Monohybrid Inheritance — AQA A-level Biology

SPEC 3.7.1 - GENETICS, POPULATIONS, EVOLUTION & ECOSYSTEMS

Inheritance: Monohybrid & Dihybrid Crosses

Punnett squares, codominance, the 9:3:3:1 ratio and chi-squared — the genetics toolkit the examiner tests every year

The Language of Genetics

Before crossing anything, the examiner expects precise definitions. These terms carry marks in their own right.

Key Definition

Genotype — the genetic constitution of an organism; the specific alleles present at a locus or loci.

Phenotype — the observable characteristics of an organism, resulting from the interaction of its genotype with the environment.

Key Definition

Dominant allele — expressed in the phenotype whether the organism is homozygous or heterozygous.

Recessive allele — only expressed in the phenotype when the organism is homozygous.

Codominant alleles — both alleles are expressed in the heterozygous phenotype.

Homozygous vs Heterozygous

Homozygous — two identical alleles at a given locus (e.g. TT or tt).

Heterozygous — two different alleles at a given locus (e.g. Tt).

Monohybrid Cross — Notation

Key Definition

Monohybrid cross — a genetic cross examining the inheritance of one gene.

AQA expects a specific sequence. Show every stage for full marks:

1
Parental phenotypes — describe what each parent looks like (e.g. "Tall plant × Dwarf plant")
2
Parental genotypes — use standard notation: capital = dominant (T), lowercase = recessive (t)
3
Gametes — circle them; show which allele(s) each gamete can carry
4
Punnett square — lay out the 4 possible offspring genotypes
5
Offspring genotypes and phenotypes — give both ratios (e.g. 1TT : 2Tt : 1tt; 3 tall : 1 dwarf)

Examiner Alert

Do not skip the gametes stage. Even if you get the Punnett square right, omitting gametes loses marks. Always circle or list the gametes explicitly.

Monohybrid Cross — Worked Example

Tall plant (Tt) × Tall plant (Tt) — both heterozygous.

Exam Tip

The 3:1 phenotype ratio only applies to a heterozygous × heterozygous cross. A homozygous dominant × heterozygous gives 1:1. Know which cross produces which ratio.

The Test Cross

If a plant shows the dominant phenotype, you cannot tell from appearance alone whether it is TT or Tt. A test cross resolves this by crossing the unknown with a homozygous recessive (tt).

Result A — all tall offspring

Unknown parent must be TT (homozygous dominant)

All offspring Tt — all tall

TT × tt → all Tt

Result B — 1:1 offspring

Unknown parent must be Tt (heterozygous)

Half Tt (tall) : half tt (dwarf)

Tt × tt → Tt : tt (1:1)

Mark-Scheme Wording

"Cross the organism of unknown genotype with a homozygous recessive individual. If all offspring show the dominant phenotype, the unknown is homozygous dominant. If offspring show a 1:1 ratio of dominant to recessive phenotypes, the unknown is heterozygous."

Codominance

Key Definition

Codominant alleles — two alleles that are both expressed in the heterozygous phenotype. Neither masks the other.

Example: cattle coat colour. Use superscript notation — both alleles are capital (neither is dominant over the other).

Roan Coat Explained

C^RC^W cattle have both red and white hairs — a roan appearance. Neither allele is masked. This is not blending; the individual hairs are still distinct red or white.

Codominance — F2 Cross

When two roan cattle (C^RC^W) are crossed, the offspring show three distinct phenotypes.

Examiner Alert

AQA does not require ABO blood group as a codominance example. Use coat colour (cattle) or any clearly labelled codominant example. The notation C^RC^W is correct — do not write C^RC^r for codominance as that implies dominance.

Dihybrid Cross

Key Definition

Dihybrid cross — a genetic cross examining the inheritance of two genes located on different chromosomes.

Because the genes are on different chromosomes, they assort independently during meiosis (Mendel's Law of Independent Assortment). This produces 4 types of gamete from a double heterozygote.

Gametes from TtRr parent

A TtRr parent produces 4 gamete types in equal proportions: TR Tr tR tr

each at frequency ¼

Why 4 gamete types?

At meiosis I, the T/t pair and the R/r pair separate independently. Each combination of one T-allele and one R-allele is equally likely. This independence only holds when genes are on different chromosomes — linked genes violate this.

Dihybrid Cross — The 9:3:3:1 Ratio

TtRr × TtRr — cross of two double heterozygotes. The 16-square Punnett grid gives the expected ratio.

Expected ratio: 9 T_R_ : 3 T_rr : 3 ttR_ : 1 ttrr

Chi-Squared Test (χ²)

Key Definition

Chi-squared test — a statistical test used to determine whether the difference between observed and expected phenotypic ratios is due to chance.

χ² = Σ (O − E)² / E

Where O = observed frequency; E = expected frequency; Σ = sum of all classes

1
State the null hypothesis: "The difference between observed and expected ratios is due to chance."
2
Calculate expected frequencies from the predicted ratio (e.g. 3:1 → ¾ × total : ¼ × total).
3
Calculate χ² = Σ (O − E)² / E for each phenotype class, then sum.
4
Find degrees of freedom: df = (number of phenotype classes − 1).
5
Compare χ² to the critical value at p = 0.05 from the table.
6
Conclude: if χ² < critical value → accept null hypothesis (difference is due to chance). If χ² > critical value → reject null hypothesis (ratio is not as expected).

Chi-Squared — Worked Example

A monohybrid cross predicts a 3:1 ratio. 80 offspring are counted: 54 tall, 26 dwarf.

