1. Overview
Monohybrid inheritance is the study of how a single characteristic is passed from parents to offspring through the transmission of genes. By understanding the relationship between alleles, we can use genetic diagrams to predict the probability of specific physical traits appearing in the next generation, which is essential for fields ranging from medicine to agriculture.
2. Key Definitions
- Inheritance: The transmission of genetic information from generation to generation.
- Genotype: The genetic make-up of an organism in terms of the alleles present (e.g., $Tt$ or $GG$).
- Phenotype: The observable features of an organism (e.g., tall height, green seeds).
- Allele: An alternative form of a gene.
- Homozygous: Having two identical alleles of a particular gene (e.g., $TT$ or $tt$).
- Heterozygous: Having two different alleles of a particular gene (e.g., $Tt$).
- Dominant Allele: An allele that is expressed if it is present in the genotype (represented by a capital letter).
- Recessive Allele: An allele that is only expressed when there is no dominant allele of the gene present in the genotype (represented by a lowercase letter).
- Pure-breeding: When two identical homozygous individuals breed together, all offspring will have the same phenotype as the parents.
3. Core Content
Pure-breeding vs. Heterozygous
- Homozygous (Pure-breeding): Individuals with genotypes like $AA$ or $aa$ are pure-breeding because they can only pass on one type of allele to their gametes.
- Heterozygous: Individuals with the genotype $Aa$ are not pure-breeding because they can pass on either the dominant ($A$) or the recessive ($a$) allele to their offspring.
Predicting Inheritance: Genetic Diagrams & Punnett Squares
To predict the outcome of a cross, follow these steps:
- Define the symbols: e.g., $B$ = Brown eyes (dominant), $b$ = Blue eyes (recessive).
- State the parental phenotypes: e.g., Brown eyes x Blue eyes.
- State the parental genotypes: e.g., $Bb$ x $bb$.
- Identify the gametes: Separate the alleles and circle them.
- Use a Punnett Square: Place the gametes of one parent on the top and the other on the side.
Common Phenotypic Ratios
- 3:1 Ratio: Occurs when two heterozygous parents ($Aa$ x $Aa$) are crossed. 75% show the dominant trait, 25% show the recessive trait.
- 1:1 Ratio: Occurs when a heterozygous parent ($Aa$) is crossed with a homozygous recessive parent ($aa$). 50% dominant trait, 50% recessive trait.
Pedigree Diagrams
Pedigree charts show the inheritance of a trait over several generations.
- Squares represent males; Circles represent females.
- Shaded shapes represent individuals who express the trait being studied.
- Horizontal lines between a square and circle represent a mating pair.
- Vertical lines leading down to a bracket represent the offspring.
4. Supplement Content (Extended Only)
The Test Cross
A test cross is used to identify the unknown genotype of an organism with a dominant phenotype (is it $AA$ or $Aa$?).
- Process: Cross the unknown individual with a homozygous recessive ($aa$) individual.
- Result A: If any offspring show the recessive phenotype, the unknown parent must be heterozygous ($Aa$).
- Result B: If 100% of the offspring show the dominant phenotype, the unknown parent is likely homozygous dominant ($AA$).
Codominance
Codominance occurs when both alleles in a heterozygous organism contribute to the phenotype. Neither allele is dominant over the other.
- ABO Blood Groups: Controlled by three alleles: $I^A$, $I^B$, and $I^o$.
- $I^A$ and $I^B$ are codominant.
- $I^o$ is recessive to both $I^A$ and $I^B$.
- Phenotypes and Genotypes:
- Group A: $I^AI^A$ or $I^AI^o$
- Group B: $I^BI^B$ or $I^BI^o$
- Group AB: $I^AI^B$ (Both A and B antigens are expressed)
- Group O: $I^oI^o$
Sex-linked Characteristics
A sex-linked characteristic is a feature where the gene responsible is located on a sex chromosome (usually the X chromosome).
- Red-green colour blindness: The gene is carried on the X chromosome.
- Mechanism: Because males are $XY$, they only have one X chromosome. If they inherit the faulty allele, they will be colour blind. Females ($XX$) have two X chromosomes; if they have one faulty allele, they are "carriers" but have normal vision. They only express the trait if they have two faulty alleles.
- Genetic Notation: Use $X^R$ (Normal) and $X^r$ (Colour blind). A carrier female is $X^RX^r$. A colour blind male is $X^rY$.
5. Key Equations
- Phenotypic Ratio: $\text{Number of offspring with phenotype A} : \text{Number of offspring with phenotype B}$
- Probability: $\frac{\text{Number of specific outcomes}}{\text{Total number of possible outcomes}} \times 100%$
- Note: In biology, ratios should always be simplified (e.g., 3:1 rather than 75:25).
