Inheritance

In a nutshell

Inheritance is how alleles pass from parents to offspring, and how the alleles an organism inherits combine to give its characteristics.

This subtopic is really two exam skills: building and reading fully labelled genetic diagrams to predict the offspring of a cross, and using the chi-squared test to decide whether the results you actually get fit the ratio you expected.

Almost every mark here is won or lost on precise terminology, so nail the vocabulary before the crosses.

Assumed knowledge: DNA, genes and chromosomes, How genetic diversity arises (meiosis).

Core content

The language of inheritance

Get these terms exactly right. Examiners routinely reject "gene" used for "allele", and reject "the visible characteristics" as a definition of phenotype.

• Gene: a sequence of DNA that codes for a specific polypeptide, found at a fixed position called a locus on a chromosome.
• Allele: a different version of a gene.
• Genotype: the genetic constitution of an organism (the alleles it has).
• Phenotype: the expression of this genetic constitution and its interaction with the environment.

The phenotype mark nearly always needs both halves: the characteristic is due to the genotype and due to the environment. An answer that only says "the observable characteristics" scores nothing.

At a single locus in a diploid organism, the two alleles are either:

• Homozygous: the two alleles are the same (for example BB or bb).
• Heterozygous: the two alleles are different (for example Bb).

Alleles interact in three ways you must be able to name and use:

• Dominant: always expressed in the phenotype, even in a heterozygote (one copy is enough).
• Recessive: only expressed when homozygous, that is when no dominant allele is present.
• Codominant: both alleles are expressed in the phenotype of a heterozygote (neither is masked).

There may also be multiple alleles: a gene with more than two alleles existing in the population, although any one diploid individual still carries only two of them.

Monohybrid crosses (one gene)

A monohybrid cross follows the inheritance of a single gene with two alleles.

A fully labelled genetic diagram must show, in order: the parental phenotypes, their genotypes, the gametes (each carrying one allele, so it is often circled), the offspring genotypes, and the offspring phenotypes with the ratio. Missing out the gametes or the genotype-to-phenotype link is the most common way to lose marks.

For two heterozygous parents (Bb × Bb), the renderer works the gametes and grid out for you:

	B	b
B	BB	Bb
b	Bb	bb

Phenotype ratio 3 black : 1 white

This gives a genotypic ratio of 1 BB : 2 Bb : 1 bb and a phenotypic ratio of 3 dominant : 1 recessive. The probability of a dominant-phenotype offspring is 3/4 (75%).

Ratios describe probabilities, not guaranteed numbers. The next four offspring will not always be exactly 3 black and 1 white, because fertilisation is random, gametes are not always made in equal numbers, and real samples are small.

Dihybrid crosses (two genes)

A dihybrid cross follows two genes at the same time. It works because of independent assortment in meiosis: either allele of one gene can end up in a gamete with either allele of the other gene (when the genes are on separate chromosomes).

A parent that is heterozygous for both genes (RrYy) therefore makes four equally likely gamete types: RY, Ry, rY and ry.

	RY	Ry	rY	ry
RY	RRYY	RRYy	RrYY	RrYy
Ry	RRYy	RRyy	RrYy	Rryy
rY	RrYY	RrYy	rrYY	rrYy
ry	RrYy	Rryy	rrYy	rryy

Phenotype ratio 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green

Crossing two double heterozygotes gives the classic 9 : 3 : 3 : 1 phenotypic ratio (9 both-dominant : 3 dominant-A recessive-B : 3 recessive-A dominant-B : 1 both-recessive). Learn this ratio: spotting when a real result departs from it is how you detect linkage and epistasis later.

Codominance and multiple alleles

In codominance, the heterozygote expresses both alleles in its phenotype at the same time. It is not blending: both allele products appear.

Codominant alleles are written as one capital letter for the gene with a different superscript for each allele (both superscripts capital, because neither is recessive), for example C^R and C^W for coat colour in cattle:

The clearest example of multiple alleles plus codominance is the human ABO blood group gene, which has three alleles in the population: I^A, I^B and I^O. I^A and I^B are codominant with each other; I^O is recessive to both.

Any one person carries only two of the three alleles, one on each chromosome of the homologous pair.

Sex linkage

A gene is sex-linked when it is located on a sex chromosome, almost always the X chromosome. The X chromosome is large and carries many genes; the much smaller Y chromosome carries very few, so most X-linked genes have no matching allele on the Y.

Because a male is XY, he has only one allele for any X-linked gene. A single recessive allele on his X chromosome is therefore expressed, since there is no second X to mask it. This is why X-linked recessive conditions (such as haemophilia and red-green colour blindness) are more common in males.

A female is XX, so she needs two copies of the recessive allele (she must be homozygous recessive) to show the condition. With one copy she is an unaffected carrier. A male cannot be a carrier: he either has the condition or he does not.

Always write sex-linked alleles on the X chromosome, and never put an allele on the Y. Using haemophilia (X^H = normal clotting, X^h = haemophilia):

• Females: X^HX^H, X^HX^h (carrier) or X^hX^h (affected).
• Males: X^HY (normal) or X^hY (affected).

