Genetics: Monohybrid Crosses and Probability

What you'll learn

Monohybrid crosses form the foundation of inheritance patterns tested in CXC CSEC Integrated Science papers. This topic examines how a single characteristic passes from parents to offspring through alleles, using probability to predict offspring ratios. Exam questions typically require you to construct Punnett squares, calculate genetic ratios, and interpret inheritance patterns using correct terminology.

Key terms and definitions

Gene — a section of DNA that codes for a specific characteristic (e.g., flower colour, blood group)

Allele — alternative forms of the same gene; for example, the allele for tall stems (T) versus the allele for short stems (t)

Dominant allele — an allele that expresses its characteristic even when only one copy is present; represented by a capital letter (e.g., T)

Recessive allele — an allele that only expresses its characteristic when two copies are present; represented by a lowercase letter (e.g., t)

Genotype — the genetic makeup of an organism for a particular characteristic, written as a pair of alleles (e.g., TT, Tt, tt)

Phenotype — the observable characteristic resulting from the genotype (e.g., tall stem, short stem)

Homozygous — having two identical alleles for a gene (e.g., TT or tt); also called pure-breeding

Heterozygous — having two different alleles for a gene (e.g., Tt); also called hybrid

Core concepts

Mendel's principles of inheritance

Gregor Mendel established the laws of inheritance through experiments with pea plants. His work revealed that:

• Characteristics are controlled by pairs of factors (now called alleles)
• Each parent contributes one allele to the offspring
• Some alleles are dominant over others
• Alleles separate during gamete formation (sex cell production)

The monohybrid cross examines the inheritance of one characteristic controlled by one gene with two alleles. This contrasts with dihybrid crosses (two characteristics), which exceed CSEC requirements.

Understanding genotypes and phenotypes

Every organism inherits two alleles for each gene — one from each parent. The combination determines both genotype and phenotype.

Example using stem height in pigeon pea plants (common in Caribbean agriculture):

• Gene: Stem height
• Alleles: T (tall) and t (short)
• Possible genotypes: TT, Tt, tt
• Phenotypes: TT and Tt both produce tall plants; only tt produces short plants

The dominant allele (T) masks the recessive allele (t) in heterozygous individuals. This means Tt plants appear identical to TT plants, even though their genotypes differ.

Constructing Punnett squares

The Punnett square is the standard method for predicting offspring ratios in genetic crosses. CXC examiners expect you to construct these accurately and interpret results correctly.

Step-by-step method:

1. Identify the parent genotypes
2. Determine the possible gametes each parent can produce (each gamete carries one allele)
3. Draw a 2×2 grid for a monohybrid cross
4. Write one parent's gametes across the top
5. Write the other parent's gametes down the left side
6. Fill each box by combining the gametes that meet
7. Count the genotype and phenotype ratios

Cross 1: Homozygous dominant × Homozygous recessive (TT × tt)

Parent genotypes: TT × tt Gametes: T, T from first parent; t, t from second parent

     T    T
t   Tt   Tt
t   Tt   Tt

Result: All offspring (100%) are Tt — heterozygous genotype, tall phenotype

Cross 2: Heterozygous × Heterozygous (Tt × Tt)

This cross produces the classic 3:1 phenotypic ratio frequently tested at CSEC level.

Parent genotypes: Tt × Tt Gametes: T, t from each parent

     T    t
T   TT   Tt
t   Tt   tt

Results:

• Genotypes: 1 TT : 2 Tt : 1 tt (ratio 1:2:1)
• Phenotypes: 3 tall : 1 short (ratio 3:1)

Probability in genetics

Genetic inheritance follows probability rules. Each fertilization event is independent, meaning previous offspring do not affect the probability of future combinations.

Key probability principles:

• Probability of any outcome = (number of ways it can occur) ÷ (total possible outcomes)
• Express as fractions, decimals, or percentages
• For a Tt × Tt cross: probability of tall offspring = 3/4 = 0.75 = 75%
• Probability of short offspring = 1/4 = 0.25 = 25%

Caribbean example — Chicken feather colour:

In some Caribbean poultry breeds, black feathers (B) are dominant over white feathers (b).

If two heterozygous black chickens (Bb × Bb) breed:

• Probability of black chick = 3/4 (75%)
• Probability of white chick = 1/4 (25%)
• If the farmer hatches 8 chicks, the expected number of white chicks = 8 × 0.25 = 2 chicks

Remember: these are expected ratios based on probability. Actual results may vary, especially with small sample sizes.

Test crosses and determining unknown genotypes

A tall plant could be either TT or Tt genotype — both produce the same tall phenotype. A test cross reveals the unknown genotype by crossing the organism with a homozygous recessive individual (tt).

Scenario: A tall tomato plant in a Jamaican market garden has unknown genotype

Test cross: Unknown tall plant × tt (short plant)

If the tall plant is TT:

• All offspring will be Tt (100% tall)

If the tall plant is Tt:

• Offspring will be 50% Tt (tall) and 50% tt (short)

By examining offspring phenotypes, you can deduce the parent's genotype. This technique applies to plant breeding programmes across the Caribbean agricultural sector.

Applications in Caribbean agriculture and livestock

Understanding monohybrid inheritance helps Caribbean farmers improve crops and livestock:

Plant breeding:

• Selecting tomato varieties resistant to bacterial wilt (common in Trinidad)
• Developing cocoa trees with disease resistance in Grenada
• Producing sweet pepper varieties with specific fruit colours for export

Livestock breeding:

• Selecting cattle with polled (hornless) characteristics in Jamaican beef production
• Breeding chickens with desired feather patterns
• Maintaining pure breeds versus creating hybrids for vigor

Worked examples

Example 1: Basic monohybrid cross

Question: In guinea pigs, rough coat (R) is dominant over smooth coat (r). A homozygous rough-coated guinea pig is crossed with a smooth-coated guinea pig.