6. Common Mistakes to Avoid
- ❌ Wrong: Thinking the dominant allele is the "strongest" or "most common" in a population.
- ✅ Right: The dominant allele is simply the one expressed in the phenotype when the genotype is heterozygous.
- ❌ Wrong: Using different letters for the same gene (e.g., $B$ for brown and $g$ for green).
- ✅ Right: Always use the same letter in upper and lower case (e.g., $B$ and $b$).
- ❌ Wrong: Forgetting to use the $I^A / I^B / I^o$ notation for blood groups.
- ✅ Right: Always write the superscript to show the specific alleles.
7. Exam Tips
- Command Words: If asked to "Define" a term like genotype, use the exact syllabus wording: "the genetic make-up of an organism in terms of the alleles present."
- Show Your Work: In Punnett square questions, always write out the gametes and the final ratio. Even if your final answer is wrong, you can earn marks for the correct process.
- Contexts: Expect questions about plant height, seed colour, or human diseases like cystic fibrosis (recessive) or Huntington's (dominant).
- Numerical Values: Be prepared to calculate percentages. For example, in a 3:1 ratio, the probability of the recessive phenotype is 0.25 or 25%.
Exam-Style Questions
Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0610 Theory papers.
Exam-Style Question 1 — Short Answer [5 marks]
Question:
In pea plants, the allele for tall stems (T) is dominant to the allele for short stems (t).
(a) Define the terms genotype and phenotype. [2]
(b) A tall pea plant with an unknown genotype is crossed with a short pea plant. Of the offspring, half are tall and half are short. State the genotype of:
(i) the tall parent [1]
(ii) the short parent [1]
(c) State the ratio of tall to short plants in the offspring. [1]
Worked Solution:
(a)
Genotype is the genetic makeup of an organism. This refers to the alleles present in the organism's cells.
Phenotype is the observable characteristics of an organism. This is determined by the genotype and the environment.
How to earn full marks:
- For genotype: mention genetic makeup or alleles present.
- For phenotype: mention observable characteristics or features.
(b) (i) The tall parent must be heterozygous to produce both tall and short offspring. $\boxed{Tt}$
How to earn full marks:
- State the correct genotype.
(ii) The short parent must be homozygous recessive to express the short phenotype. $\boxed{tt}$
How to earn full marks:
- State the correct genotype.
(c)
- The question states that half the offspring are tall and half are short. $\boxed{1:1}$
How to earn full marks:
- State the correct ratio.
Common Pitfall: Students sometimes confuse genotype and phenotype. Remember that genotype refers to the actual alleles, while phenotype is the physical expression of those alleles. Also, make sure you understand what homozygous and heterozygous mean.
Exam-Style Question 2 — Short Answer [6 marks]
Question:
Coat colour in a certain breed of dog is controlled by a single gene with two alleles: black (B) and brown (b). Black coat colour is dominant to brown coat colour.
(a) State what is meant by the term dominant allele. [1]
(b) A breeder mates two dogs with black coats. They produce a litter of puppies, some of which have black coats and some of which have brown coats.
(i) State the genotypes of the two parent dogs. [2]
(ii) Complete a Punnett square to show the possible genotypes of the offspring. [2]
(iii) State the probability of the offspring having a brown coat. [1]
Worked Solution:
(a)
- A dominant allele is expressed even if only one copy is present in the genotype. This means the phenotype will show the characteristic associated with the dominant allele.
How to earn full marks:
- Mention that the allele is expressed even if only one copy is present, OR that it masks the recessive allele.
(b) (i) Since both parents have black coats but produce brown puppies, they must both be heterozygous. $\boxed{Bb}$
How to earn full marks:
- State the correct genotype for both parents.
(ii)
- Set up the Punnett square with the parent genotypes.
| B | b | |
|---|---|---|
| B | BB | Bb |
| b | Bb | bb |
How to earn full marks:
- Correctly place the alleles from each parent on the axes.
- Correctly fill in all four boxes with the resulting genotypes.
(iii)
- From the Punnett square, one out of four offspring have the genotype bb, which results in a brown coat. $\boxed{25 %}$
How to earn full marks:
- State the correct percentage.
Common Pitfall: When working with Punnett squares, double-check that you've correctly transferred the parent genotypes to the top and side of the square. A small error there can throw off the entire calculation. Also, remember that a percentage needs the % symbol.
Exam-Style Question 3 — Extended Response [9 marks]
Question:
A farmer breeds chickens. In chickens, the allele for black feathers (B) is dominant to the allele for white feathers (b). The farmer wants to produce only chickens with black feathers.
