Notes · BiologyUK · A-Levels

Genetics, populations, evolution and ecosystems

This A-level chapter follows genes from the family to the population and the ecosystem. It covers patterns of inheritance and the genetic crosses used to predict them, the chi-squared test, the Hardy-Weinberg principle for allele frequencies, natural selection and speciation, and the dynamics of populations in ecosystems including succession and the methods used to study them.

6 sections~19 min reading time3 competenciesLevel Standard 1 · Advanced 5

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Exam profile

AO1 · Describe patterns of inheritance, selection, speciation and population dynamicsAO2 · Carry out genetic-cross, Hardy-Weinberg and mark-release-recapture calculationsAO3 · Analyse inheritance and ecological data with the chi-squared test and evaluate conservation strategies

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basic level

This whole topic is A-level (A2) only; it is not part of the AS qualification.

higher level

The full A-Level expects confident genetic-cross work, statistical testing with chi-squared, Hardy-Weinberg calculations and quantitative ecology.

Depth

Text

Contents · 6 sections

Genetics, populations, evolution and ecosystems

Monohybrid inheritance and codominance#

StandardAQA 7402 3.7.1DfE GCE Biology - inheritance

A monohybrid Punnett square

Fig. 1Bb x Bb gives a 3:1 phenotype ratio (3 dominant : 1 recessive).

Inheritance is described with precise terms. A gene is a length of DNA at a locus that codes for a characteristic; its different versions are alleles. The genotype is the alleles an organism has, and the phenotype is the characteristic it shows. An allele is dominant if it is expressed in the phenotype even when only one copy is present, and recessive if it is expressed only when two copies are present; an organism with two identical alleles is homozygous and with two different alleles is heterozygous.

A monohybrid cross follows the inheritance of a single gene. A genetic diagram sets out the parental genotypes, the gametes (each carrying one allele, because meiosis separates the alleles), and the offspring produced by all combinations, usually shown in a Punnett square. Crossing two heterozygotes (

\text{Aa} \times \text{Aa}

) gives a genotype ratio of 1 AA : 2 Aa : 1 aa and a phenotype ratio of 3 dominant : 1 recessive - the classic 3:1 that signals a heterozygous monohybrid cross.

Not all alleles are simply dominant or recessive. In codominance both alleles are expressed in the heterozygote, so the heterozygote shows a distinct third phenotype rather than one masking the other - for example, in some flowers a red allele and a white allele give pink heterozygotes, and in human blood groups the A and B alleles are codominant (giving group AB). Codominance is written with a common capital letter for the gene and superscripts for the alleles (for example

\text{C}^\text{R}

and

\text{C}^\text{W}

) to show neither is recessive.

Ratios are predictions of probability, not guarantees. A 3:1 ratio means each offspring has a 3/4 chance of the dominant phenotype, and the observed numbers in a real cross scatter around the expected ratio, especially with small samples. This is exactly why a statistical test - the chi-squared test, met shortly - is needed to decide whether an observed set of numbers is close enough to a predicted ratio to support the proposed pattern of inheritance.

A codominance cross

Fig. 2With codominance the heterozygote is a distinct phenotype, giving a 1:2:1 phenotype ratio.

Worked example

A monohybrid cross

In pea plants, tall (T) is dominant to dwarf (t). Two heterozygous tall plants are crossed. Determine the genotype and phenotype ratios of the offspring.

Parental genotypes and gametes
Both parents are Tt; each produces gametes T and t.
Combine the gametes
The Punnett square gives TT, Tt, Tt, tt - a genotype ratio of 1 TT : 2 Tt : 1 tt.
Phenotype ratio
TT and Tt are tall, tt is dwarf, so the phenotype ratio is 3 tall : 1 dwarf.

Result: Genotype 1:2:1 (TT:Tt:tt); phenotype 3 tall : 1 dwarf.

Typical mistakes

Confusing genotype ratio (1:2:1) with phenotype ratio (3:1) in a heterozygous cross.
Writing a codominant heterozygote as if one allele were dominant (e.g. Rr rather than C-R C-W).

