HomeLearning HubA Level BiologyA2 16: Inheritance
A2 16

Inheritance

A Level · Topic 16 · Papers 4 and 5

🎯What you need to be able to do

  • Explain haploid and diploid, homologous chromosomes, and the need for a reduction division.
  • Describe chromosome behaviour in meiosis and identify the main stages from micrographs.
  • Explain how crossing over, independent assortment and random fertilisation produce genetic variation.
  • Use the terms gene, locus, allele, dominant, recessive, codominant, linkage, test cross, F1, F2, phenotype, genotype, homozygous and heterozygous.
  • Construct and interpret genetic diagrams for monohybrid and dihybrid crosses involving dominance, codominance, multiple alleles and sex linkage.
  • Construct and interpret genetic diagrams involving autosomal linkage, epistasis and test crosses.
  • Use the chi-squared test to assess the significance of a difference between observed and expected results.
  • Explain the relationship between genes, proteins and phenotype for TYR, HBB, F8 and HTT, and the role of gibberellin and the Le allele.
  • Explain gene control: structural and regulatory genes, repressible and inducible enzymes, the lac operon, transcription factors and DELLA proteins.

📚The biology

Meiosis

Haploid (n) cells have one set of chromosomes; diploid (2n) cells have two. Homologous chromosomes are a matching pair, one from each parent, with the same genes at the same loci — though not necessarily the same alleles.

A reduction division is necessary because fertilisation doubles the chromosome number. Without halving it during gamete formation, the number would double every generation. Meiosis therefore produces haploid gametes from diploid cells, in two divisions.

Names of the stages are required — prophase I, metaphase I, anaphase I, telophase I, then prophase II, metaphase II, anaphase II, telophase II — but not the sub-divisions of prophase I. The essential difference from mitosis:

  • In prophase I, homologous chromosomes pair up and crossing over occurs at chiasmata: sections of chromatid are exchanged between homologues, producing new combinations of alleles on a chromosome.
  • In metaphase I, homologous pairs line up on the equator, and the orientation of each pair is random and independent of the others — independent assortment.
  • In anaphase I, whole chromosomes (still two chromatids) are separated, one of each pair to each pole. This is the reduction to haploid.
  • The second division separates sister chromatids, as in mitosis.

Three sources of variation, and questions usually want all three: crossing over, independent assortment, and the random fusion of gametes at fertilisation, which combines two already-varied haploid sets.

The vocabulary

  • Gene — a length of DNA coding for a polypeptide. Locus — its position on a chromosome. Allele — a version of a gene.
  • Genotype — the alleles present. Phenotype — the observable characteristics, produced by genotype and environment.
  • Homozygous — two identical alleles. Heterozygous — two different alleles.
  • Dominant — expressed in the heterozygote. Recessive — expressed only in the homozygote. Codominant — both alleles are expressed in the heterozygote, so the phenotype shows both.
  • Linkage — two genes on the same chromosome, so their alleles tend to be inherited together. Test cross — crossing an individual of unknown genotype with a homozygous recessive, to reveal the unknown from the offspring.

Writing a genetic diagram

Marks are given for the layout as well as the answer, so use the same structure every time: parental phenotypes, parental genotypes, gametes (circled), a Punnett square, offspring genotypes, offspring phenotypes and ratio. Define your symbols first. Use a capital and its own lower case for dominance (A, a); for codominance and multiple alleles use a base letter with superscripts, as in CR and CW; for sex linkage always write the alleles on the X, as XHY.

Ratios to recognise, and what they signal:

  • 3:1 — monohybrid F₂, both parents heterozygous.
  • 1:2:1 — monohybrid F₂ with codominance, since the heterozygote is distinguishable.
  • 1:1 — a test cross, or a heterozygote crossed with a homozygous recessive.
  • 9:3:3:1 — dihybrid F₂, both parents heterozygous for two unlinked genes.
  • 1:1:1:1 — dihybrid test cross with unlinked genes.
  • Anything close to 9:7, 12:3:1, 13:3, 15:1 — a modified dihybrid ratio, which means epistasis: one gene is affecting the expression of another. You are not required to know which ratio goes with which type of epistasis.
  • A dihybrid cross giving far more parental types than expected and very few recombinants — autosomal linkage. The genes are on the same chromosome, so they do not assort independently; the few recombinants arise by crossing over.

