AS & A Level Biology · 9700 · Topic 16B · 2025–2027 Exam

Genetics & Hardy-Weinberg

Mendelian genetics predicts offspring ratios from parental genotypes. But alleles interact in more ways than simple dominant-recessive: codominance, incomplete dominance, and multiple alleles at a single locus each produce characteristic phenotypic patterns. Epistasis adds complexity when two genes interact to produce one trait. Statistically, chi-squared tests whether observed offspring ratios really differ from expected ones. And the Hardy-Weinberg principle calculates allele and genotype frequencies in a population — revealing how selection, mutation, and migration drive evolution away from equilibrium.

Topic 16.2 (Part B) A Level Papers 4–5 Dihybrid · Codominance · ABO · χ² · Hardy-Weinberg

My Study Progress — Topic 16B

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Topic 16.2a · A Level

Genetic crosses & dominance patterns

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Genetic terminology

Term	Definition	Example
Gene	A section of DNA that codes for one (or more) polypeptides	Gene for ABO blood group
Allele	One of the alternative forms of a gene at a particular locus	I^A, I^B, i
Locus	The specific position of a gene on a chromosome	ABO gene is at a specific locus on chromosome 9
Genotype	The alleles an organism possesses for a particular gene	I^Ai (heterozygous blood group A)
Phenotype	The observable characteristics produced by the genotype and environment	Blood group A
Homozygous	Two identical alleles at a locus	AA or aa
Heterozygous	Two different alleles at a locus	Aa
Dominant	Allele that is expressed in the phenotype whether in heterozygous or homozygous form	A in Aa genotype gives dominant phenotype
Recessive	Allele only expressed in phenotype when homozygous	a expressed only in aa
Test cross	Cross of an unknown dominant phenotype with a homozygous recessive individual to determine genotype	AA × aa gives all dominant; Aa × aa gives 1:1

Dihybrid crosses

A dihybrid cross follows the inheritance of two genes simultaneously. When both parents are dihybrid heterozygous (AaBb × AaBb), and the two genes are on different chromosomes (unlinked), Mendel's laws predict a 9:3:3:1 ratio among offspring:

9:3:3:1 ratio — why this arises

For each gene independently: Aa × Aa → 3 A_ : 1 aa. Since genes assort independently:

9 A_B_ (dominant for both)
3 A_bb (dominant A, recessive b)
3 aaB_ (recessive a, dominant B)
1 aabb (recessive for both)

This only holds when: (1) both genes are unlinked (on different chromosomes), (2) both show simple dominant-recessive inheritance, (3) there is no epistasis.

Test cross in dihybrid

To determine whether an unknown dominant phenotype (A_B_) is homozygous or heterozygous for each gene, cross with the double-recessive (aabb). Possible outcomes: if AaBb × aabb → 1 AaBb : 1 Aabb : 1 aaBb : 1 aabb (1:1:1:1). If AABB × aabb → all AaBb (one phenotype only).

Codominance and incomplete dominance

Pattern 1

Codominance

Both alleles are fully expressed simultaneously in the heterozygote — neither dominates the other. The heterozygous phenotype shows features of both alleles, not a blend. Use the same upper-case letter with different superscripts.

Classic example — ABO blood groups (see below): I^AI^B heterozygotes show blood group AB — both A and B antigens on red blood cells.

Classic example — Roan cattle: C^RC^R = red; C^WC^W = white; C^RC^W = roan (mix of red and white hairs, NOT pink).

Pattern 2

Incomplete dominance

The heterozygote shows an intermediate or blended phenotype between the two homozygotes. The alleles are often written as superscripts on a single letter.

Classic example — snapdragon flower colour: C^RC^R = red; C^WC^W = white; C^RC^W = pink (intermediate).

Key distinction from codominance: incomplete dominance produces a new blended phenotype; codominance expresses both parental phenotypes simultaneously (two-phenotype mosaic).