Active revision

In cattle, the coat-colour alleles for red and white are codominant, with heterozygotes being roan. Predict the offspring ratio from a cross between two roan cattle.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Dihybrid crosses, sex linkage and epistasis#

AdvancedAQA 7402 3.7.1DfE GCE Biology - dihybrid inheritance and linkage

The dihybrid 9:3:3:1 ratio

Fig. 3Two unlinked genes assorting independently give a 9:3:3:1 phenotype ratio.

A dihybrid cross follows two genes at once. If the two genes are on different chromosomes, they are inherited independently (independent assortment), so a cross between two double heterozygotes (

\text{RrYy} \times \text{RrYy}

) gives four phenotypes in the ratio 9:3:3:1. Each double heterozygote produces four kinds of gamete (RY, Ry, rY, ry) in equal numbers, and the 16 combinations in the Punnett square give the characteristic dihybrid ratio.

Sex linkage arises when a gene is carried on a sex chromosome, almost always the X chromosome (the Y is small and carries few genes). Because males have only one X (

\text{XY}

), a single recessive allele on their X is expressed, so X-linked recessive conditions such as haemophilia and red-green colour blindness are far more common in males; females (

\text{XX}

) need two copies to be affected but can be unaffected carriers. Sex-linked alleles are written as superscripts on the X, for example

\text{X}^\text{H}

and

\text{X}^\text{h}

Two departures from the simple dihybrid ratio are examined. Autosomal linkage occurs when two genes are on the same autosome (non-sex chromosome), so they tend to be inherited together and do not assort independently; this gives more of the parental combinations than expected and fewer recombinants (which arise only by crossing over). Epistasis occurs when one gene affects the expression of another - for example, a gene that determines whether any pigment is made at all masks a second gene for the colour of that pigment - which alters the expected ratio, often to forms such as 9:3:4 or 9:7.

The key skill is to read a ratio as evidence of a mechanism. A clean 9:3:3:1 indicates two unlinked genes assorting independently; an excess of parental types indicates autosomal linkage; a modified ratio indicates epistasis; and a difference between the sexes indicates sex linkage. Working carefully through the gametes and the Punnett square, and being ready to explain a departure from the expected ratio, is where the marks lie.

A sex-linked cross

Fig. 4A carrier mother (X-H X-h) and unaffected father (X-H Y): half the sons are affected.

Worked example

A sex-linked cross

For red-green colour blindness (X-linked recessive), a carrier mother (X-B X-b) has children with a father with normal vision (X-B Y). Work out the proportion of sons and of daughters expected to be colour blind.

Parental gametes
Mother: X-B and X-b. Father: X-B and Y.
Offspring
Daughters: X-B X-B (normal) and X-B X-b (carrier). Sons: X-B Y (normal) and X-b Y (colour blind).
Proportions
Half the sons are colour blind; no daughters are colour blind (though half are carriers).

Result: 1/2 of sons colour blind; 0 of daughters colour blind (1/2 of daughters carriers).

Typical mistakes

Writing only two gamete types for a double heterozygote instead of four (RY, Ry, rY, ry).
Omitting the Y chromosome, or the sex, when setting out a sex-linked cross.

Active revision

Haemophilia is X-linked recessive. A woman who is a carrier has children with an unaffected man. State the probability that a son is affected and that a daughter is affected.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

The chi-squared test on genetic ratios#

AdvancedAQA 7402 3.7.1DfE GCE Biology - statistical tests

A chi-squared calculation

Fig. 5Sum the (O - E) squared / E over all categories to find the chi-squared statistic.

The chi-squared (

\chi^2

) test decides whether an observed set of categorical results differs significantly from the numbers expected on a hypothesis - typically a predicted genetic ratio. It is used because the observed offspring of a real cross always scatter a little around the expected ratio, and the test provides an objective rule for deciding whether the difference is small enough to be due to chance or large enough to reject the hypothesis.

The starting point is a null hypothesis, which states that there is no significant difference between the observed and expected results (any difference is due to chance). The expected numbers are calculated from the predicted ratio applied to the actual total number of offspring, so that observed and expected are compared on the same total. The statistic is then computed as

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

, summed over every category.