(a) State what is meant by the term homozygous. [1]
(b) Explain how the farmer could use a test cross to determine the genotype of a black-feathered chicken. [4]
(c) The farmer has two black-feathered chickens. He breeds them together and they produce 12 offspring. 9 of the offspring have black feathers and 3 have white feathers. Calculate the ratio of black-feathered to white-feathered offspring. [1]
(d) Predict, using a Punnett square, the genotypes of the parents. [3]
Worked Solution:
(a)
- Homozygous means having two identical alleles for a particular gene. For example, BB or bb.
How to earn full marks:
- Mention two identical alleles.
(b)
Cross the black-feathered chicken with a white-feathered chicken. A test cross always involves crossing with a homozygous recessive individual.
If all offspring have black feathers, the black-feathered chicken is homozygous dominant (BB). This is because even if the black-feathered chicken contributes a B allele, the white-feathered chicken will contribute only b alleles, resulting in all offspring with the genotype Bb.
If any offspring have white feathers, the black-feathered chicken is heterozygous (Bb). This is because the black-feathered chicken can contribute either a B or b allele. If it contributes a b allele, and the white-feathered chicken contributes a b allele, the resulting genotype will be bb, resulting in a white-feathered chicken.
How to earn full marks:
- State that the black-feathered chicken is crossed with a white-feathered chicken.
- Explain that if all offspring are black, the black-feathered chicken is homozygous dominant.
- Explain that if any offspring are white, the black-feathered chicken is heterozygous.
(c)
- Divide the number of black-feathered offspring by the number of white-feathered offspring. $9/3 = 3$ $\boxed{3:1}$
How to earn full marks:
- State the correct ratio.
(d)
- Since the offspring show a 3:1 ratio of black to white feathers, both parents must be heterozygous (Bb). This is a classic monohybrid cross ratio.
| B | b | |
|---|---|---|
| B | BB | Bb |
| b | Bb | bb |
$\boxed{Bb \times Bb}$
How to earn full marks:
- State the correct genotype for each parent.
- Justify your answer with the correct Punnett square.
Common Pitfall: Many students forget that a test cross involves crossing the unknown genotype with a homozygous recessive individual. Also, when calculating ratios, make sure you simplify them to their lowest terms.
Exam-Style Question 4 — Extended Response [10 marks]
Question:
In humans, the ABO blood group system is controlled by three alleles: $I^A$, $I^B$, and $I^O$. $I^A$ and $I^B$ are codominant, while $I^O$ is recessive to both $I^A$ and $I^B$.
(a) Define the term codominance. [2]
(b) State the possible genotypes for a person with:
(i) blood group A [2]
(ii) blood group O [1]
(c) A man with blood group AB marries a woman with blood group A. Her mother had blood group O.
(i) State the genotype of the man. [1]
(ii) State the possible genotypes of the woman. [1]
(iii) Complete a Punnett square to show the possible genotypes of their children. [2]
(iv) State the probability of their child having blood group O. [1]
Worked Solution:
(a)
- Codominance is when both alleles in a heterozygous individual are expressed. This means that neither allele is dominant or recessive, and both contribute to the phenotype.
How to earn full marks:
- Mention that both alleles are expressed.
- Mention that neither allele is dominant or recessive.
(b) (i) Blood group A can be achieved by either two $I^A$ alleles, or one $I^A$ allele and one $I^O$ allele. $\boxed{I^AI^A \text{ or } I^AI^O}$
How to earn full marks:
- State both possible genotypes.
(ii) Blood group O can only be achieved by two $I^O$ alleles. $\boxed{I^OI^O}$
How to earn full marks:
- State the correct genotype.
(c) (i) Blood group AB is only achieved by having one $I^A$ allele and one $I^B$ allele. $\boxed{I^AI^B}$
How to earn full marks:
- State the correct genotype.
(ii) The woman has blood group A. Since her mother had blood group O ($I^OI^O$), the woman must have inherited one $I^O$ allele from her mother. $\boxed{I^AI^O}$
How to earn full marks:
- State the correct genotype.
(iii)
- Set up the Punnett square with the parent genotypes.
| $I^A$ | $I^B$ | |
|---|---|---|
| $I^A$ | $I^AI^A$ | $I^AI^B$ |
| $I^O$ | $I^AI^O$ | $I^BI^O$ |
How to earn full marks:
- Correctly place the alleles from each parent on the axes.
- Correctly fill in all four boxes with the resulting genotypes.
(iv)
- There are no possible offspring with the $I^OI^O$ genotype. $\boxed{0 %}$
How to earn full marks:
- State the correct probability.
Common Pitfall: With codominance problems like blood groups, it's crucial to remember that both alleles are expressed. Also, pay close attention to the information given about family history, as this can help you narrow down the possible genotypes. Don't forget the % symbol for percentage answers.