Ch 2 · 16.2 The roles of genes in determining the phenotype · AS & A Level Biology 9700
Topic 2 of 3 in Inheritance — AS & A Level Biology 9700 — 108 lessons.
Introduction
A phenotype is rarely controlled by a single gene acting alone: genes may be linked, may interact, may be sex-linked, or may exist as several alleles in a population. Examiners also expect the chi-squared test to check whether observed cross results fit an expected ratio. This lesson works through the key terms, monohybrid and dihybrid crosses, the chi-squared test, linkage, test crosses, epistasis, sex linkage and multiple alleles, then four human examples linking a gene to a protein to a phenotype. (Syllabus section 16.2.)
Key ideas — Key terms and monohybrid crosses
A monohybrid cross follows one gene. Write the parents' phenotypes and genotypes, the gametes (circled), a Punnett square, then the offspring genotypes, phenotypes and ratio. For two heterozygotes (Aa × Aa): 1 AA : 2 Aa : 1 aa, so 3 dominant : 1 recessive. With codominance, e.g. red (C^R C^R) × white (C^W C^W) flowers giving pink (C^R C^W) F1, crossing the F1 gives 1 red : 2 pink : 1 white.
🧑🏫 Teacher's explanation
Think of your DNA as a massive library of instruction manuals. A 'gene' is a single manual that tells the cell how to build a specific protein. However, there can be different versions of that manual—one version might instruct the cell to make blue eyes, while another version instructs it to make brown eyes. These different versions are called 'alleles'.
📖 Definitions
🔍 Worked example — Easy
In pea plants, tall stems (T) are dominant to short stems (t). A heterozygous tall plant is crossed with a homozygous recessive short plant. Determine the phenotypic ratio of the offspring.
🔍 Worked example — Standard
In certain cattle, coat color is determined by codominance. Red coat (C^R C^R) and white coat (C^W C^W) result in roan (spotted red and white) offspring (C^R C^W). If two roan cattle are crossed, what is the expected phenotypic ratio in the F_2 generation?
🔍 Worked example — Exam-style
A researcher performs a test cross on a plant with an unknown phenotype. The plant is crossed with a homozygous recessive plant. The resulting F_1 generation shows 50% tall plants and 50% short plants. What was the genotype of the original plant?
⚠️ Common errors
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🎯 Exam question
(a) State the genotypes of the parents. [1] (b) Using a genetic diagram, predict the phenotypic ratio of the offspring. [3] [4 marks]
Key ideas — Dihybrid crosses
A dihybrid cross follows the inheritance of two genes at once. If the two genes are on different (non-homologous) chromosomes, they assort independently during meiosis. A cross between two organisms heterozygous for both genes (AaBb × AaBb) produces gametes AB, Ab, aB and ab in equal proportions; combining these in a 4 × 4 Punnett square gives offspring in the ratio 9 : 3 : 3 : 1 for the four possible phenotypes.
🧑🏫 Teacher's explanation
Think of genes as individual recipes in a cookbook. A monohybrid cross is like looking at how one recipe (e.g., how much salt to add) is passed down. A dihybrid cross is like looking at how two different recipes (e.g., salt content AND sugar content) are passed down at the same time. If the recipes are on different pages (different chromosomes), they are independent; choosing one doesn't affect the other. This is called 'independent assortment'.
📖 Definitions
🔍 Worked example — Easy
In pea plants, yellow seeds (Y) are dominant to green seeds (y), and wrinkled seeds (W) are dominant to smooth seeds (w). If a plant with genotype YyWw is crossed with another YyWw plant, what is the expected phenotypic ratio of the F_2 generation?
🔍 Worked example — Standard
A researcher crosses a plant with genotype Aabb with a plant with genotype aaBb. What is the probability that an offspring will have the phenotype for both dominant alleles?
🔍 Worked example — Exam-style
In a dihybrid cross between two double heterozygotes (AaBb × AaBb), the observed number of offspring with the double recessive phenotype (aabb) is 15, while the expected number is 16. Calculate the chi^2 value.
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🎯 Exam question
Key ideas — The chi-squared test
The chi-squared test (chi^2) determines whether a difference between observed and expected results is statistically significant or simply due to chance. The formula is
🧑🏫 Teacher's explanation
Imagine you are flipping a coin. You expect 50 heads and 50 tails in 100 flips. If you actually get 52 heads and 48 tails, you wouldn't be shocked; it's likely just random chance. But if you get 90 heads and 10 tails, you'd suspect the coin is rigged. The chi-squared test is the mathematical way of deciding exactly when a 'weird' result is actually 'significant' (rigged) rather than just 'random' (chance).
