Difficulty: Intermediate | Prerequisites: Mendelian genetics, basic probability, understanding of dominance. Epistasis notes (companion doc) are helpful but not required.
Polygenic inheritance is how most real-world traits work: height, skin colour, crop yield, and disease risk are all controlled by multiple genes acting together rather than a single Mendelian locus. This topic bridges classical genetics and modern quantitative genetics. Where Mendelian genetics gives you discrete ratios (3:1, 9:3:3:1), polygenic inheritance gives you a continuous distribution, which is why traits like human height form a bell curve rather than falling into neat categories. If you understand additive allele contributions, you can solve every problem in this set.
Polygenic traits are controlled by two or more genes whose alleles contribute additively to the phenotype. Each dominant (or "contributing") allele adds a fixed increment. To find how many genes control a trait, use the frequency of an extreme phenotype in the population and the formula (1/4)^n = frequency of one extreme class.
Polygenic inheritance
A pattern of inheritance in which two or more genes (often many more) contribute to a single phenotypic trait. In simple terms, multiple genes team up to produce one trait, and each gene adds a small piece to the final result.
Additive alleles (contributing alleles)
Alleles that each contribute an equal, fixed amount to the phenotype. Capital letters are often used for contributing alleles and lowercase for non-contributing alleles. Think of each contributing allele as adding one "unit" of effect.
Quantitative trait
A trait that varies continuously across a range rather than falling into distinct categories. Height, weight, skin colour, and crop yield are all quantitative traits. In simple terms, these are "how much" traits rather than "which type" traits.
Continuous variation
The smooth range of phenotypes produced by polygenic inheritance (plus environmental effects). With enough genes, the distribution of phenotypes in a population approaches a bell curve (normal distribution).
Contribution per allele
Calculated as: (maximum phenotype value - minimum phenotype value) / (2n), where n is the number of genes. This tells you how much one contributing allele adds to the baseline.
Determining number of genes from population data
Use the formula: proportion of one extreme phenotype = (1/4)^n, where n = number of gene pairs. Alternatively: 1/(4^n) = frequency of the extreme class. Solve for n. In simple terms, the rarer the extreme phenotype in an F2 population, the more genes are involved.
Threshold trait
A trait that appears to be all-or-nothing (you either show it or you do not) but is actually underlain by a polygenic, continuous distribution. The trait only appears when enough contributing or non-contributing alleles accumulate past a threshold. Q6 in this problem set is an example.
Each contributing (dominant) allele adds an equal increment to the phenotype.
The baseline phenotype is set by the homozygous recessive individual (all lowercase alleles).
The maximum phenotype is set by the homozygous dominant individual (all uppercase alleles).
Formula: Contribution per allele = (max - min) / total number of alleles = (max - min) / 2n, where n = number of genes.
Count the number of contributing (uppercase) alleles in the genotype.
Multiply that count by the contribution per allele.
Add the result to the baseline (minimum) value.
Example: If 2 genes control height, min = 16 ft, max = 24 ft, then each allele contributes (24 - 16) / 4 = 2 ft. An individual with genotype AaBb has 2 contributing alleles, so height = 16 + (2 x 2) = 20 ft.
Cross two heterozygotes (e.g. AaBb x AaBb) and list all possible genotypes.
For each genotype, count contributing alleles and assign phenotype.
The distribution of phenotypes follows the coefficients of the binomial expansion.
For n genes, the F2 generation has (2n + 1) phenotypic classes.
With 2 genes: the classes are 0, 1, 2, 3, 4 contributing alleles, in the ratio 1:4:6:4:1.
With 3 genes: 0 through 6 contributing alleles, in the ratio 1:6:15:20:15:6:1.
Two methods are commonly used:
Method 1 (from extreme-class frequency): The fraction of the F2 that shows one extreme phenotype = (1/4)^n. So if 1/256 of offspring show the extreme, (1/4)^n = 1/256, and n = 4.
Method 2 (from population survey data): If you know the total surveyed and how many show the extreme, compute the proportion and solve (1/4)^n = proportion. This is less precise because real populations include environmental variation, so you look for the closest match.
Human skin colour in this simplified model uses 3 gene pairs (A, B, C). Each contributing allele adds one "unit of pigment." The phenotype classes are:
6 contributing alleles (AABBCC) = extremely dark
5 contributing alleles (e.g. AaBBCC) = very dark
4 contributing alleles (e.g. AaBbCC) = dark
3 contributing alleles (e.g. AaBbCc) = intermediate
2 contributing alleles (e.g. aaBbCc) = light
1 contributing allele (e.g. aabbCc) = very light
0 contributing alleles (aabbcc) = extremely light
Some problems define a cutoff: the trait only appears when you have at least X non-contributing (or contributing) alleles.
Treat these the same way: list all possible offspring genotypes, count the relevant alleles, then identify which genotypes meet the threshold.
Q6 in the practice set is this type: the trait shows only when an individual has 3 or more lowercase alleles.