The calculated value is compared with a critical value read from a table at the appropriate number of degrees of freedom (the number of categories minus one) and, by convention in biology, the 0.05 (5%) probability level. If the calculated

\chi^2

is greater than the critical value, the difference is significant, so the null hypothesis is rejected (the results do not fit the predicted ratio); if it is less than the critical value, the null hypothesis is accepted (the results are consistent with the ratio).

The 0.05 level means there is only a 5% probability that a difference as large as the one observed would arise by chance alone if the null hypothesis were true. Being able to state the null hypothesis, calculate the expected numbers and the statistic, determine the degrees of freedom, and interpret the comparison correctly against the critical value is a complete and frequently examined AO3 skill.

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

The chi-squared statistic

$O$ = observed, $E$ = expected; compare with the critical value at (categories - 1) degrees of freedom, p = 0.05.

Worked example

Testing a 9:3:3:1 ratio

A dihybrid cross of 160 offspring gives 94 round yellow, 28 round green, 30 wrinkled yellow and 8 wrinkled green. Test the fit to a 9:3:3:1 ratio (critical value 7.82 at 3 degrees of freedom).

Expected numbers
160 in a 9:3:3:1 ratio (total 16 parts) gives $90, 30, 30, 10$ (each part = 10).
Compute each term
$(94-90)^2/90 = 0.18$ ; $(28-30)^2/30 = 0.13$ ; $(30-30)^2/30 = 0$ ; $(8-10)^2/10 = 0.40$ .
Sum and compare
$\chi^2 = 0.18 + 0.13 + 0 + 0.40 = 0.71$ ; degrees of freedom = 3.
$\chi^2 = 0.71 < 7.82$
Conclude
0.71 is less than the critical value 7.82, so we accept the null hypothesis: the results fit a 9:3:3:1 ratio.

Result: Chi-squared = 0.71 < 7.82, so the results are consistent with a 9:3:3:1 ratio.

Typical mistakes

Using percentages or the ratio itself instead of the actual observed and expected numbers.
Getting the interpretation backwards: a calculated value greater than the critical value means the difference is significant (reject the null hypothesis).

Active revision

A cross predicts a 3:1 ratio. Of 200 offspring, 138 show the dominant phenotype and 62 the recessive. Calculate chi-squared and, using a critical value of 3.84 at 1 degree of freedom, state your conclusion.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

The Hardy-Weinberg principle#

AdvancedAQA 7402 3.7.2DfE GCE Biology - populations and allele frequencies

Hardy-Weinberg genotype frequencies

Fig. 6Genotype frequencies are the terms of the expansion p-squared + 2pq + q-squared = 1.

A population shares a gene pool - all the alleles of all the genes in the population - and the proportion of a particular allele in it is its allele frequency. The Hardy-Weinberg principle predicts that these allele frequencies will stay constant from generation to generation, provided a set of conditions is met, and it gives equations linking allele frequencies to genotype frequencies. It is the null model of population genetics: if frequencies do change, one of its conditions has been broken (for example, by selection).

For a gene with two alleles, let

p

be the frequency of the dominant allele and

q

the frequency of the recessive allele. Because these are the only two alleles, their frequencies sum to one:

p + q = 1

. The genotype frequencies are then given by

p^2 + 2pq + q^2 = 1

, where

p^2

is the frequency of the homozygous dominant genotype,

2pq

the heterozygous (carrier) frequency, and

q^2

the homozygous recessive frequency.

The equations are powerful because the recessive homozygote is the one genotype that can be identified from the phenotype alone. So, from the observed frequency of the recessive phenotype (

q^2

), you can find

q

(its square root), then

p

(from

p + q = 1

), and hence the frequency of carriers (

2pq

) and of the dominant homozygote (

p^2

) - information you could not get by inspection, because carriers and dominant homozygotes look identical.