Incomplete dominance cross example — F₂ from pink × pink

Parents: C^RC^W (pink) × C^RC^W (pink)

Gametes: C^R and C^W from each parent

Offspring: ¼ C^RC^R (red) : ½ C^RC^W (pink) : ¼ C^WC^W (white)

Phenotype ratio: 1 red : 2 pink : 1 white — the heterozygote is distinct from both homozygotes

Multiple alleles — ABO blood groups

Most genes have only two alleles in a population. But some genes have three or more alleles (multiple alleles). Only two can be present in any one individual (one per homologous chromosome), but the population has more. The ABO blood group system is the classic example:

Allele	Dominance relationship	Antigen on RBCs
I^A	Dominant over i; codominant with I^B	Antigen A
I^B	Dominant over i; codominant with I^A	Antigen B
i	Recessive to both I^A and I^B	No antigen (O)

Blood group (phenotype)	Possible genotypes
A	I^AI^A or I^Ai
B	I^BI^B or I^Bi
AB	I^AI^B only (codominance)
O	ii only (homozygous recessive)

ABO exam question types

Determine possible blood types of children: construct a Punnett square with all gamete allele combinations; list all resulting genotypes and their phenotypes
Determine parental genotypes from children's blood types: work backwards; e.g. if a child is blood group O (ii), they must have received one i allele from each parent; therefore each parent must carry at least one i allele
Exclude parentage: if neither parent carries the i allele, they cannot have a blood group O child; if one parent is group AB, they must pass either I^A or I^B — they cannot pass i

MCQ · Multiple alleles · Paper 4

A man with blood group A (genotype I^Ai) and a woman with blood group B (genotype I^Bi) have four children. Which blood types are possible among their children?

A. A and B only
B. A, B and O only
C. A, B, AB and O
D. AB and O only

Answer: C
Gametes from father: I^A or i · Gametes from mother: I^B or i
Possible offspring: I^AI^B (AB) : I^Ai (A) : I^Bi (B) : ii (O)
All four blood groups are possible in a 1:1:1:1 ratio.

Topic 16.2b · A Level

Epistasis & chi-squared test

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Epistasis — one gene masking another

Epistasis occurs when the alleles of one gene (epistatic gene) mask or modify the expression of alleles at a second gene (hypostatic gene). It modifies the classic 9:3:3:1 dihybrid ratio into characteristic epistatic ratios. Two types are commonly tested:

Type 1

Recessive epistasis — 9:3:4

A recessive homozygote (aa) at gene A masks the expression of gene B entirely.

Classic example — Labrador coat colour:

Gene B (black/yellow): B_ = black pigment deposited; bb = yellow (no black pigment)
Gene E (expression): E_ = pigment expressed; ee = no pigment expressed at all (epistatic)
Genotypes B_E_ = black; B_ee = yellow; bbE_ = chocolate; bbee = yellow (ee masks B)

Cross BbEe × BbEe: 9 B_E_ : 3 B_ee : 3 bbE_ : 1 bbee
Phenotype ratio: 9 black : 3 chocolate : 4 yellow

Type 2

Dominant epistasis — 12:3:1

A single dominant allele at gene A masks expression of gene B.

Example — squash fruit colour:

Gene W: W_ = white (dominant, epistatic over colour gene C)
Gene C: C_ = yellow; cc = green
W_ (any) = white; wwC_ = yellow; wwcc = green

Cross WwCc × WwCc: 9 W_C_ + 3 W_cc = 12 white : 3 yellow (wwC_) : 1 green (wwcc)
Phenotype ratio: 12 white : 3 yellow : 1 green

How to identify epistasis in an exam question

If a dihybrid cross gives a modified ratio (not 9:3:3:1), check the total: the parts still add to 16 (the 4×4 gamete matrix always produces 16 squares). Common modified ratios:

9:3:4 = recessive epistasis (9+3+4 = 16)
12:3:1 = dominant epistasis (12+3+1 = 16)
9:7 = duplicate recessive epistasis (9+7 = 16)
15:1 = duplicate dominant epistasis (15+1 = 16)

When the ratio adds to 16, it's always a dihybrid situation. Count the distinct phenotype groups in the 16 squares to find which epistasis type applies.

Structured · Epistasis · Paper 4 · 6 marks

In mice, coat colour is controlled by two genes. Gene A (alleles A and a) controls whether colour is produced; A_ allows colour, aa prevents all colour (white). Gene B (alleles B and b) controls colour: B_ = black, bb = brown. Both genes are on different chromosomes.

