Inheritance

Every objective in this topic is printed under "A Level subject content" in the 9700 syllabus, so all of it is A Level and none of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper and does not reach this topic, so no multiple-choice practice on this site is tagged to it.

Syllabus points

16.1 Passage of information from parents to offspring

• Explain the meanings of the terms haploid (n) and diploid (2n).
• Explain what is meant by homologous pairs of chromosomes.
• Explain the need for a reduction division during meiosis in the production of gametes.
• Describe the behaviour of chromosomes in plant and animal cells during meiosis and the associated behaviour of the nuclear envelope, the cell surface membrane and the spindle (names of the main stages of meiosis, but not the sub-divisions of prophase I, are expected: prophase I, metaphase I, anaphase I, telophase I, prophase II, metaphase II, anaphase II and telophase II).
• Interpret photomicrographs and diagrams of cells in different stages of meiosis and identify the main stages of meiosis.
• Explain that crossing over and random orientation (independent assortment) of pairs of homologous chromosomes and sister chromatids during meiosis produces genetically different gametes.
• Explain that the random fusion of gametes at fertilisation produces genetically different individuals.

16.2 The roles of genes in determining the phenotype

• Explain the terms gene, locus, allele, dominant, recessive, codominant, linkage, test cross, F1, F2, phenotype, genotype, homozygous and heterozygous.
• Interpret and construct genetic diagrams, including Punnett squares, to explain and predict the results of monohybrid crosses and dihybrid crosses that involve dominance, codominance, multiple alleles and sex linkage.
• Interpret and construct genetic diagrams, including Punnett squares, to explain and predict the results of dihybrid crosses that involve autosomal linkage and epistasis (knowledge of the expected ratios for different types of epistasis is not expected).
• Interpret and construct genetic diagrams, including Punnett squares, to explain and predict the results of test crosses.
• Use the chi-squared test to test the significance of differences between observed and expected results (the formula for the chi-squared test will be provided, as shown in the Mathematical requirements).
• Explain the relationship between genes, proteins and phenotype with respect to the: TYR gene, tyrosinase and albinism; HBB gene, haemoglobin and sickle cell anaemia; F8 gene, factor VIII and haemophilia; HTT gene, huntingtin and Huntington's disease.
• Explain the role of gibberellin in stem elongation including the role of the dominant allele, Le, that codes for a functional enzyme in the gibberellin synthesis pathway, and the recessive allele, le, that codes for a non-functional enzyme.

16.3 Gene control

• Describe the differences between structural genes and regulatory genes and the differences between repressible enzymes and inducible enzymes.
• Explain genetic control of protein production in a prokaryote using the lac operon (knowledge of the role of cAMP is not expected).
• State that transcription factors are proteins that bind to DNA and are involved in the control of gene expression in eukaryotes by decreasing or increasing the rate of transcription.
• Explain how gibberellin activates genes by causing the breakdown of DELLA protein repressors, which normally inhibit factors that promote transcription.

Haploid, diploid and homologous pairs

A diploid (2n) cell has two complete sets of chromosomes, one from each parent. A haploid (n) cell has one. In humans 2n is 46 and n is 23.

A homologous pair is two chromosomes of the same size and shape carrying the same genes at the same loci, one from each parent. They carry the same genes but not necessarily the same alleles, and that distinction is the whole basis of genetics.

The need for a reduction division follows arithmetically. If gametes were diploid, fertilisation would double the chromosome number every generation. Meiosis halves it, so fertilisation restores it, and the number stays constant from one generation to the next.

Meiosis

Meiosis is two divisions after one round of DNA replication, which is why one diploid cell gives four haploid cells.

Meiosis I, the reduction division:

• Prophase I. Chromosomes condense and become visible, each as two sister chromatids. Homologous chromosomes pair up to form bivalents. Crossing over occurs at points called chiasmata, where non-sister chromatids exchange sections. The nuclear envelope breaks down and the spindle forms.
• Metaphase I. Bivalents line up on the equator, and each pair orients at random with respect to every other pair.
• Anaphase I. Spindle fibres shorten and whole chromosomes, each still two chromatids, are pulled to opposite poles. The homologous pairs are separated, and this is where the chromosome number halves.
• Telophase I. Chromosomes reach the poles, and in animal cells the nuclear envelopes reform and the cell surface membrane pulls inwards to cleave the cell. In plant cells no cleavage occurs; a cell plate forms instead, and many plants go straight on into meiosis II without reforming a nuclear envelope.

Meiosis II looks like mitosis. Chromosomes line up singly on a new equator at metaphase II, and at anaphase II the sister chromatids are separated. Four haploid cells result.

The single most examined distinction: anaphase I separates homologous chromosomes; anaphase II separates sister chromatids. A diagram showing pairs of chromosomes moving apart is anaphase I; one showing single chromatids is anaphase II.

Where genetic variation comes from

Three sources, and questions frequently ask for all three.

