Monohybrid Inheritance and Pedigrees
This topic covers monohybrid inheritance, which is the prediction of genetic outcomes for a single trait. You will use Punnett squares and family pedigrees to calculate probabilities and ratios, which are essential for understanding how traits and genetic conditions are passed down through generations.
Part of the ESAT Biology syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- A monohybrid cross involves tracking a single gene with two different versions, or alleles (e.g., 'T' and 't').
- For a standard cross between two heterozygous parents (e.g., Tt x Tt), the expected phenotypic ratio in offspring is 3 dominant: 1 recessive, and the genotypic ratio is 1 homozygous dominant: 2 heterozygous: 1 homozygous recessive (1:2:1).
- In pedigree analysis, if two unaffected parents have an affected child, the trait MUST be recessive, and both parents must be heterozygous carriers.
- Inherited conditions can be dominant, requiring only one copy of the allele to be expressed, or recessive, requiring two copies.
- While single-gene inheritance is a key concept, most real-world phenotypes (e.g., height) are polygenic, meaning they are influenced by multiple genes.
- The outcome of each new offspring is a statistically independent event; previous outcomes do not influence the next.
Formulae
P(outcome) = (Favourable outcomes) / (Total possible outcomes) To calculate the probability of a specific genotype or phenotype from a Punnett square.
P(A and B) = P(A) × P(B) To find the probability of two or more independent events occurring together, such as multiple children inheriting the same trait.
Definitions
- Monohybrid Cross
- A genetic cross that considers the inheritance of a single characteristic controlled by one gene.
- Genotype
- The specific combination of alleles an organism has for a gene, represented by letters (e.g., FF, Ff, ff).
- Phenotype
- The observable physical or biochemical characteristic of an organism, resulting from its genotype and environmental factors (e.g., having cystic fibrosis).
- Carrier
- A heterozygous individual (e.g., Ff) who does not show symptoms of a recessive genetic disorder but can pass the recessive allele to their offspring.
- Pedigree
- A diagram that tracks the inheritance of a specific trait through several generations of a family.
Worked example
Cystic fibrosis is an inherited condition caused by a recessive allele, 'f'. An unaffected man, whose father had cystic fibrosis, has a child with an unaffected woman who is a known carrier. What is the probability that their first child will be an unaffected carrier?
- 1
Determine the parents' genotypes.
The woman is a known carrier, so her genotype is Ff.
- 2
The man is unaffected, so he has at least one 'F' allele.
His father had cystic fibrosis (ff), so must have passed an 'f' allele to him.
Therefore, the man's genotype must be Ff.
- 3
Set up a Punnett square for the cross Ff x Ff.
- 4
The possible offspring genotypes are FF, Ff, Ff, and ff, in a 1:2:1 ratio.
- 5
Identify the 'unaffected carrier' genotype, which is Ff.
- 6
From the Punnett square, 2 out of the 4 possible outcomes result in the Ff genotype.
- 7
Calculate the probability:
P(Ff) = 2 / 4 = 1/2
Answer: 1/2 or 50%
Common mistakes
- ×Confusing ratios and fractions: A 3:1 ratio means the probabilities are 3/4 and 1/4, not 1/3. Remember to divide by the total number of parts (3+1=4). This avoids 'off-by-factor' errors.
- ×Forgetting about carriers in pedigree problems: If a condition is recessive, an unaffected individual can be homozygous dominant (e.g., FF) or heterozygous (Ff). Failing to consider the carrier possibility is a common 'missing constraint' mistake.
- ×Mixing up genotype and phenotype ratios: For an Ff x Ff cross, the phenotype ratio is 3:1 (dominant:recessive) but the genotype ratio is 1:2:1 (FF:Ff:ff). Use the correct one for the question asked.
- ×Assuming dependence between offspring: The probability for each child is independent. The chance of a second child having a condition is the same as the first, regardless of the first child's outcome.
- ×Misinterpreting 'unaffected': For a dominant condition, an unaffected individual MUST be homozygous recessive. For a recessive condition, they could be homozygous dominant or heterozygous.
No-calculator tips
- ✓Think in fractions, not decimals. It's much easier to multiply 1/2 × 3/4 (which is 3/8) than 0.5 × 0.75 in your head.
- ✓For large offspring numbers, use the ratio. If a 1:1 phenotypic ratio is expected from 150 offspring, you expect about 75 of each without needing to write anything down.
- ✓For multi-child probabilities (e.g., 'three children in a row'), calculate the probability for one child first (e.g., 1/4) and then power it (e.g., (1/4)3 = 1/64). Remember simple powers: 23=8, 43=64.