(a) State what type of epistasis gene A shows over gene B. [1]
(b) Complete the cross AaBb × AaBb and show the expected phenotype ratio. [4]
(c) A student observes 13 black, 4 brown, and 3 white mice. Use chi-squared to test whether the ratio 9:3:4 fits the observed data. [1]

(a) Epistasis type [1 mark]

Recessive epistasis: the recessive homozygote aa masks (is epistatic to) gene B [1]

(b) Cross AaBb × AaBb [4 marks]

Gametes: AB, Ab, aB, ab (each with frequency ¼) from both parents [1]

From 4×4 Punnett square [1]:

9 A_B_ = black (colour present, black) [1]
3 A_bb = brown (colour present, brown) [1]
3 aaB_ + 1 aabb = 4 white (aa masks, no colour)

Phenotype ratio: 9 black : 3 brown : 4 white

(c) Chi-squared check [1 mark]

Total = 20; expected under 9:3:4 ratio: black = 9/16 × 20 = 11.25; brown = 3/16 × 20 = 3.75; white = 4/16 × 20 = 5.0

χ² = (13−11.25)²/11.25 + (4−3.75)²/3.75 + (3−5.0)²/5.0 = 0.27 + 0.02 + 0.80 = 1.09

With df = 2 (3 categories − 1) and p = 0.05, critical value = 5.99. Since 1.09 < 5.99, the deviation is not significant; the 9:3:4 ratio is supported [1]

Chi-squared test (χ²)

The χ² test is a statistical test that determines whether the difference between observed and expected results is due to chance (random sampling variation) or is statistically significant (indicating the expected ratio is wrong).

Chi-squared formula and procedure

χ² = ∑ (O − E)² / E

where O = observed frequency, E = expected frequency for each category

State the null hypothesis: e.g. "there is no significant difference between observed and expected results; any deviation is due to chance"
Calculate expected numbers for each phenotype class from the predicted ratio × total
Calculate χ²: for each category, compute (O−E)²/E; sum all categories
Determine degrees of freedom: df = number of phenotype categories − 1
Read critical value from the table at p = 0.05 and the appropriate df
Compare: if χ² > critical value → reject null hypothesis (significant deviation); if χ² ≤ critical value → accept null hypothesis (deviation due to chance; expected ratio is supported)

Chi-squared worked example — monohybrid ratio test

A student crosses two heterozygous pea plants (Tt × Tt) and observes: 78 tall, 22 short (total = 100). Expected ratio 3:1 → expected: 75 tall, 25 short.

Category	O	E	(O−E)²/E
Tall	78	75	(78−75)²/75 = 9/75 = 0.12
Short	22	25	(22−25)²/25 = 9/25 = 0.36
χ²			0.48

df = 2−1 = 1; critical value at p=0.05 is 3.84. Since 0.48 < 3.84: null hypothesis accepted. The deviation is not significant; the 3:1 ratio is supported.

Degrees of freedom (df)	Critical value at p = 0.05
1	3.84
2	5.99
3	7.82
4	9.49

Chi-squared: the conclusion language

If χ² ≤ critical value: "The deviation is not statistically significant (p > 0.05). There is no significant difference between observed and expected results. The null hypothesis is accepted. The data are consistent with the expected [3:1 / 9:3:3:1 / etc.] ratio."
If χ² > critical value: "The deviation is statistically significant (p < 0.05). The null hypothesis is rejected. The observed ratio differs significantly from the expected ratio, suggesting [linkage / epistasis / selection / other factor]."

Always state whether the null hypothesis is accepted or rejected and link to a biological interpretation.

Topic 16.2c · A Level

Hardy-Weinberg principle

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The Hardy-Weinberg principle predicts the expected frequencies of genotypes and alleles in a population that is not evolving. If allele frequencies are changing, one or more of the assumptions is being violated — indicating that evolution is occurring.

The Hardy-Weinberg equations

The two equations

For a gene with two alleles A (frequency p) and a (frequency q):

p + q = 1 (allele frequencies sum to 1)

p² + 2pq + q² = 1 (genotype frequencies sum to 1)

p² = frequency of AA (homozygous dominant)
2pq = frequency of Aa (heterozygous)
q² = frequency of aa (homozygous recessive)

The second equation is simply an expansion of (p + q)² = 1, using the assumption of random mating.