Sex linkage

Genes on the X chromosome have no equivalent on the much shorter Y. A male has only one allele, so he expresses it whether dominant or recessive — he cannot be heterozygous, and the terms homozygous and heterozygous do not apply to him for X-linked genes. This is why X-linked recessive conditions such as haemophilia and red-green colour blindness are far commoner in males, and why an affected male passes the allele to all his daughters (who become carriers) and none of his sons.

The chi-squared test

Use it to test whether a difference between observed and expected frequencies is significant, or is within what chance alone could produce.

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

Set out a table with columns for O, E, O − E, (O − E)² and (O − E)²/E, and total the last column. Degrees of freedom \( \nu = c - 1 \), where \( c \) is the number of classes — you are expected to know this formula, as it is not provided. Compare your value with the critical value at p = 0.05 for that number of degrees of freedom.

  • χ² less than the critical value → the difference is not significant, is due to chance, and the null hypothesis is accepted.
  • χ² greater than the critical value → the difference is significant, chance alone is an inadequate explanation, and the null hypothesis is rejected — so something else, such as linkage or differential mortality, is operating.

Chi-squared uses counts, never percentages or ratios, and questions will only ask it on one row or column of data.

Genes, proteins and phenotype

Four examples are named, and each should be learnt as gene → protein → phenotype:

  • TYRtyrosinasealbinism. A recessive allele codes for a non-functional tyrosinase, so the pathway converting tyrosine to melanin is blocked and no pigment is made.
  • HBB → the β-globin of haemoglobinsickle cell anaemia. A single base substitution changes one amino acid; the altered haemoglobin polymerises when oxygen is low, distorting red blood cells into a sickle shape. The alleles are codominant: heterozygotes have both normal and sickle haemoglobin, and have some resistance to malaria.
  • F8factor VIIIhaemophilia. An X-linked recessive allele gives non-functional factor VIII, so the blood clotting cascade fails.
  • HTThuntingtinHuntington’s disease. A dominant allele with an expanded repeat produces an altered huntingtin protein that damages neurones; because onset is usually after reproductive age, the allele is not removed by selection.

Gibberellin and stem length. The dominant allele Le codes for a functional enzyme in the gibberellin synthesis pathway, so gibberellin is made and the stem elongates — tall plants. The recessive le codes for a non-functional enzyme, so little gibberellin is made and plants are dwarf. Applying gibberellin to a dwarf plant makes it grow tall, which is neat evidence that the gene acts through the hormone.

Gene control

Structural genes code for proteins used in the cell’s structure or metabolism; regulatory genes code for proteins that control the expression of other genes. An inducible enzyme is made only when its substrate is present; a repressible enzyme is made continuously until something switches it off.

The lac operon is the syllabus’s prokaryotic example. A regulatory gene codes for a repressor protein.

  • Lactose absent: the repressor binds to the operator. RNA polymerase cannot bind at the promoter and move along, so the structural genes are not transcribed and no enzymes are made — no waste.
  • Lactose present: lactose binds to the repressor and changes its shape, so it can no longer bind the operator. RNA polymerase now transcribes the structural genes, and the enzymes for taking up and hydrolysing lactose are made. They are therefore inducible enzymes.

Knowledge of the role of cAMP in the operon is not required.

In eukaryotes, transcription factors are proteins that bind to DNA and decrease or increase the rate of transcription of particular genes. The gibberellin example completes the story from Topic 15: gibberellin causes the breakdown of DELLA protein repressors, which normally inhibit the transcription factors that promote transcription. Removing the repressor releases the factors, so the amylase gene is switched on.