(a) State the genotypes of the two parents. [2 marks] (b) Using a Punnett square, determine the genotypes of the offspring. [3 marks] (c) What percentage of offspring will have rough coats? [1 mark]

Solution:

(a) Homozygous rough coat: RR Smooth coat: rr (must be homozygous recessive to show the recessive trait)

(b) Punnett square:

Parent genotypes: RR × rr
Gametes: R, R and r, r

     R    R
r   Rr   Rr
r   Rr   Rr

All offspring genotypes: Rr [3 marks for correct square, gametes, and offspring]

(c) 100% of offspring will have rough coats (all are Rr heterozygous with the dominant allele expressed) [1 mark]

Example 2: Heterozygous cross with probability

Question: In a St. Lucian vegetable farm, purple stem colour (P) in eggplant is dominant over green stem colour (p). Two heterozygous purple-stemmed plants are crossed.

(a) Draw a Punnett square to show this cross. [3 marks] (b) State the phenotypic ratio of the offspring. [2 marks] (c) If 60 seeds from this cross are planted, how many are expected to produce green-stemmed plants? [2 marks]

Solution:

(a) Punnett square:

Parent genotypes: Pp × Pp
Gametes: P, p from each parent

     P    p
P   PP   Pp
p   Pp   pp

[3 marks: 1 for correct parent genotypes, 1 for gametes, 1 for completed square]

(b) Phenotypic ratio: 3 purple : 1 green (or 3:1) [2 marks: 1 for correct ratio, 1 for correct phenotype labels]

(c) Probability of green stem = 1/4 = 0.25 Expected number = 60 × 0.25 = 15 green-stemmed plants [2 marks: 1 for calculation method, 1 for correct answer]

Example 3: Test cross application

Question: A farmer in Barbados has a tomato plant producing red fruits. Red fruit colour (R) is dominant over yellow fruit colour (r). The farmer wants to know if the plant is homozygous (RR) or heterozygous (Rr).

(a) Describe how the farmer could use a test cross to determine the plant's genotype. [2 marks] (b) If the plant is heterozygous, what ratio of offspring would result from the test cross? [2 marks]

Solution:

(a) Cross the red-fruited plant with a homozygous recessive (rr) yellow-fruited plant. Examine the phenotypes of the offspring to determine the parent's genotype. [2 marks]

(b) If the plant is Rr:

     R    r
r   Rr   rr
r   Rr   rr

Ratio: 1 red : 1 yellow (or 50% red, 50% yellow) [2 marks]

(Alternative: If all offspring are red, the parent was RR)

Common mistakes and how to avoid them

• Mistake: Using the same letter in different cases for unrelated genes (e.g., T for tall and t for texture) Correction: Each gene uses its own letter. Different alleles of the same gene use capital/lowercase versions of the same letter (T/t for height, R/r for colour)

• Mistake: Writing phenotypes instead of genotypes in Punnett squares (writing "tall" instead of "TT") Correction: Punnett squares contain only allele symbols (letters). Phenotypes are descriptive words used in conclusions

• Mistake: Confusing genotypic and phenotypic ratios; stating 1:2:1 when asked for phenotypic ratio Correction: Genotypic ratio includes all genetic combinations (1 TT : 2 Tt : 1 tt). Phenotypic ratio groups by appearance (3 tall : 1 short)

• Mistake: Forgetting that probability applies to each individual offspring, not guaranteeing exact ratios in small samples Correction: Use words like "expected," "probability," or "on average." A 3:1 ratio means each offspring has a 3/4 chance of the dominant phenotype, not that exactly 3 out of every 4 will show it

• Mistake: Placing parent alleles incorrectly in the Punnett square (mixing up which parent's gametes go on top versus side) Correction: Be consistent — typically place one parent's gametes across the top row and the other's down the left column. The specific arrangement doesn't matter as long as you're systematic

• Mistake: Assuming a dominant trait is "better" or more common than recessive Correction: Dominance refers only to expression in heterozygotes, not quality or frequency. Many recessive alleles are common in populations

Exam technique for Genetics: Monohybrid Crosses and Probability

• Command word "State": Give genotypes using correct notation without explanation (e.g., "TT and Tt"). Brief answers earn full marks; avoid lengthy explanations that waste time.

• Command word "Using a Punnett square": You must draw the grid and show gametes even if you can work out ratios mentally. Examiners award marks for the method, not just the final answer. Label parent genotypes and gametes clearly.

• Command word "Calculate" or "Determine": Show your working. For probability calculations, write the formula: (probability × total number). Even if your final answer is incorrect, you can earn method marks.

• Marks allocation patterns: A typical 6-mark question might include: parent genotypes (1 mark), gametes identified (1 mark), Punnett square completed (2 marks), offspring ratio (1 mark), probability calculation or interpretation (1 mark). Allocate your time proportionally — don't spend 5 minutes on a 2-mark question.

Quick revision summary

Monohybrid crosses examine single-gene inheritance patterns. Each parent contributes one allele; dominant alleles (capitals) mask recessive alleles (lowercase) in heterozygotes. Construct Punnett squares by writing gametes on the grid's edges and combining them in boxes. Homozygous crosses (TT × tt) produce all heterozygous offspring. Heterozygous crosses (Tt × Tt) yield the classic 3:1 phenotypic ratio and 1:2:1 genotypic ratio. Calculate probability as favourable outcomes divided by total outcomes. Test crosses with homozygous recessive individuals reveal unknown genotypes. Master Punnett square construction and probability calculations — these skills appear on every CSEC Integrated Science genetics question.