The prediction of constant frequencies holds only if there is no mutation, no natural selection, no migration, random mating, and a large population (so chance has little effect). In reality one or more conditions is usually broken, so allele frequencies do change - which is evolution. The Hardy-Weinberg model is therefore most useful as a baseline: a measured departure from its prediction is evidence that an evolutionary process, most often selection, is acting.

p + q = 1

Allele frequencies

$p$ = frequency of the dominant allele, $q$ = frequency of the recessive allele.

p^2 + 2pq + q^2 = 1

Genotype frequencies

$p^2$ homozygous dominant, $2pq$ heterozygous, $q^2$ homozygous recessive.

Worked example

Finding carrier frequency

A recessive genetic condition affects 1 in 400 people (0.25%). Assuming Hardy-Weinberg equilibrium, calculate the frequency of carriers.

Recessive genotype frequency
$q^2 = \dfrac{1}{400} = 0.0025$ .
Recessive allele frequency
$q = \sqrt{0.0025} = 0.05$ .
Dominant allele frequency
$p = 1 - q = 1 - 0.05 = 0.95$ .
Carrier frequency
$2pq = 2 \times 0.95 \times 0.05 = 0.095$ .
$2pq = 2(0.95)(0.05) = 0.095$

Result: The carrier frequency is 0.095, or 9.5% of the population.

Typical mistakes

Taking the frequency of the recessive allele straight from the recessive phenotype frequency (that is $q^2$ , not $q$ - you must square-root it).
Confusing the carrier frequency ( $2pq$ ) with the recessive allele frequency ( $q$ ).

Active revision

In a population, 9% of people show a recessive condition. Assuming Hardy-Weinberg equilibrium, calculate the frequency of the recessive allele and the percentage of the population who are carriers.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Natural selection, evolution and speciation#

AdvancedAQA 7402 3.7.3AQA 7402 3.7.4DfE GCE Biology - selection and speciation

The three types of selection

Fig. 7Disruptive selection, favouring both extremes, is the pattern most likely to drive speciation.

Natural selection changes allele frequencies over generations, and there are three patterns to distinguish by their effect on a continuously varying character. Stabilising selection favours the intermediate phenotype and selects against both extremes, so the mean stays the same but variation is reduced. Directional selection favours one extreme, shifting the mean towards it (as in antibiotic resistance). Disruptive selection favours both extremes over the intermediate, which can split a population into two groups and is the pattern most likely to lead to new species.

Speciation is the formation of new species from existing ones, and it requires reproductive isolation - some barrier that stops two groups interbreeding, so that their gene pools become separate and diverge under different selection and mutation. Once isolated, the groups accumulate genetic differences until they can no longer interbreed to produce fertile offspring, at which point they are separate species. There are two geographical routes.

In allopatric speciation the barrier is geographical: a physical barrier (a mountain range, a river, an ocean) separates two populations, which then experience different environments, mutations and selection pressures, so their gene pools diverge until they can no longer interbreed. In sympatric speciation the populations live in the same area but become reproductively isolated by other means - for example, differences in behaviour, breeding season or (in plants) chromosome changes such as polyploidy - so gene flow between them stops even without a physical barrier.

Small populations are also shaped by genetic drift and its special cases, which act alongside selection. Genetic drift is the change in allele frequencies due to chance, and it has a large effect in small populations, where a few random events can markedly alter frequencies. The founder effect (a new population started by a few individuals carries only a sample of the original alleles) and the genetic bottleneck (a sharp reduction in population size, for example after a disaster, leaves a small, less diverse gene pool) both reduce genetic diversity and can speed divergence.

Worked example

Explaining allopatric speciation

A population of insects is divided by the formation of a new mountain range. Explain how two species could eventually form.

Isolation
The mountain range is a geographical barrier that stops the two populations interbreeding, so their gene pools are separated (no gene flow).
Divergence
The two sides have different environments, so different mutations arise and different alleles are selected; the allele frequencies of the two gene pools diverge over many generations.
Reproductive isolation becomes permanent
Eventually the groups differ so much that they can no longer interbreed to produce fertile offspring even if reunited - they are now separate species.

Result: Geographical isolation prevents gene flow; divergent selection and mutation lead to reproductive isolation and two species.

Typical mistakes

Confusing the three types of selection, especially stabilising (intermediate) with disruptive (both extremes).
Saying speciation happens 'because the environment changes' without the essential idea of reproductive isolation preventing gene flow.