A worked sex-linked cross is in Worked examples below.

Autosomal linkage

An autosome is any chromosome that is not a sex chromosome. Autosomal linkage is when two or more genes are on the same autosome, so they tend to be inherited together rather than assorting independently.

Linked genes mostly pass into gametes in their parental combinations, so a dihybrid cross of two heterozygotes does not give the expected 9:3:3:1 ratio. Instead the parental phenotypes appear in excess and the recombinant (new) combinations are rare.

The recombinants exist at all only because crossing over in meiosis can separate linked alleles. The closer two genes are on the chromosome, the less likely a crossover falls between them, so the fewer recombinants appear.

Still don't get it? · why linkage breaks the 9:3:3:1 ratio

Think of two genes as two beads threaded on the same string (the chromosome). When you hand the string on, both beads go together, because they are tied to the same string. Only if the string is cut between them (a crossover) do the beads get swapped onto different strings.

Now the biology, one step at a time:

1. When genes are on different chromosomes, meiosis shuffles them freely (independent assortment), so a double heterozygote makes all four gamete types equally: this is what builds the 9:3:3:1 ratio.
2. When the two genes sit on the same chromosome (linked), they are tied to the same "string", so they mostly travel together. A double heterozygote makes mainly the two parental gamete types and only a few of the swapped (recombinant) types.
3. Fewer gamete types means fewer offspring combinations. The offspring that look like the parents are far more common than 9:3:3:1 predicts, and the recombinant types are rare.

Exam version: linked genes do not assort independently, so the offspring ratio departs from 9:3:3:1, with an excess of parental-type phenotypes and only a small number of recombinants produced by crossing over.

Epistasis

Epistasis is when one gene at one locus affects or masks the expression of another gene at a different locus. It changes the expected dihybrid ratio into a modified one.

Recessive epistasis: the masking gene only hides the other gene when it is homozygous recessive. For example, coat colour in mice needs a functioning pigment-deposition gene (C) before the colour gene (B, black vs b, brown) can show. Any cc mouse is albino whatever its B alleles.

Crossing two BbCc parents therefore gives 9 black : 3 brown : 4 albino (a 9 : 3 : 4 ratio), because the two "cc" classes both come out albino.

Dominant epistasis: the masking gene hides the other when just one dominant allele is present. Here two double heterozygotes give a 12 : 3 : 1 ratio. A non-Mendelian ratio like 9:3:4 or 12:3:1 is your signal that epistasis is happening.

The chi-squared test (data-handling skill)

The chi-squared (χ²) test compares an observed set of results with the expected results from a predicted ratio, to decide whether the difference between them is significant or just due to chance.

Use it when you have categoric data (counts of phenotypes) and a ratio to test against, such as 3:1 or 9:3:3:1.

$\chi^2 = \sum \frac{(O - E)^2}{E}$

where O is the observed count and E is the expected count for each phenotype.

The method:

1. State the null hypothesis (H₀): there is no significant difference between the observed and expected results.
2. Work out the expected counts from the ratio and the total number of offspring.
3. Put O and E into the formula and add up each (O - E)²/E to get χ².
4. Work out the degrees of freedom: df = (number of categories) - 1.
5. Read the critical value at p = 0.05 for that df.
6. Compare and conclude:
   • If χ² is less than the critical value: accept H₀. The difference is not significant, and there is a greater than 5% probability it is due to chance.
   • If χ² is greater than or equal to the critical value: reject H₀. The difference is significant, and there is a less than 5% probability it is due to chance.

Critical values at p = 0.05:

Still don't get it? · what "accept or reject the null hypothesis" actually means

Imagine you toss a coin 100 times and get 46 heads. You expected 50. Is the coin bent, or is 46 just normal luck? You need a rule that separates "normal luck" from "something real is going on". Chi-squared is that rule for genetics counts.

Step by step:

1. The null hypothesis is the boring explanation: "nothing special, the results match the expected ratio and any gap is just chance."
2. Chi-squared turns the gaps between observed and expected into a single number. Small gaps give a small number; big gaps give a big number.
3. The critical value is the "too much to be luck" line for your number of categories. Below the line, the gap is the kind of wobble chance produces all the time, so you keep the boring explanation (accept H0). On or above the line, the gap is bigger than chance would usually give, so you drop it (reject H0) and say something real is causing the difference (for example the genes are linked).

Exam version: if chi-squared is below the critical value at p = 0.05, accept the null hypothesis, the difference is not significant and is probably due to chance; if it is equal to or above the critical value, reject the null hypothesis, the difference is significant and there is a less than 5% probability it is due to chance.

Worked examples

Model answer: a fully labelled sex-linked cross. "A carrier woman and a man with normal blood clotting have a child. Show the possible offspring and the chance the child has haemophilia." (X^H = normal clotting, X^h = haemophilia.)