• Crossing over in prophase I exchanges sections between non-sister chromatids of a homologous pair, producing chromatids with new combinations of alleles that neither parent had on a single chromosome.
• Random orientation (independent assortment) at metaphase I means each bivalent lines up independently of the others. With 23 pairs, the number of different combinations from orientation alone is 2²³, which is over eight million.
• Random fusion of gametes at fertilisation means any one of millions of possible sperm may fuse with any one of millions of possible eggs.

Mutation is a fourth source, but it is not a product of meiosis.

The vocabulary, used precisely

• Gene: a length of DNA coding for a polypeptide.
• Locus: the position of a gene on a chromosome.
• Allele: an alternative form of a gene.
• Dominant: an allele expressed in the phenotype when present on only one chromosome of a homologous pair.
• Recessive: an allele expressed only when present on both.
• Codominant: both alleles are expressed in the phenotype of a heterozygote.
• Linkage: two or more genes at different loci on the same chromosome, so they tend to be inherited together.
• Test cross: a cross between an individual of unknown genotype showing the dominant phenotype and a homozygous recessive individual, to find out which it is.
• F1: the offspring of a cross between two homozygous parents. F2: the offspring of a cross between F1 individuals.
• Genotype: the alleles an organism carries. Phenotype: the observable characteristics, produced by the genotype interacting with the environment.
• Homozygous: two identical alleles at a locus. Heterozygous: two different alleles.

Monohybrid crosses

A heterozygous cross, Aa × Aa, gives the familiar 3:1 phenotypic ratio and a 1:2:1 genotypic ratio.

Codominance. In snapdragons, C^R C^R is red, C^W C^W is white and C^R C^W is pink. Crossing two pink plants gives 1 red : 2 pink : 1 white. Note the phenotypic ratio now equals the genotypic ratio, because every genotype looks different. Use superscript letters rather than upper and lower case, since neither allele is dominant.

Multiple alleles. The human ABO blood group gene has three alleles: I^A and I^B are codominant with each other and both are dominant to I^O. Four phenotypes result from six genotypes. An individual still carries only two alleles; the population carries three.

Sex linkage. Genes on the X chromosome have no equivalent on the much shorter Y. A male is therefore hemizygous: one allele decides his phenotype. Write the genotypes with the chromosomes shown, X^H Y and X^H X^h, never as bare letters, because the mark scheme wants the sex chromosomes visible.

Worked example, haemophilia. A carrier mother, X^H X^h, and an unaffected father, X^H Y.

Of four offspring: one unaffected daughter, one carrier daughter, one unaffected son, one affected son. So half the sons are affected, none of the daughters is affected, and the probability that any given child is affected is 1 in 4.

This is why haemophilia is far commoner in males. A male needs only the one X^h he inherits from his mother, whereas a female would need an affected father and a carrier or affected mother.

Dihybrid crosses

Two genes on different chromosomes assort independently, so a cross between two double heterozygotes, AaBb × AaBb, gives the 9:3:3:1 ratio.

A test cross on a double heterozygote, AaBb × aabb, gives 1:1:1:1.

Autosomal linkage

If the two genes sit on the same chromosome, they do not assort independently. They are inherited as a unit except where crossing over separates them.

The signature in data is a ratio that departs strongly from the expected one, with a large excess of the parental phenotypes and only a small number of recombinants. A test cross on a linked double heterozygote gives something like 42 : 8 : 9 : 41 rather than 1:1:1:1: the two large classes are the parental combinations, and the two small classes are the offspring produced by crossing over.

Worked example. A test cross gives 100 offspring: 42, 8, 9 and 41 in the four classes.

recombinants = 8 + 9 = 17

recombination frequency = 17 / 100 × 100 = 17%

The closer two loci are on a chromosome, the less often a chiasma falls between them, and the lower the recombination frequency.

Epistasis

Epistasis is one gene affecting the expression of another. Where the first gene's product is needed before the second can have any effect, the recessive homozygote at the first locus masks whatever is at the second.

The syllabus does not require you to memorise the ratios for the different types. It requires you to read a set of results, notice that a 9:3:3:1 has collapsed into something like 9:7 or 12:3:1, and explain it as one gene blocking a pathway that the other gene acts later in.

The chi-squared test

Use it when you have categories and you want to know whether the difference between observed and expected numbers is due to chance or is significant. The formula is provided in the exam.

χ² = Σ (O - E)² / E

Worked example. A dihybrid F2 cross gives 160 offspring. On a 9:3:3:1 ratio the expected numbers are 90, 30, 30 and 10. The observed numbers are 100, 26, 28 and 6.

χ² = 1.111 + 0.533 + 0.133 + 1.600 = 3.38

Degrees of freedom are the number of classes minus one:

degrees of freedom = 4 - 1 = 3

At three degrees of freedom the critical value at p equals 0.05 is 7.82. The calculated value of 3.38 is less than the critical value, so the difference between observed and expected is not significant and could have arisen by chance. The results support the 9:3:3:1 hypothesis.