✏️Worked example

In a dihybrid cross between two heterozygous plants, 320 offspring were counted: 190 tall red, 52 tall white, 64 short red, 14 short white. (a) State the expected ratio and calculate the expected numbers. (b) Carry out a chi-squared test and state the degrees of freedom. (c) The critical value at p = 0.05 for 3 degrees of freedom is 7.82. State and explain your conclusion.

(a) Two heterozygous parents crossed for two genes assorting independently give the F₂ ratio 9 : 3 : 3 : 1. With 320 offspring, one part is \( 320 \div 16 = 20 \), so the expected numbers are

tall red: \( 9 \times 20 = 180 \)
tall white: \( 3 \times 20 = 60 \)
short red: \( 3 \times 20 = 60 \)
short white: \( 1 \times 20 = 20 \)

(b) Working class by class:

\( \dfrac{(190-180)^{2}}{180} = \dfrac{100}{180} = 0.556 \)
\( \dfrac{(52-60)^{2}}{60} = \dfrac{64}{60} = 1.067 \)
\( \dfrac{(64-60)^{2}}{60} = \dfrac{16}{60} = 0.267 \)
\( \dfrac{(14-20)^{2}}{20} = \dfrac{36}{20} = 1.800 \)
\[ \chi^{2} = 0.556 + 1.067 + 0.267 + 1.800 = 3.69 \]

There are four classes, so the degrees of freedom are \( \nu = c - 1 = 4 - 1 = \mathbf{3} \).

(c) The calculated value, 3.69, is less than the critical value of 7.82. This means that the probability of a difference this large arising by chance alone is greater than 0.05 — that is, greater than 5%.

The difference between observed and expected results is therefore not significant, and the null hypothesis is accepted: there is no significant difference between the observed results and a 9:3:3:1 ratio. The data are consistent with the two genes being carried on different chromosomes and assorting independently, with the deviations attributable to chance.