Assumptions for Hardy-Weinberg equilibrium

The allele frequencies will remain constant generation after generation only if all of the following assumptions hold:

Assumption 1

Very large (effectively infinite) population

In small populations, genetic drift causes random changes in allele frequency not due to selection. Large populations average out these random fluctuations.

Assumption 2

Random mating

Individuals mate without regard to genotype. Non-random mating (e.g. assortative mating, where similar phenotypes preferentially mate) changes genotype frequencies without necessarily changing allele frequencies.

Assumption 3

No natural selection

All genotypes have equal survival and reproductive success (equal fitness). If certain genotypes survive or reproduce better, allele frequencies change.

Assumption 4

No mutation

No new alleles are introduced by mutation, and no alleles are lost by mutation. Mutations change allele frequencies, however slowly.

Assumption 5

No migration (gene flow)

No individuals enter (immigration) or leave (emigration) the population, bringing different allele frequencies. Gene flow between populations changes allele frequencies.

Why Hardy-Weinberg is useful despite unrealistic assumptions

No real population perfectly meets these conditions, but H-W serves as a null model: a baseline against which real populations are compared. If observed genotype frequencies differ significantly from H-W predictions (using χ²), at least one evolutionary force (selection, drift, gene flow, non-random mating) is operating. H-W also lets us calculate allele frequencies from phenotype frequencies alone, which is useful in clinical genetics.

Hardy-Weinberg calculations — step by step

Worked example 1 — find allele frequency from phenotype frequency

In a population, 9% of individuals have cystic fibrosis (autosomal recessive, genotype aa). Find the frequency of the carrier genotype (Aa).

Frequency of aa = q² = 0.09 → q = √0.09 = 0.30 (frequency of allele a)
p + q = 1 → p = 1 − 0.30 = 0.70 (frequency of allele A)
Frequency of carriers (Aa) = 2pq = 2 × 0.70 × 0.30 = 0.42 (42%)
Frequency of unaffected homozygotes (AA) = p² = 0.70² = 0.49 (49%)
Check: p² + 2pq + q² = 0.49 + 0.42 + 0.09 = 1.00 ✓

Worked example 2 — find number of carriers in a population

In a population of 10,000, 1 in 2500 individuals has phenylketonuria (PKU, autosomal recessive). How many carriers are expected?

Frequency of affected (aa) = q² = 1/2500 = 0.0004 → q = √0.0004 = 0.02
p = 1 − 0.02 = 0.98
Frequency of carriers (Aa) = 2pq = 2 × 0.98 × 0.02 = 0.0392
Number of carriers = 0.0392 × 10,000 = 392 carriers

Common calculation shortcuts and checks

Always start with q² (the homozygous recessive frequency) because it is the only genotype frequency you can determine directly from phenotype data for a recessive allele
q = √(q²); then p = 1 − q; then compute 2pq and p²
Always check your answer: p² + 2pq + q² must = 1.00
For X-linked genes, different formulas apply for the two sexes (males are hemizygous); this is rarely tested in standard H-W calculations

Calculation · Hardy-Weinberg · Paper 4 · 5 marks

In a population of wild-type and albino mice, 16% of mice are albino (homozygous recessive, aa). Albinism is caused by a recessive allele a of gene A.

(a) Calculate the frequency of allele a in the population. [1]
(b) Calculate the frequency of allele A. [1]
(c) Calculate the expected frequency of heterozygous mice (Aa). [1]
(d) State TWO assumptions that must hold for your calculations to be valid. [2]

(a) Frequency of allele a [1 mark]

q² = 0.16 → q = √0.16 = 0.40 [1]

(b) Frequency of allele A [1 mark]

p = 1 − q = 1 − 0.40 = 0.60 [1]

(c) Frequency of Aa [1 mark]

2pq = 2 × 0.60 × 0.40 = 0.48 (48%) [1]

(d) Two assumptions [2 marks; any two]

Large (effectively infinite) population size — to eliminate genetic drift [1]
Random mating within the population [1]
No natural selection (equal fitness of all genotypes) [1]
No mutation [1]
No migration (gene flow) [1]

Exam Prep

Topic 16B Practice — Comprehensive

Mixed practice covering dihybrid crosses, epistasis, chi-squared and Hardy-Weinberg.