PhenotypeObserved (O)Expected (E)(O − E)²(O − E)² / E

Tall5460360.60

Dwarf2620361.80

χ² =2.40

Conclusion

Degrees of freedom = 2 − 1 = 1

Critical value at p = 0.05, df = 1: 3.84

χ² (2.40) < critical value (3.84) → Accept null hypothesis

"The difference between observed and expected results is not significant at p = 0.05. The difference is due to chance. The results are consistent with the predicted 3:1 ratio."

Examiner Traps — Inheritance

Trap 1 — Chi-Squared Conclusion

Never just say "accept" or "reject." Always link back: "Because χ² < critical value, we accept the null hypothesis — the difference is due to chance — supporting the expected 3:1 ratio." The AQA mark scheme always requires both the statistical decision AND the biological interpretation.

Trap 2 — Dihybrid only works for unlinked genes

The 9:3:3:1 ratio assumes genes on different chromosomes. If genes are on the same chromosome (linked), ratios deviate. A question may give you unexpected ratios and ask you to explain — the answer is autosomal linkage (covered in Lesson 7.02).

Trap 3 — Codominance notation

In codominance, both alleles are written as capitals (e.g. C^R and C^W). Writing C^R and C^r implies dominance and will be penalised. The heterozygote C^RC^W has a phenotype different from both homozygotes.

Trap 4 — p = 0.05 significance level

AQA uses p = 0.05 as the conventional significance threshold. This means there is a 5% probability the difference is due to chance even if it is real. Always state this level in your chi-squared conclusion.

DEEP Chain — Genetics Extended Response

Question: "Explain how a 3:1 phenotype ratio is produced in the offspring of a monohybrid cross between two heterozygous parents."

DEEP Answer Framework

D — Define

"A monohybrid cross involves the inheritance of a single gene. Heterozygous parents carry two different alleles at that locus, one dominant (T) and one recessive (t)."

E — Explain the Mechanism

"During meiosis, homologous chromosomes separate, so each gamete carries only one allele — either T or t. Random fertilisation means any gamete can fuse with any other. The four possible combinations are TT, Tt, Tt, and tt in equal proportions."

E — Evidence / Example

"Because T is dominant, both TT and Tt individuals display the dominant phenotype — only the tt genotype produces the recessive phenotype. Three out of four offspring show the dominant phenotype, giving a 3:1 ratio."

P — Practical / Quantitative

"A chi-squared test can assess whether deviation from the expected 3:1 ratio is due to chance. If χ² < 3.84 (critical value at p = 0.05, df = 1), the null hypothesis is accepted and the results are consistent with a 3:1 ratio."

Synoptic Links

Link to Meiosis (4.04)

The 3:1 and 9:3:3:1 ratios depend on independent assortment during meiosis I. Crossing over during prophase I can also create new allele combinations (recombinants). In Lesson 7.02 you will see how linked genes on the same chromosome alter these expected ratios.

Link to Hardy-Weinberg (7.03)

Chi-squared is the same statistical tool used to test whether a population's genotype frequencies deviate from Hardy-Weinberg equilibrium. Mastering the chi-squared procedure here will directly feed Lesson 7.03.

Link to Gene Mutations (8.01)

Mutations that create new alleles — especially recessive lethal alleles — will change the expected offspring ratios. A 2:1 ratio instead of 3:1 is a classic examiner signal that a homozygous genotype is lethal.

Practice Question

In rabbits, coat colour shows codominance. C^B allele produces black coat; C^W allele produces white coat. Heterozygous rabbits (C^BC^W) are grey.

(a) A grey rabbit is crossed with a white rabbit. Draw a Punnett square and predict the phenotype ratio of the offspring. [3 marks]

(b) 40 offspring are produced from this cross. 25 are grey and 15 are white. A student claims this result is consistent with the expected ratio. Describe how you would use the chi-squared test to evaluate this claim. Include the null hypothesis and explain how to reach a conclusion. [4 marks]

D

Define: Codominant alleles — both expressed in the heterozygous phenotype. Cross: C^BC^W × C^WC^W

E

Punnett square gametes: C^B and C^W from grey; both C^W from white. Offspring: C^BC^W (grey) : C^WC^W (white) — 1:1 ratio

E

Chi-squared: Null hypothesis: "The difference between observed (25 grey, 15 white) and expected (20 grey, 20 white) results is due to chance." Calculate χ² = (25−20)²/20 + (15−20)²/20 = 1.25 + 1.25 = 2.50

P

df = 1; critical value at p = 0.05 = 3.84. χ² (2.50) < 3.84 → Accept null hypothesis. Difference is not significant — consistent with expected 1:1 ratio. The student's claim is supported.

Lesson Summary

Core Content — Spec 3.7.1

Genotype = alleles present; Phenotype = observable characteristics

Monohybrid cross (Tt × Tt): 3:1 phenotype ratio; 1:2:1 genotype ratio

Test cross : cross unknown dominant with homozygous recessive — all dominant → homozygous; 1:1 → heterozygous

Codominance : both alleles expressed; C^RC^W notation; 1:2:1 phenotype ratio in F2

Dihybrid cross (TtRr × TtRr): 9:3:3:1 ratio — only valid for unlinked genes

Chi-squared : χ² = Σ(O−E)²/E; compare to critical value at p = 0.05; always state conclusion biologically

Key Ratios

Monohybrid Tt × Tt → 3:1

Monohybrid Tt × tt → 1:1

Codominance F2 → 1:2:1

Dihybrid TtRr × TtRr → 9:3:3:1

Chi-Squared Steps

State null hypothesis

Calculate E values

Apply Σ(O−E)²/E

df = classes − 1

Compare to critical value (p=0.05)

Conclude biologically