Monohybrid inheritance
LO 1: Inheritance is the transmission of genetic information from generation to generation. This information is carried on chromosomes within the nucleus of cells.
LO 2 & 3: Genotype vs Phenotype
- Genotype: The genetic make-up of an organism, specifically the combination of alleles present. An allele is an alternative form of a gene.
- Phenotype: The observable features of an organism (e.g., eye colour, blood group). The phenotype results from the interaction of the genotype with the environment.
LO 4 & 6: Homozygous vs Heterozygous
- Homozygous: An individual having two identical alleles of a particular gene (e.g., BB or bb).
- Heterozygous: An individual having two different alleles of a particular gene (e.g., Bb).
LO 5 & 7: Pure-breeding
- Two identical homozygous individuals that breed together will be pure-breeding. This means all offspring will have the same genotype and phenotype as the parents for that trait.
- A heterozygous individual is not pure-breeding because it carries a hidden recessive allele that can be passed to offspring, potentially resulting in different phenotypes in the next generation.
LO 8 & 9: Dominant and Recessive Alleles
- Dominant allele: An allele that is expressed if it is present in the genotype. It masks the effect of a recessive allele. Represented by a capital letter (e.g., B).
- Recessive allele: An allele that is only expressed when there is no dominant allele of the gene present in the genotype. It is only visible in the phenotype if the individual is homozygous recessive (e.g., bb). Represented by a lowercase letter (e.g., b).
Consider a cross between two heterozygous parents (Bb \times Bb) for a trait where B is dominant and b is recessive.
Step 1: Parental Genotypes
Parent 1: Bb
Parent 2: Bb
Step 2: Gametes
Each parent produces gametes with only one allele. The alleles segregate during gamete formation.
Gametes from Parent 1: B and b
Gametes from Parent 2: B and b
Step 3: Punnett Square
| B | b | |
|---|---|---|
| B | BB | Bb |
| b | Bb | bb |
Step 4: Results
- Genotypes: 1 BB : 2 Bb : 1 bb
- Phenotypes: Since B is dominant, both BB and Bb show the dominant trait. Only bb shows the recessive trait.
- Dominant phenotype: 3
- Recessive phenotype: 1
- Phenotypic Ratio: 3 : 1
This demonstrates how a heterozygous cross yields a 3:1 phenotypic ratio.
Correction: Always check the command word. If asked for genotype, use letters (BB, Bb, bb). If asked for phenotype, describe the observable feature ('brown eyes', 'white flowers').
Correction: Gametes are haploid. They carry only one allele for each gene. If the parent is Bb, the gametes are either B OR b, never both.
Tip: Look for the 'key' to recessive traits. If two unaffected parents produce an affected child, the trait must be recessive. The parents are therefore heterozygous carriers.
Reasoning: Examiners accept this logic because a dominant trait cannot skip generations in this manner; if it were dominant, at least one parent would have to show the phenotype. This directly addresses the definition of recessive inheritance.
Tip: State that 'the individual must carry the recessive allele because they have an affected parent/offspring or produced an affected offspring.'
Reasoning: This shows you understand that the phenotype does not reveal the hidden genotype in dominant traits, but the transmission pattern reveals it.
LO 13: Using a Test Cross
A test cross is used to identify the unknown genotype of an individual showing the dominant phenotype. Since both BB and Bb show the dominant trait, we cannot tell them apart by looking.
Method: Cross the unknown dominant individual with a homozygous recessive individual (bb).
Interpretation:
- If all offspring show the dominant phenotype, the unknown parent is homozygous dominant (BB). (Cross: BB \times bb \rightarrow all Bb)
- If the offspring show a 1:1 ratio of dominant to recessive phenotypes, the unknown parent is heterozygous (Bb). (Cross: Bb \times bb \rightarrow Bb : bb)
Codominance occurs when both alleles in a heterozygous organism contribute to the phenotype. Neither allele is dominant or recessive; both are fully expressed.
Example: Blood groups or flower colour.
If R codes for red pigment and W codes for white pigment, a heterozygote (RW) has both red and white patches (or pink if blending, but codominance specifically means distinct expression of both). In chickens, feather colour might be speckled black and white.
The ABO blood group system involves three alleles: I^A, I^B, and I^O.
- I^A and I^B are codominant to each other.
- I^O is recessive to both I^A and I^B.
Genotype to Phenotype Mapping:
| Genotype | Phenotype (Blood Group) |
|---|---|
| I^A I^A or I^A I^O | A |
| I^B I^B or I^B I^O | B |
| I^A I^B | AB (Codominant) |
| I^O I^O | O (Recessive) |
Example Cross: Parent 1 (I^A I^O) x Parent 2 (I^B I^O)
Gametes: I^A, I^O and I^B, I^O
Offspring Genotypes: I^A I^B, I^A I^O, I^B I^O, I^O I^O
Offspring Phenotypes: AB, A, B, O (each with probability 0.25 or 25%).
LO 16: Sex-Linkage
A sex-linked characteristic is a feature where the gene responsible is located on a sex chromosome (usually the X chromosome). This makes the characteristic more common in one sex than the other.
Why is it more common in males?
Males have one X and one Y chromosome (XY). Females have two X chromosomes (XX).
- A female needs two copies of a recessive allele to show the trait (X^b X^b).
- A male only needs one copy (he is hemizygous) because the Y chromosome does not carry the corresponding gene (X^b Y). Therefore, it is statistically easier for a male to inherit and express the condition.
LO 17: Red-Green Colour Blindness
Red-green colour blindness is an example of sex-linked inheritance. It is caused by a recessive allele on the X chromosome.
Let X^B = normal vision (dominant)
Let X^b = colour blind (recessive)
Genotypes and Phenotypes:
- X^B X^B: Normal female
- X^B X^b: Carrier female (normal vision but carries the allele)
- X^b X^b: Colour blind female
- X^B Y: Normal male
- X^b Y: Colour blind male
Cross: Carrier female (X^B X^b) x Normal male (X^B Y)
Gametes:
Female: X^B, X^b
Male: X^B, Y
Punnett Square:
| X^B (Male) | Y (Male) | |
|---|---|---|
| X^B (Female) | X^B X^B (Normal Female) | X^B Y (Normal Male) |
| X^b (Female) | X^B X^b (Carrier Female) | X^b Y (Colour Blind Male) |
Results:
- Females: 50% Normal (X^B X^B), 50% Carrier (X^B X^b). None are colour blind.
- Males: 50% Normal (X^B Y), 50% Colour Blind (X^b Y).
- Overall probability of a colour blind child: 25% (or 1 in 4).
Key Insight: All daughters are phenotypically normal, but half are carriers. Half the sons are colour blind.
Correction: Sex-linked traits are almost always on the X chromosome. The Y chromosome is much smaller and carries very few genes. Do not write Y^B or Y^b.
Correction: Because the normal allele (X^B) is dominant, a carrier female has normal vision. She only 'carries' the gene. She is phenotypically normal.
Tip: You must show all steps clearly: Parental genotypes, Gametes (with arrows or brackets), Punnett square with offspring genotypes, and finally the Phenotypes/Ratios. Label each step.
Reasoning: Examiners award marks for each stage. If you only give the final answer, you may lose marks for not showing your working. Explicitly stating 'Gametes:' helps separate this step from the parental genotype.
Tip: Be careful with the denominator. If asked for the probability of a child being affected, use all 4 squares (e.g., 1/4). If asked for the probability that a son is affected, use only the male squares as the denominator (e.g., 1/2).
Reasoning: The question specifies the sample space. 'A child' implies the total population of offspring. 'A son' restricts the population to males only.