Active revision

Explain how a new river cutting through the range of a species of small mammal could, over a long time, lead to the formation of two separate species.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Populations in ecosystems and succession#

AdvancedAQA 7402 3.7.4DfE GCE Biology - populations in ecosystems

Logistic (S-shaped) population growth

Fig. 8Population growth slows as limiting factors act, levelling off at the carrying capacity.

An ecosystem is all the organisms (the community) in an area together with the non-living (abiotic) factors, and within it each species occupies a niche - its role, defined by how it exploits its environment. A population's size is not fixed: it tends to grow rapidly when resources are plentiful, then level off around the carrying capacity - the maximum size the environment can support - giving an S-shaped (sigmoidal) growth curve. Growth slows as limiting factors bite.

Population size is controlled by limiting factors, which may be density-dependent (their effect increases as the population gets denser, such as competition for food, predation and disease) or density-independent (such as a drought or fire, which affect the same proportion regardless of density). Competition may be intraspecific (between members of the same species for the same resources, the main factor setting the carrying capacity) or interspecific (between different species, which can restrict the distribution and size of each). Predator and prey populations often cycle out of phase: a rise in prey allows predators to increase, which then reduces the prey, which then reduces the predators, and so on.

Communities change over time through succession - a directional change in the community over time. Primary succession begins on bare ground with no soil (for example, bare rock or new volcanic land); pioneer species (such as lichens) colonise, and as they die and decompose they build soil, allowing larger plants to establish, each stage (a seral stage) changing the environment so that the next can move in. Succession usually ends in a stable climax community. Secondary succession follows a disturbance (such as a fire) where soil already exists, so it is faster.

Ecologists estimate population sizes by sampling. For slow-moving or non-motile organisms, quadrats (frames placed randomly, or along a transect to study a gradient) give abundance or percentage cover; for mobile animals, the mark-release-recapture method is used, with the population estimated from the Lincoln index. Conservation - the active management of ecosystems to maintain biodiversity - often works by managing succession, for example by grazing or cutting to prevent a grassland reaching its climax woodland, thereby preserving the species of the earlier stage.

N = \dfrac{n_1 \times n_2}{n_3}

Mark-release-recapture (Lincoln index)

$n_1$ = number first marked, $n_2$ = number in the second sample, $n_3$ = number in the second sample that are marked.

A predator-prey cycle

Fig. 9Predator numbers lag behind prey numbers, giving out-of-phase oscillations.

Worked example

A mark-release-recapture estimate

In a study of beetles, 60 are captured, marked and released. Later, 80 beetles are captured, of which 24 are marked. Estimate the population size.

Identify the values
$n_1 = 60$ (first marked), $n_2 = 80$ (second sample), $n_3 = 24$ (marked in second sample).
Apply the Lincoln index
$N = \dfrac{n_1 \times n_2}{n_3} = \dfrac{60 \times 80}{24}$ .
$N = \frac{60 \times 80}{24} = 200$

Result: The estimated population is 200 beetles.

Typical mistakes

Confusing intraspecific (same species) with interspecific (different species) competition.
In mark-release-recapture, dividing by the wrong figure - the denominator is the number of marked individuals recaptured in the second sample.

Active revision

Describe how you would use quadrats to estimate the percentage cover of a plant species in a field, and explain how the quadrats should be positioned.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Biology 7402 specification

Genetics, populations, evolution and ecosystems

6 sections~19 min reading time3 competenciesLevel Standard 1 · Advanced 5

Genetics, populations, evolution and ecosystems

Key points

A monohybrid cross

Key points

A sex-linked cross

Key points

Testing a 9:3:3:1 ratio

Key points

Finding carrier frequency

Key points

Explaining allopatric speciation

Key points

A mark-release-recapture estimate

Sources

Genetics, populations, evolution and ecosystems

Key points

A monohybrid cross

Key points

A sex-linked cross

Key points

Testing a 9:3:3:1 ratio

Key points

Finding carrier frequency

Key points

Explaining allopatric speciation

Key points

A mark-release-recapture estimate

Sources