Three points that decide marks:

• Always use the raw numbers, never percentages or ratios.
• Compare with the critical value at p = 0.05, and say what that means: a probability of 0.05 that a difference this large is due to chance alone.
• A calculated value greater than the critical value means the difference is significant, and the hypothesis, which is usually independent assortment, must be rejected. Linkage is the commonest reason.

Genes, proteins and phenotype

The chain always runs the same way: an allele codes for a polypeptide, the polypeptide is or forms a protein, and the protein produces the phenotype. Four examples are specified.

• TYR and albinism. The TYR gene codes for tyrosinase, the enzyme that catalyses the first step in converting tyrosine to melanin. A recessive mutation gives a non-functional enzyme, no melanin is made, and the individual has white hair, very pale skin and unpigmented irises.
• HBB and sickle cell anaemia. A substitution in the HBB gene changes one base, so the sixth amino acid of the beta-globin chain is valine instead of glutamic acid. Valine is hydrophobic, so at low oxygen concentrations the haemoglobin molecules stick together into long fibres that distort the red blood cell into a sickle shape. Codominant: heterozygotes have both normal and sickle haemoglobin, and are protected against malaria.
• F8 and haemophilia A. The F8 gene codes for factor VIII, a clotting factor. A mutation gives no functional factor VIII, so the clotting cascade stalls and bleeding is prolonged. The gene is on the X chromosome, hence the sex linkage above.
• HTT and Huntington's disease. The HTT gene codes for huntingtin. The mutant allele has an expanded repeat of the triplet CAG, so the protein carries a long run of glutamines, misfolds and accumulates, killing neurones. It is dominant and its onset is usually after the age of 35, which is why it persists: sufferers have often had children before any symptom appears.

Gibberellin and stem length. The Le allele codes for a functional enzyme in the gibberellin synthesis pathway, so plants with at least one Le allele make gibberellin and grow tall. The le allele codes for a non-functional enzyme, so le le plants make little gibberellin and are dwarf. Applying gibberellin to a dwarf plant makes it grow tall, which is the neat demonstration that the gene acts through the hormone.

Gene control in prokaryotes: the lac operon

• A structural gene codes for a protein used by the cell, such as an enzyme.
• A regulatory gene codes for a protein that controls 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 of E. coli is the standard example of induction. It consists of a promoter, an operator, and the structural genes lacZ and lacY, coding for beta-galactosidase and lactose permease. A separate regulatory gene, lacI, codes for the repressor protein.

No lactose present. The repressor protein binds to the operator. RNA polymerase cannot move past it from the promoter, so the structural genes are not transcribed and the enzymes are not made.

Lactose present. Lactose binds to the repressor protein, changing its shape so that it can no longer bind to the operator. RNA polymerase transcribes lacZ and lacY, the enzymes are made, and the lactose is taken up and hydrolysed.

The logic is economy. Making enzymes costs amino acids and ATP, so a bacterium that only makes them when the substrate is there grows faster than one that makes them all the time.

Gene control in eukaryotes

Transcription factors are proteins that bind to DNA and either increase or decrease the rate of transcription of particular genes. They are how a eukaryotic cell decides which of its genes to express, and therefore how one genome produces a liver cell and a neurone.

Gibberellin works through this system in a germinating barley grain. Normally, DELLA proteins act as repressors, binding to and inhibiting the transcription factors that would switch on the amylase gene. Gibberellin causes the breakdown of the DELLA proteins. The transcription factors are released, they promote transcription of the amylase gene, amylase is made and secreted from the aleurone layer, and starch in the endosperm is hydrolysed to maltose for the embryo.

Note the shape of this. Gibberellin does not switch the gene on; it removes the thing that was holding it off. That is exactly the same logic as lactose removing the repressor from the operator, which is why the two examples sit in the same section of the syllabus.

Common mistakes

• Saying homologous chromosomes carry the same alleles. They carry the same genes at the same loci, and often different alleles.
• Placing crossing over in meiosis II or in mitosis. It is prophase I.
• Saying anaphase I separates chromatids. It separates homologous chromosomes; anaphase II separates chromatids.
• Giving only crossing over as a source of variation, and leaving out random orientation and random fertilisation.
• Writing sex-linked genotypes as bare letters rather than showing the X and Y chromosomes.
• Saying a male carrier of haemophilia. A male has one X, so he is affected or he is not; only females can be carriers.
• Treating a departure from 9:3:3:1 as an experimental error rather than as evidence of linkage or epistasis.
• Using percentages or ratios in a chi-squared calculation instead of raw numbers.
• Concluding that a chi-squared value below the critical value proves the hypothesis. It means the difference is not significant, which is weaker.
• Getting the degrees of freedom wrong by counting the classes rather than the classes minus one.
• Saying an operator or a promoter codes for a protein. Neither is transcribed into a polypeptide.
• Saying lactose switches the lac genes on directly. It binds the repressor, which is what stops the repressor binding the operator.