Crop breeding relies heavily on quantitative genetics. Farmers cross high-yield and disease-resistant lines, then select offspring with the best combination of contributing alleles from both parents. The tomato skin-thickness problem (Q10) is a direct analogy to this kind of breeding programme.
Q5: Genes A and B interact to determine seed colour in wheat. AABB = dark red, aabb = white, AaBb = medium red. Colours depend on the number of contributing alleles (0 to 4).
a) Predict the phenotype of each genotype:
Count the contributing (uppercase) alleles:
i. AaBb = 2 contributing alleles = medium red
ii. AaBB = 3 contributing alleles = medium-dark red
iii. aaBb = 1 contributing allele = light red
iv. Aabb = 1 contributing allele = light red
b) If medium red plants (AaBb) are bred, out of 1200 offspring, how many would you expect to have:
AaBb x AaBb gives a 1:4:6:4:1 distribution across 16 outcomes:
4 contributing alleles (AABB) = 1/16 = dark red
3 contributing alleles = 4/16 = medium-dark red
2 contributing alleles = 6/16 = medium red
1 contributing allele = 4/16 = light red
0 contributing alleles (aabb) = 1/16 = white
i. Dark red seeds: 1/16 x 1200 = 75
ii. Medium red seeds: 6/16 x 1200 = 450
iii. White seeds: 1/16 x 1200 = 75
Q6: A trait depends on two gene pairs (A/a and B/b). Only individuals with 3 or more lowercase alleles show the trait. What is the chance the first child shows the trait?
The possible numbers of lowercase alleles range from 0 to 4. "Three or more" means 3 or 4 lowercase alleles.
a) AaBb x AABb
For the A locus: Aa x AA gives 1/2 AA, 1/2 Aa. Maximum lowercase A alleles = 1 (from Aa).
For the B locus: Bb x Bb gives 1/4 BB, 2/4 Bb, 1/4 bb.
Possible offspring genotypes and their lowercase counts:
AABb (1 lowercase) = 1/2 x 2/4 = 2/8
AABB (0) = 1/2 x 1/4 = 1/8
AAbb (2) = 1/2 x 1/4 = 1/8
AaBB (1) = 1/2 x 1/4 = 1/8
AaBb (2) = 1/2 x 2/4 = 2/8
Aabb (3) = 1/2 x 1/4 = 1/8
Only Aabb has 3 or more lowercase alleles. Probability = 1/8 (12.5%).
b) AaBb x AaBb
This is a standard dihybrid. The distribution of lowercase allele counts across 16 outcomes:
0 lowercase (AABB) = 1/16
1 lowercase (AABb or AaBB) = 4/16
2 lowercase (AAbb, AaBb, aaBB) = 6/16
3 lowercase (Aabb, aaBb) = 4/16
4 lowercase (aabb) = 1/16
Three or more lowercase = 4/16 + 1/16 = 5/16 (31.25%).
c) AaBb x AABB
For the A locus: Aa x AA gives 1/2 AA, 1/2 Aa (max 1 lowercase a).
For the B locus: Bb x BB gives 1/2 BB, 1/2 Bb (max 1 lowercase b).
Maximum lowercase alleles in any offspring = 1 (a) + 1 (b) = 2. No offspring can have 3 or more lowercase alleles. Probability = 0 (0%).
Q7: Using the skin colour table (3 gene pairs, 7 phenotype classes). A light individual (aaBbCc, 2 contributing alleles) mates with a dark individual (AaBbCC, 4 contributing alleles). What is the chance of very light offspring (1 contributing allele)?
Parents: aaBbCc x AaBbCC.
For very light offspring, we need exactly 1 contributing allele total.
A locus: aa x Aa gives 1/2 Aa, 1/2 aa. Contributing A alleles: Aa gives 1, aa gives 0.
B locus: Bb x Bb gives 1/4 BB (2), 2/4 Bb (1), 1/4 bb (0).
C locus: Cc x CC gives 1/2 CC (2), 1/2 Cc (1).
We need combinations that sum to exactly 1 contributing allele:
aa (0) + bb (0) + Cc (1) = 1. Probability: 1/2 x 1/4 x 1/2 = 1/16.
aa (0) + Bb (1) + cc (0) = impossible, because C locus is Cc x CC, so minimum 1 C allele.
Aa (1) + bb (0) + cc (0) = impossible for the same reason.
The only valid combination is aa, bb, Cc. Probability = 1/16 (6.25%).
Q9: Two genes control giraffe height additively. 4 dominant alleles = 24 ft, 4 recessive alleles = 16 ft. If these two giraffes mate, what percentage of their grandchildren would be at least 22 ft tall?
Contribution per allele = (24 - 16) / 4 = 2 ft per allele.
P cross: AABB (24 ft) x aabb (16 ft). F1 = all AaBb (20 ft).