MCQ · Codominance vs incomplete dominance · Paper 4

In Andalusian fowl, black (C^BC^B) crossed with white (C^WC^W) produces blue F₁ offspring (C^BC^W). What type of inheritance is this?

A. Simple dominant-recessive; black is dominant
B. Incomplete dominance; the heterozygote has a new intermediate phenotype (blue)
C. Codominance; the heterozygote shows both black and white features simultaneously
D. Multiple alleles; three alleles are present

Answer: B — Incomplete dominance. Blue Andalusian fowl are an intermediate phenotype — the heterozygote (C^BC^W) is blue, a new colour not seen in either parent. This is incomplete dominance. For codominance (C), the heterozygote would show both black AND white simultaneously (e.g. patches of both colours), which is not the case here. If blue birds have intermingled black and white feathers rather than a true blue colour, some textbooks classify it differently, but the CIE 9700 context uses this as incomplete dominance.

MCQ · Hardy-Weinberg · Paper 4

In a population, the frequency of allele a is 0.3 and allele A is 0.7. What is the expected frequency of heterozygous individuals (Aa)?

A. 0.09
B. 0.49
C. 0.42
D. 0.21

Answer: C — Frequency of Aa = 2pq = 2 × 0.7 × 0.3 = 0.42. (A) = q² = frequency of aa. (B) = p² = frequency of AA. (D) = pq (missing the factor of 2 for the heterozygous combination). The factor of 2 arises because Aa can form in two ways: A from father and a from mother, OR a from father and A from mother.

Structured · Chi-squared · Paper 4 · 8 marks

A student crosses two dihybrid pea plants (AaBb × AaBb) and obtains the following offspring: 88 A_B_, 32 A_bb, 24 aaB_, 16 aabb. Total = 160.

(a) State the expected phenotype ratio from this cross if genes A and B assort independently. [1]
(b) Calculate the expected number in each phenotype class. [2]
(c) Calculate the chi-squared value. Show all working. [3]
(d) With 3 degrees of freedom, the critical value at p=0.05 is 7.82. Interpret your result and draw a conclusion. [2]

(a) Expected ratio [1 mark]

9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb [1]

(b) Expected numbers [2 marks]

Total = 160; out of 16 parts:

A_B_ : 9/16 × 160 = 90 [1]
A_bb : 3/16 × 160 = 30; aaB_ : 3/16 × 160 = 30; aabb : 1/16 × 160 = 10 [1]

(c) Chi-squared calculation [3 marks]

χ² = ∑(O−E)²/E:

A_B_: (88−90)²/90 = 4/90 = 0.044 [1]
A_bb: (32−30)²/30 = 4/30 = 0.133 [1]
aaB_: (24−30)²/30 = 36/30 = 1.200 [1]
aabb: (16−10)²/10 = 36/10 = 3.600
χ² = 0.044 + 0.133 + 1.200 + 3.600 = 4.977

(d) Interpretation [2 marks]

The calculated χ² (4.977) is less than the critical value (7.82 at df=3, p=0.05) [1]
The difference between observed and expected results is not statistically significant; the null hypothesis is accepted. The data are consistent with a 9:3:3:1 ratio from independent assortment of two genes [1]

Synoptic · Topics 16A + 16B + 17 · Paper 4 · 7 marks

In a large population of beetles, a recessive allele (m) causes reduced pigmentation. Currently, 4% of beetles are affected (mm).

(a) Calculate the frequency of the M allele and the expected frequency of carriers (Mm). [3]
(b) A new predator begins targeting pale-coloured (mm) beetles. Explain the expected change in the frequency of allele m over several generations. [2]
(c) State which Hardy-Weinberg assumption is being violated and explain why allele frequencies will change. [2]

(a) Allele frequencies and carrier frequency [3 marks]

q² = 0.04 → q = √0.04 = 0.20 (frequency of allele m) [1]
p = 1 − 0.20 = 0.80 (frequency of allele M) [1]
Frequency of carriers (Mm) = 2pq = 2 × 0.80 × 0.20 = 0.32 (32%) [1]

(b) Change in frequency of allele m [2 marks]

mm beetles are selectively removed by predation; they survive and reproduce at a lower rate than MM and Mm beetles [1]
Over generations, the mm genotype becomes rarer; since mm beetles are the main source of allele m combinations, the frequency of allele m will decrease; allele M will increase in frequency [1]