📖 Definitions
🔍 Worked example — Easy
In a monohybrid cross, the expected ratio of phenotypes is 3:1. In a sample of 40 offspring, you observe 32 with the dominant phenotype and 8 with the recessive phenotype. Calculate the chi^2 value.
🔍 Worked example — Standard
A researcher observes 25 red flowers and 15 white flowers. The expected ratio is 1:1. Calculate the chi^2 value.
🔍 Worked example — Exam-style
In a dihybrid cross, the expected ratio is 9:3:3:1. In a sample of 100 plants, the observed values are: Yellow/Round (50), Yellow/Wrinkled (10), Green/Round (10), Green/Wrinkled (30). Calculate chi^2 and state if the result is significant at p=0.05 (Critical value for df=3 is 7.81).
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🎯 Exam question
Key ideas — Autosomal linkage
Genes located close together on the same autosome (a chromosome that is not a sex chromosome) tend to be inherited together because they are far less likely to be separated by crossing over; this is autosomal linkage. A test cross between a linked double heterozygote and a double recessive produces mostly parental (non-recombinant) phenotypes, close to 1 : 1, with only a small proportion of recombinant phenotypes from crossing over between the linked loci, rather than the 1 : 1 : 1 : 1 ratio expected for unlinked genes.
🧑🏫 Teacher's explanation
Imagine you are packing a suitcase. If you put your socks and your shoes in the same small compartment, they will almost always travel together. In biology, genes are like items in that compartment. If two genes are located very close to each other on the same chromosome (an autosome), they are physically 'stuck' together during meiosis. This is called linkage.
📖 Definitions
🔍 Worked example — Easy
In a dihybrid test cross involving unlinked genes, the expected phenotypic ratio is 1:1:1:1. If a test cross between a double heterozygote (AaBb) and a double recessive (aabb) yields 50 individuals with phenotype A\_B\_, 50 with aabb, 0 with A\_bb, and 0 with aaB\_, determine if the genes are linked.
🔍 Worked example — Standard
A researcher performs a test cross on a dihybrid organism. The observed phenotypes in the offspring are: 450 A\_B\_, 450 aabb, 10 A\_bb, and 10 aaB\_. Calculate the percentage of recombinant phenotypes.
🔍 Worked example — Exam-style
A test cross of a dihybrid (AaBb) results in 800 A\_B\_, 800 aabb, 50 A\_bb, and 50 aaB\_. Calculate the chi-squared (chi^2) value to determine if the deviation from the 1:1:1:1 ratio is significant. Use the formula chi^2 = Σ (O-E)^2/E where E = 425.
⚠️ Common errors
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🎯 Exam question
(a) Explain why the number of A\_bb and aaB\_ individuals is so low compared to the A\_B\_ and aabb individuals. [3] (b) State the term used to describe the A\_bb and aaB\_ phenotypes. [1] [4 marks]
Key ideas — Test crosses
A test cross determines the genotype of an individual showing the dominant phenotype, which could be homozygous or heterozygous, by crossing it with a homozygous recessive individual. If all offspring show the dominant phenotype, the test individual is homozygous dominant; if about half show the recessive phenotype, it is heterozygous, because the recessive allele carried by half its gametes is unmasked in the offspring.
🧑🏫 Teacher's explanation
Imagine you are looking at a light switch that is currently 'ON'. You know the switch is in the 'ON' position, but you don't know if it is a permanent 'ON' switch (homozygous dominant) or a toggle switch that can be flipped to 'OFF' (heterozygous). To find out, you perform a 'test': you connect it to a device that only works if the switch is in the 'OFF' position. If the device works every single time, the switch is permanently 'ON'. If it only works half the time, you know the switch has the potential to be 'OFF'.
📖 Definitions
🔍 Worked example — Easy
A pea plant has purple flowers (dominant allele P) and is crossed with a white-flowered plant (recessive allele p). If all offspring are purple, what is the genotype of the purple parent?
🔍 Worked example — Standard
In a species of moth, black fur (B) is dominant to white fur (b). A black moth is test-crossed with a white moth. The resulting F_1 generation consists of 48 black moths and 52 white moths. Determine the genotype of the black parent.
🔍 Worked example — Exam-style
A researcher performs a test cross on a plant showing a dominant phenotype. The offspring show a ratio of 15 dominant phenotypes to 1 recessive phenotype. Explain why this result suggests the parent was NOT a simple monohybrid heterozygous individual.