F2 (grandchildren) = AaBb x AaBb. Distribution of contributing alleles (out of 16):
4 alleles (AABB) = 1/16 = 24 ft
3 alleles = 4/16 = 22 ft
2 alleles = 6/16 = 20 ft
1 allele = 4/16 = 18 ft
0 alleles = 1/16 = 16 ft
At least 22 ft means 3 or 4 contributing alleles = 4/16 + 1/16 = 5/16 = 31.25%.
Q10: Three genes determine tomato skin thickness. Homozygous recessive = 0.06 mm. Homozygous dominant = 0.18 mm. Can the farmer produce 0.10 mm skin thickness?
Contribution per allele = (0.18 - 0.06) / 6 = 0.02 mm per allele.
Phenotype for each number of contributing alleles:
0 alleles: 0.06 mm
1 allele: 0.08 mm
2 alleles: 0.10 mm
3 alleles: 0.12 mm
4 alleles: 0.14 mm
5 alleles: 0.16 mm
6 alleles: 0.18 mm
Yes, 0.10 mm is possible (2 contributing alleles).
P cross: AABBCC x aabbcc. F1 = AaBbCc (0.12 mm). F2 = AaBbCc x AaBbCc.
With 3 genes, the distribution follows the 6th row of Pascal's triangle: 1:6:15:20:15:6:1 out of 64 total.
2 contributing alleles corresponds to the coefficient 15.
Percentage with 0.10 mm = 15/64 = 23.4%.
Q11: Alligator tails range from 3.5 ft to 9 ft. Out of 125,000 alligators, 120 had tails of 9 ft. How many genes control tail length?
Proportion showing the maximum extreme = 120 / 125,000 = 0.00096.
Using (1/4)^n = proportion of one extreme:
(1/4)^1 = 0.25
(1/4)^2 = 0.0625
(1/4)^3 = 0.015625
(1/4)^4 = 0.00390625
(1/4)^5 = 0.000977
0.00096 is closest to (1/4)^5 = 0.000977. So approximately 5 genes control alligator tail length.
Q12: Octopus tentacle length ranges from 0.42 m to 0.89 m. Out of 319 measured, 20 had tentacles of 0.42 m. How many genes control tentacle length?
Proportion of the minimum extreme = 20 / 319 = 0.0627.
Using (1/4)^n:
(1/4)^1 = 0.25
(1/4)^2 = 0.0625
0.0627 is very close to (1/4)^2 = 0.0625. So approximately 2 genes control tentacle length.
Students often assume that "polygenic" means the trait must involve dozens of genes. Two genes with additive effects already count as polygenic.
The (1/4)^n formula gives the expected proportion of one extreme class only. The other extreme class has the same frequency.
Environmental variation makes real data messier than textbook ratios. When a problem says "excluding environmental influence," it is telling you to use the clean additive model.
Contribution per allele is always calculated from the total range divided by the total number of alleles (2n), not divided by the number of genes (n). This is a common arithmetic slip.
Expect questions that give you a range and an extreme-class frequency and ask you to solve for n. The (1/4)^n formula is the fastest route.
Questions on phenotype prediction from genotype are straightforward if you remember to count contributing alleles and multiply by the per-allele increment.
Three-gene additive problems use a 64-square Punnett or Pascal's triangle row 6 (1:6:15:20:15:6:1). Know this distribution or be prepared to derive it.
Threshold-trait problems ("3 or more lowercase alleles") combine polygenic counting with a simple inequality. List genotypes, count alleles, check the threshold.
True or False: In an additive model with 3 genes, there are 7 possible phenotypic classes.
True. The classes correspond to 0 through 6 contributing alleles.
Fill in the blank: The contribution per allele is calculated as (max - min) / ______.
2n, where n is the number of genes.
True or False: If (1/4)^n = 1/64, then n = 3.
True. (1/4)^3 = 1/64.
Fill in the blank: The F2 phenotypic ratio for a 2-gene additive cross is 1:4:______:4:1.
True or False: A genotype AaBBcc has 3 contributing alleles.
True. Aa contributes 1, BB contributes 2, cc contributes 0. Total = 3 contributing alleles.
Polygenic inheritance connects to epistasis (see the companion doc) because both involve multiple genes affecting one trait, but in different ways: epistasis involves masking, while polygenic inheritance involves additive contributions. Understanding both is essential for interpreting real genetic data where both phenomena may operate simultaneously.
This topic also connects to population genetics and evolution. Natural selection acts on quantitative traits by shifting the population mean, which is why traits like beak depth in Darwin's finches can change measurably across generations.
In medical genetics, most common diseases (diabetes, heart disease, certain cancers) are polygenic. Genome-wide association studies (GWAS) are essentially large-scale versions of the "determine number of genes" problems in this set.
polygenic inheritance, quantitative genetics, additive alleles, contributing alleles, continuous variation, bell curve, normal distribution, polygenic trait, skin colour genetics, wheat seed colour, threshold trait, number of genes formula, (1/4)^n, Pascal's triangle, binomial distribution, phenotypic classes, additive model, crop breeding, quantitative trait loci, QTL, BIO 138, polygenic epistatic quantitative practice