A full-mark genetic diagram shows every stage, clearly labelled:

1. Parental phenotypes: carrier female × normal male
2. Parental genotypes: X^HX^h × X^HY
3. Gametes: X^H and X^h (from the mother); X^H and Y (from the father)

4. Offspring genotypes and phenotypes: X^HX^H normal female, X^HX^h carrier female, X^HY normal male, X^hY haemophiliac male.
5. Answer: the probability of a child with haemophilia is 1 in 4 (0.25 or 25%), and every affected child is male.

Notice how many separate things earn marks: the genotypes, the gametes, the offspring genotypes linked to the right phenotypes (including the sex), and the final probability. Drop any one and you drop a mark.

Model answer: a chi-squared calculation. A dihybrid cross is expected to give a 9:3:3:1 ratio. Of 320 offspring, the observed counts are 178, 62, 56 and 24. Do the results fit the expected ratio?

Expected counts from 9:3:3:1 out of 320:

$E = \frac{9}{16}\times 320 = 180, \quad \frac{3}{16}\times 320 = 60, \quad \frac{3}{16}\times 320 = 60, \quad \frac{1}{16}\times 320 = 20$

• Degrees of freedom = 4 categories - 1 = 3.
• Critical value at p = 0.05 and df = 3 = 7.81.
• Conclusion: 1.16 is less than 7.81, so accept the null hypothesis. There is no significant difference between the observed and expected results; the difference is due to chance and the results fit the expected 9:3:3:1 ratio.

If χ² had come out above 7.81, you would reject the null hypothesis and conclude the difference is significant, which for a dihybrid cross would suggest the genes are linked.

Common exam mistakes

• Defining the phenotype as only "the observable characteristics" and omitting the environment; the mark needs genotype and its interaction with the environment.
• Using "gene" when you mean allele, or "homologous" when you mean homozygous. Answers about inheritance must be in terms of alleles; the wrong word is not credited.
• Writing gamete genotypes as diploid (for example writing AaBb as a gamete, or writing AB as a whole genotype). A gamete carries one allele of each gene (haploid), so from AaBb the gametes are AB, Ab, aB, ab.
• Building a 4 × 4 dihybrid square and assuming the answer must be 9:3:3:1. With codominance, multiple alleles, sex linkage, linkage or epistasis the ratio is different; work it out from the genotypes.
• In sex-linkage crosses, leaving out the X and Y chromosomes and doing an autosomal cross instead, or drawing an allele on the Y chromosome. Always attach the allele to the X.
• Calling a male a carrier for an X-linked recessive condition. A male has one X, so he is either affected or unaffected, never a carrier.
• Describing codominance as "blending". In codominance both alleles are expressed (AB blood has both antigens; roan cattle have both red and white hairs), not a mixture.
• Writing epistasis vaguely. The mark needs "one gene affects or masks the expression of another gene (at a different locus)", not just "genes interact".
• In chi-squared, giving the ratio "142:50" style rather than calculating it, or forgetting the units of the conclusion. State whether the difference is significant, not whether "the results" are significant.
• Chi-squared conclusions that stop at "the difference is significant" without rejecting or accepting the null hypothesis and without linking back to the biology (for example "the genes may be linked").
• Getting degrees of freedom wrong: df is (number of categories) - 1, not the number of organisms.

Key definitions

• Genotype: the genetic constitution of an organism (the alleles it has).
• Phenotype: the expression of the genotype and its interaction with the environment.
• Allele: a version of a gene.
• Dominant allele: an allele that is always expressed in the phenotype (even in a heterozygote).
• Recessive allele: an allele that is only expressed in the phenotype when homozygous, that is when no dominant allele is present.
• Codominant alleles: alleles that are both expressed in the phenotype of a heterozygote.
• Homozygous: having two identical alleles at a locus.
• Heterozygous: having two different alleles at a locus.
• Multiple alleles: a gene for which there are more than two alleles in the population (though a diploid individual has only two).
• Sex linkage: a gene (and its alleles) located on a sex chromosome, usually the X chromosome.
• Autosomal linkage: two or more genes located on the same autosome, so they are inherited together and do not assort independently.
• Epistasis: where one gene affects or masks the expression of another gene at a different locus.
• Null hypothesis: the hypothesis that there is no significant difference between the observed and expected results.

Specification

• I can define genotype as the genetic constitution of an organism, and phenotype as the expression of the genotype and its interaction with the environment.
• I can state that there may be many alleles of a single gene, and that alleles may be dominant, recessive or codominant.
• I can state that the alleles at a locus in a diploid organism may be homozygous or heterozygous.
• I can use fully labelled genetic diagrams to interpret and predict the results of monohybrid and dihybrid crosses involving dominant, recessive and codominant alleles.
• I can use fully labelled genetic diagrams for crosses involving sex linkage, autosomal linkage, multiple alleles and epistasis.
• I can use the chi-squared test to compare the goodness of fit of observed phenotypic ratios with expected ratios.

Related notes

• DNA, genes and chromosomes
• How genetic diversity arises (meiosis)
• Populations (allele frequencies and Hardy-Weinberg)
• Evolution and speciation