⚠️ Common errors
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🎯 Exam question
(a) State the genotype of the purple-flowered parent. [1] (b) Explain how the results of this test cross allow you to determine the genotype. [2] [3 marks]
Key ideas — Interaction of gene loci
Sometimes two gene loci interact to produce a single characteristic, a phenomenon called epistasis, in which the allele at one locus masks or modifies the expression of alleles at a second locus. For example, in some dog breeds, one gene controls whether pigment is produced at all, and a second gene controls the colour of that pigment; if the first gene is homozygous recessive (no pigment produced), coat colour is determined by the absence of pigment regardless of the second locus, since there is no pigment for its alleles to act on.
🧑🏫 Teacher's explanation
Imagine you are building a house. The first blueprint (Gene A) determines if you have any walls at all. The second blueprint (Gene B) determines what color to paint those walls. If the first blueprint fails and you have no walls, it doesn't matter if the second blueprint says 'blue' or 'red'—the house will still look like a pile of bricks. This is epistasis.
📖 Definitions
🔍 Worked example — Easy
In a specific plant species, gene A determines the production of a precursor pigment, and gene B determines the color of that pigment (e.g., A\_ results in color, aa results in white). If a plant has the genotype aaBB, what is its phenotype?
🔍 Worked example — Standard
A researcher performs a dihybrid cross between two plants with genotypes AaBb and AaBb. In this system, A is epistatic to B. If A is required for any color to appear, calculate the expected phenotypic ratio in the F_2 generation.
🔍 Worked example — Exam-style
In a dihybrid cross involving epistasis, the observed results for 160 offspring are: 90 colored, 30 colored, and 40 white. Use the chi-squared (chi^2) test to determine if this fits the expected 9:3:1 ratio. chi^2 = Σ (O-E)^2/E.
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🎯 Exam question
Key ideas — Sex determination and sex linkage
In humans, sex is determined by one pair of chromosomes: XX produces a female, XY produces a male. All eggs carry an X chromosome, while sperm carry either an X or a Y chromosome in equal proportions, so a genetic diagram of egg (X) × sperm (X or Y) gives an expected 1 : 1 ratio of female to male offspring. A sex-linked characteristic is controlled by a gene on a sex chromosome, usually X, since Y carries very few genes. Because males (XY) have only one X chromosome, a single recessive allele there is always expressed, whereas a female (XX) must be homozygous recessive to show the phenotype; this explains why X-linked recessive conditions such as red-green colour blindness and haemophilia are far more common in males.
🧑🏫 Teacher's explanation
Think of your genome as a massive library of instruction manuals. Each manual is a gene, and the specific shelf where it sits is the locus. Sometimes, a manual has a typo (an allele). If the typo is 'dominant', it overrides the correct version; if it is 'recessive', you only see the error if there is no correct version to fix it.
📖 Definitions
🔍 Worked example — Easy
A female is heterozygous for red-green colour blindness (an X-linked recessive trait). A male with normal vision marries her. Predict the phenotypic ratio of their offspring.
🔍 Worked example — Standard
In a cross between a haemophiliac male (X^h Y) and a homozygous dominant female (X^H X^H), calculate the percentage of male offspring that will be haemophiliacs.
🔍 Worked example — Exam-style
A researcher performs a test cross on an individual showing a dominant phenotype for a sex-linked trait. The individual is crossed with a homozygous recessive individual. If the individual is heterozygous, what is the expected phenotypic ratio of the offspring?
⚠️ Common errors
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🎯 Exam question
(a) Write down the genotypes of the woman and the man. [2] (b) Using a genetic diagram, predict the probability that they will have a son who is colour blind. [3] [5 marks]
Key ideas — Multiple alleles: the ABO blood group system
Some genes have more than two alleles in a population, though any one diploid individual carries only two; this is multiple alleles. The ABO blood group gene has three alleles, I^A, I^B and I^O. I^A and I^B are codominant, and both are dominant to I^O. The genotypes and phenotypes are: I^AI^A or I^AI^O (group A), I^BI^B or I^BI^O (group B), I^AI^B (group AB, both antigens expressed) and I^OI^O (group O).
🧑🏫 Teacher's explanation
Think of a gene as a recipe book for making a specific protein. Usually, we think of genes having just two versions (alleles), like 'salty' or 'sweet'. However, 'multiple alleles' means there are more than two versions of that recipe available in the whole population, even though you only ever inherit one from your mother and one from your father.
📖 Definitions
🔍 Worked example — Easy
A man with blood group AB marries a woman with blood group O. Predict the possible phenotypes of their children using a Punnett square.
🔍 Worked example — Standard
In a population, the frequency of the I^O allele is 0.4. Calculate the frequency of individuals with blood group O, assuming the population is in Hardy-Weinberg equilibrium (where p + q + r = 1 for three alleles).