Check it. Two checks before you trust the number. First, the expected values must sum to the same total as the observed: 180 + 60 + 60 + 20 = 320 ✓. If they do not, the ratio or the arithmetic is wrong and every subsequent step inherits the error. Second, each individual contribution should be modest when the fit is good; a single class contributing 15 or 20 on its own is the signature of a real departure, and you would expect to reject. Here the largest single contribution is 1.8, which is unremarkable.
Using percentages or ratios instead of counts, and the wrong degrees of freedom. Chi-squared must be calculated on actual numbers of individuals; converting to percentages changes the value and makes the test meaningless. And ν is the number of classes minus one, which is 3 here, not the number of offspring minus one and not the number of genes. A third trap: state the conclusion in terms of the null hypothesis and the 0.05 probability level — “the results fit the ratio” on its own does not earn the mark.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Explain how meiosis produces genetically different gametes, naming three mechanisms.
(i) Crossing over in prophase I: homologous chromosomes pair up and equivalent sections of non-sister chromatids are exchanged at chiasmata, producing chromatids carrying new combinations of alleles that were not present in either parental chromosome. (ii) Independent assortment in metaphase I: each homologous pair lines up on the equator in an orientation that is random and independent of every other pair, so the combination of maternal and paternal chromosomes passed to each pole differs from gamete to gamete; with 23 pairs there are 223 possible combinations. Sister chromatids also orientate randomly at metaphase II. (iii) Random fertilisation: any one of a vast number of genetically different male gametes may fuse with any one of a large number of different female gametes, multiplying the variation again. The first two occur during meiosis; the third occurs after it, and questions usually want all three.
2. In cattle, the coat colour alleles CR (red) and CW (white) are codominant; heterozygotes are roan. Give the genetic diagram for a cross between two roan cattle and state the expected ratio.
Parental phenotypes: roan × roan. Parental genotypes: CRCW × CRCW. Gametes: (CR) and (CW) from each parent. Punnett square offspring genotypes: CRCR, CRCW, CRCW, CWCW. Offspring phenotypes and ratio: 1 red : 2 roan : 1 white. The 1:2:1 phenotypic ratio, rather than the 3:1 of simple dominance, is the signature of codominance: because both alleles are expressed in the heterozygote, the heterozygote has its own distinguishable phenotype, so the phenotypic ratio matches the genotypic ratio. Note that roan is not a blend — it is a coat with both red and white hairs, which is what codominance predicts.
3. Explain why haemophilia is much more common in males than in females.
The F8 gene is carried on the X chromosome and the allele causing haemophilia is recessive. The Y chromosome is much shorter and carries no equivalent allele. A male is XHY or XhY: he has only one allele of this gene, so if it is the recessive one he has no dominant allele to mask it and he has the condition. A female is XHXH, XHXh or XhXh: she must inherit the recessive allele from both parents to be affected, which requires an affected father and a mother who is at least a carrier — a far less likely combination. A female with one copy is a carrier with normal clotting. This is why an affected male passes the allele to all his daughters, who become carriers, and to none of his sons, since he gives them his Y.
4. A dihybrid test cross gives 210 : 15 : 18 : 205 rather than the expected 1:1:1:1. Explain what this indicates and how the two rare classes arise.
The two large classes are the parental combinations and the two rare classes are the recombinants. A 1:1:1:1 ratio would be expected if the two genes were on different chromosomes and assorted independently. The strong excess of parental types shows that the genes are linked — carried on the same chromosome — so their alleles tend to be inherited together and do not assort independently at metaphase I. The rare recombinant classes arise by crossing over in prophase I, where non-sister chromatids of homologous chromosomes exchange sections at a chiasma, separating alleles that were originally on the same chromosome. The recombinant frequency here is (15 + 18)/448 = about 7%, and the lower the frequency the closer together the two loci lie, since a chiasma is less likely to form between them.
5. Explain how the lac operon allows a bacterium to make lactose-digesting enzymes only when lactose is present.
A regulatory gene codes for a repressor protein, which is produced continuously. When lactose is absent, the repressor binds to the operator, a region of DNA next to the promoter. This prevents RNA polymerase from binding at the promoter and moving along the DNA, so the structural genes are not transcribed, no mRNA is made and no enzymes are produced — so the cell does not waste amino acids and ATP making enzymes it cannot use. When lactose is present, lactose acts as the inducer: it binds to the repressor protein and changes its shape so that it can no longer bind to the operator. The operator is free, RNA polymerase transcribes the structural genes, and the enzymes for taking up and hydrolysing lactose are synthesised. These are therefore inducible enzymes, made only when their substrate is available.
6. Explain how a single base substitution in the HBB gene produces the sickle cell phenotype, and why the heterozygote is said to show codominance.
A substitution of one base in the HBB gene changes one codon, so one amino acid in the β-globin polypeptide is different. Because the substituted amino acid has a different R group, the tertiary and quaternary structure of the haemoglobin molecule is altered: the altered molecules stick to one another and polymerise into long fibres when the oxygen concentration is low. These fibres distort the red blood cell into a sickle shape. Sickled cells carry less oxygen, are destroyed more rapidly causing anaemia, and block capillaries causing pain and tissue damage. The alleles are codominant because a heterozygote produces both normal and sickle-cell haemoglobin — both alleles are expressed and both proteins are present in the red blood cells, which can be shown by electrophoresis. Heterozygotes have sickle cell trait: usually healthy, but with some resistance to malaria, which is why the allele persists at high frequency in regions where malaria is endemic.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Any set of chi-squared critical value tables — you are given the table in the exam, but practise locating the right value for the right degrees of freedom under time pressure
  • Learn Genetics (University of Utah) — interactive Punnett squares and pedigree analysis, including sex-linked examples
  • Animations of the lac operon with and without lactose — the shape change in the repressor is the whole mechanism and is much clearer seen than described