(c) H-W assumption violated [2 marks]

The assumption of no natural selection is violated — mm beetles have reduced fitness because they are more likely to be eaten [1]
Because genotypes differ in survival/reproduction (fitness), allele frequencies change: the m allele decreases as mm individuals are selectively removed; H-W equilibrium no longer applies [1]

Exam Prep

Topic 16B — Common Mistakes

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Confusing codominance and incomplete dominanceCodominance: both alleles expressed simultaneously and distinctly in heterozygote (e.g. AB blood group has both A and B antigens; roan cattle have both red and white hairs). Incomplete dominance: heterozygote shows an intermediate blended phenotype (e.g. pink snapdragons). The test: does the heterozygote show both parental features (codominance) or a new intermediate one (incomplete dominance)?
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Forgetting that epistatic ratios still add to 169:3:4 (= 16), 12:3:1 (= 16), 9:7 (= 16), 15:1 (= 16). All modified dihybrid ratios sum to 16. If your calculated ratio doesn't add to 16, recalculate. This is your first self-check.
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Chi-squared: forgetting degrees of freedom = categories − 1If there are 4 phenotype classes (e.g. 9:3:3:1), df = 4−1 = 3, not 4. Selecting the wrong row of the critical value table gives the wrong conclusion. Always: df = number of phenotype categories − 1.
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Chi-squared: saying a significant result "proves" the hypothesis is wrongχ² shows whether deviation is statistically significant — it does not "prove" or "disprove" a hypothesis. A significant result means the null hypothesis should be rejected at p = 0.05; there is less than 5% probability that the deviation occurred by chance. Avoid the word "proves".
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Hardy-Weinberg: using q instead of q² for the recessive phenotype frequencyThe frequency of the recessive phenotype (aa) is q², not q. q is the frequency of the recessive allele. You must square-root the phenotype frequency to get the allele frequency: q = √(phenotype frequency). Starting with the wrong value cascades into all subsequent calculations.
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Hardy-Weinberg: omitting the factor of 2 in the carrier frequencyCarrier frequency = 2pq, NOT pq. The factor of 2 accounts for the two ways a heterozygote can be formed (A from father + a from mother, OR a from father + A from mother). Omitting the 2 halves the carrier frequency estimate.
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Saying a population "is evolving" as a conclusion from H-W aloneHardy-Weinberg calculations can show that allele frequencies differ from what H-W predicts, or that they have changed between generations. But to say which mechanism is causing evolution (selection, drift, gene flow, mutation) requires additional evidence. State that "at least one H-W assumption is violated" without over-specifying which one, unless data support a specific conclusion.
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ABO blood groups: saying blood group A must be I^AI^ABlood group A can be I^AI^A (homozygous) OR I^Ai (heterozygous) — both give phenotype A because I^A is dominant over i. When asked about possible children from group A parents, you must consider both possible genotypes. Only blood group AB (I^AI^B) and O (ii) have only one possible genotype.
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Dihybrid ratio 9:3:3:1 applies when genes are unlinked AND show simple dominanceThe 9:3:3:1 ratio requires (1) genes on different chromosomes (unlinked), (2) simple dominant-recessive inheritance for both genes, AND (3) no epistasis. If any of these conditions are violated, the ratio changes: linkage gives skewed parental types; epistasis gives modified ratios; codominance adds extra phenotype classes.

Topic 16B strategy

Topic 16B is the most calculation-intensive topic in the A Level biology course. Practice is essential. Highest-yield items: dihybrid cross 9:3:3:1 ratio explanation, codominance vs incomplete dominance distinction, ABO blood group genotype-phenotype table (6 genotypes / 4 phenotypes), codominance of I^AI^B, recessive epistasis 9:3:4 with Labrador example, dominant epistasis 12:3:1, chi-squared formula ∑(O−E)²/E with worked steps, null hypothesis language, H-W equations (p+q=1 and p²+2pq+q²=1), 5 H-W assumptions, H-W calculation starting from q² (recessive phenotype frequency). Synoptic links: Topic 16A (crossing over, independent assortment, linkage), Topic 17 (selection violates H-W — population genetics bridge), Topic 6 (gene structure and mutation).