🔍 Worked example — Exam-style
A heterozygous individual with blood group A has children with a person with blood group B. If they have 4 children, what is the expected ratio of blood groups among the offspring?
⚠️ Common errors
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🎯 Exam question
(a) State the genotypes of the woman and the man. [2] (b) Using a genetic diagram, predict the possible blood groups of their children. [3] [5 marks]
Key ideas — Genes, proteins and phenotype
A gene codes for a polypeptide; the protein's function produces the phenotype. A mutation that changes or removes the protein changes the phenotype:
🧑🏫 Teacher's explanation
Think of your DNA as a massive library of instruction manuals. A 'gene' is a single manual that tells the cell how to build a specific tool, which we call a 'protein'. If the manual is written correctly, the tool works perfectly and performs a job—like making pigment for your skin or clotting your blood. This visible result of the tool's job is your 'phenotype'.
📖 Definitions
🔍 Worked example — Easy
A person has the genotype H^A H^S for the HBB gene. Explain how this genotype relates to their phenotype and protein function.
🔍 Worked example — Standard
In a plant, the Le allele (dominant) codes for a functional enzyme in the gibberellin synthesis pathway, while the le allele (recessive) codes for a non-functional enzyme. If a plant is heterozygous (Le/le), describe the phenotype and the biochemical reason.
🔍 Worked example — Exam-style
A researcher observes a patient with Huntington's disease. Explain the genetic basis of this phenotype.
⚠️ Common errors
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🎯 Exam question
(a) Identify the specific gene and the protein involved in this condition. [2] (b) Explain how a mutation in this gene results in the albinism phenotype. [3] [5 marks]
Key ideas — Gibberellin and stem height: Le and le
In pea plants, stem height depends on one gene with two alleles:
🧑🏫 Teacher's explanation
Think of a gene as a blueprint for a factory machine. In this case, the machine's job is to turn a raw material (GA20) into a finished product (gibberellin). If the blueprint is correct (the Le allele), the machine works perfectly, producing plenty of gibberellin, which tells the plant to grow tall. If there is a typo in the blueprint (the le allele), the machine is broken and produces nothing.
📖 Definitions
🔍 Worked example — Easy
A homozygous tall pea plant is crossed with a dwarf pea plant. Determine the genotype and phenotype of the F_1 generation.
🔍 Worked example — Standard
In a cross between two heterozygous (Lele) pea plants, calculate the expected ratio of tall plants to dwarf plants in the F_2 generation.
🔍 Worked example — Exam-style
A researcher observes 100 pea plants. 75 are tall and 25 are dwarf. Calculate the percentage of dwarf plants observed.
⚠️ Common errors
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🎯 Exam question
(a) State the expected phenotypic ratio of tall to dwarf plants in the F_2 generation. [1] (b) Explain, in terms of enzyme function and hormone concentration, why the lele genotype results in a dwarf phenotype. [3] [4 marks]
Worked example — Model answer
Question. In a genetic cross between two double heterozygous pea plants, the expected phenotype ratio was 9 tall round-seeded : 3 tall wrinkled-seeded : 3 dwarf round-seeded : 1 dwarf wrinkled-seeded. Of 160 offspring, 86 were tall round-seeded, 32 were tall wrinkled-seeded, 33 were dwarf round-seeded and 9 were dwarf wrinkled-seeded. Use the chi-squared test to determine whether these results differ significantly from the expected 9 : 3 : 3 : 1 ratio. The critical value of chi^2 at p = 0.05 with 3 degrees of freedom is 7.815. [6]
🔍 Worked example — Easy
A heterozygous plant (Aa) is crossed with a homozygous recessive plant (aa). What is the expected ratio of phenotypes in the F_1 generation?
🔍 Worked example — Standard
In a dihybrid cross of two heterozygous individuals (AaBb × AaBb), calculate the expected number of individuals with the phenotype representing the recessive alleles for both traits if the total offspring count is 400.
🔍 Worked example — Exam-style
In a genetic cross between two double heterozygous pea plants, the expected phenotype ratio was 9 tall round-seeded : 3 tall wrinkled-seeded : 3 dwarf round-seeded : 1 dwarf wrinkled-seeded. Of 160 offspring, 86 were tall round-seeded, 32 were tall wrinkled-seeded, 33 were dwarf round-seeded and 9 were dwarf wrinkled-seeded. Use the chi-squared test to determine whether these results differ significantly from the expected 9:3:3:1 ratio. The critical value of chi^2 at p = 0.05 with 3 degrees of freedom is 7.815.
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