZ type (birds, some reptiles): females = ZW, males = ZZ. Sex determined by egg. XX-XO type (grasshopper): females = XX, males = XO (one X missing). Haplodiploidy (bees, ants, wasps): females = diploid (fertilised eggs), males = haploid (unfertilised eggs). Y-linked (holandric) traits: only in males, passed from father to ALL sons. X-linked traits: more common in males (hemizygous, only one X). Colour blindness, haemophilia, G6PD deficiency.

7. Multiple Alleles and ABO Blood Groups

ABO blood groups: controlled by gene I (chromosome 9). Three alleles: IA (produces A antigen), IB (produces B antigen), i (produces no antigen). Codominant: IA and IB. Recessive: i. Genotypes and phenotypes: IAIA or IAi → blood group A (antigen A, anti-B antibody). IBIB or IBi → blood group B (antigen B, anti-A antibody). IAIB → blood group AB (both antigens, no antibodies). ii → blood group O (no antigens, both anti-A and anti-B antibodies). Rh factor: separate gene (Rh+ dominant over Rh-). Blood transfusion: must match ABO and Rh. AB = universal recipient (no antibodies). O- = universal donor (no antigens). Haemolytic disease of newborn (HDN/erythroblastosis fetalis): Rh- mother + Rh+ father → Rh+ foetus → maternal anti-Rh antibodies (after first pregnancy) → attack second Rh+ foetus. Prevented by Rho(D) immunoglobulin injection.

8. Polygenic Inheritance

Polygenic inheritance: trait controlled by multiple genes, each with small additive effect. Continuous variation (bell curve distribution). Examples: skin colour (3-6 gene pairs), height, intelligence, blood pressure, wheat grain colour. Skin colour: approximately 6 genes contribute. Each dominant allele adds pigment. AABBCC = darkest, aabbcc = lightest. Intermediate combinations give full range of skin tones. Environment also affects phenotype. Norm of reaction: range of phenotypes a genotype can produce in different environments. Heritability: proportion of phenotypic variation due to genetic variation. H² = V_G/V_P = genetic variance / phenotypic variance. High heritability: genetics explains most variation (height). Low heritability: environment explains most (body weight, which is strongly influenced by diet and lifestyle). Quantitative trait loci (QTL) mapping: locates regions of genome containing genes affecting quantitative traits using molecular markers.

Frequently Asked Questions
1. Why is the ratio 9:3:3:1 and not some other ratio in dihybrid cross? ⌄
The 9:3:3:1 ratio comes from the mathematical product of two independent 3:1 ratios. For gene A (Aa × Aa): 3 A_ : 1 aa (monohybrid ratio). For gene B (Bb × Bb): 3 B_ : 1 bb (monohybrid ratio). Because the two genes assort independently (Mendel's 2nd law): P(A_ and B_) = (3/4)(3/4) = 9/16. P(A_ and bb) = (3/4)(1/4) = 3/16. P(aa and B_) = (1/4)(3/4) = 3/16. P(aa and bb) = (1/4)(1/4) = 1/16. Total = 16/16 = 1. Ratio = 9:3:3:1. The number 16 total combinations comes from 4 × 4 Punnett square (4 types of gametes from each parent in a dihybrid). If genes were linked with no crossing over: ratio would be 1:1:1:1 (only parental combinations) or just 3:1 (if completely linked and both genes show dominance together).
2. What are the gametes from AaBb parent and their frequencies? ⌄
Using the FOIL or branch method: from Aa: 1/2 A gametes and 1/2 a gametes. From Bb: 1/2 B gametes and 1/2 b gametes. Combining (independent assortment): AB = (1/2)(1/2) = 1/4. Ab = (1/2)(1/2) = 1/4. aB = (1/2)(1/2) = 1/4. ab = (1/2)(1/2) = 1/4. All four gamete types are equally frequent (25% each). This equal frequency of all gametes is why the 9:3:3:1 ratio results. If gene were linked: AB and ab (parental) more frequent than Ab and aB (recombinant).
3. How many genotypes and phenotypes in a trihybrid cross? ⌄
For $AaBbCc \times AaBbCc$ (trihybrid cross, each parent heterozygous for 3 genes): Number of gamete types from each parent = $2^3 = 8$ (ABC, ABc, AbC, Abc, aBC, aBc, abC, abc). Total offspring combinations = $8 \times 8 = 64$ boxes. Number of phenotypic classes = $2^3 = 8$ (with 8 different phenotypic combinations). Phenotypic ratio = $(3:1)^3 = 27:9:9:9:3:3:3:1$. Number of genotypic classes = $3^3 = 27$. General formula: for $n$ heterozygous genes: gamete types = $2^n$, phenotypic classes = $2^n$, genotypic classes = $3^n$, phenotypic ratio extends the $(3:1)^n$ pattern.
4. What was Mendel's experimental procedure and why did he choose pea plants? ⌄
Mendel's experimental design (1856-1863): (1) Started with true-breeding (homozygous) plants for each trait. (2) Made crosses (P generation). (3) Grew F1 plants (all showed dominant trait → monohybrid: Aa). (4) Allowed F1 to self-fertilise → F2 plants (showed 3:1 ratio). (5) Test cross (F1 × recessive): 1:1 ratio confirmed heterozygous F1. (6) Dihybrid crosses → 9:3:3:1. Pea plants chosen because: (a) Short life cycle (one generation/year). (b) Large number of offspring → statistically reliable. (c) Distinct, easily visible contrasting characters. (d) Easily self-fertilising (to maintain true-breeding lines) OR cross-fertilised (emasculation + manual pollination). (e) Enclosed flower structure prevents contamination. Why Mendel succeeded where others failed: chose traits controlled by single genes on different chromosomes, used large sample sizes, applied statistical analysis — unusual for 1860s biology.
5. What is a test cross and what does it reveal? ⌄
Test cross: crossing an individual with unknown genotype with a homozygous recessive individual (genotype aa or aabb). Reveals whether the individual is homozygous dominant (AA) or heterozygous (Aa). If tested individual is AA: all offspring are Aa (dominant phenotype). Ratio 0 recessive offspring. If tested individual is Aa: offspring are Aa (dominant) and aa (recessive). Ratio 1:1. For dihybrid: test cross AaBb × aabb: offspring are AaBb, Aabb, aaBb, aabb in 1:1:1:1 ratio (if independent assortment). If genes linked: parental combinations more frequent than recombinants. Recombination frequency = (number of recombinant offspring / total offspring) × 100%. Test crosses are fundamental in genetic analysis — they directly reveal gamete types produced by the tested individual.
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