3.7 Genetics, populations, evolution and ecosystems (A-level only)

Question 1

What is meant by the term genotype?

Answer 1

Genetic constitution of an organism

Question 2

What is meant by the term phenotype?

Answer 2

The expression of this genetic constitution (genotype) and its interaction with the environment

Question 3

What are alleles and how do they arise?

Answer 3

Variations of a particular gene (same locus) → arise by mutation (changes in DNA base sequence)

Question 4

How many alleles of a gene can be found in diploid organisms?

Answer 4

• 2 as diploid organisms have 2 sets of chromosomes (chromosomes are found in homologous pairs)
- But there may be many (more than 2) alleles of a single gene in a population

Question 5

Describe a dominant allele

Answer 5

Always expressed (shown in the phenotype)

Question 6

Describe a recessive allele

Answer 6

Only expressed when 2 copies present (homozygous recessive) / NOT expressed when dominant allele present (heterozygous)

Question 7

Describe a codominant allele

Answer 7

Both alleles expressed / contribute to phenotype (if inherited together)

Question 8

What is meant by the terms homozygous and heterozygous?

Answer 8

• Homozygous - Alleles at a specific locus (on each homologous chromosome) are the same

• Heterozygous - Alleles at a specific locus (on each homologous chromosome) are different

Question 9

What do monohybrid and dihybrid crosses show?

Answer 9

• Monohybrid cross - inheritance of one phenotypic characteristic coded for by a single gene

• Dihybrid cross - inheritance of two phenotypic characteristics coded for by two different genes

Question 10

Monohybrid cross (basic)

The dominant allele for the grey gene (G) in horses results in a coat colour turning progressively grey.
The recessive allele (g) results in the normal coat colour being maintained (non-grey phenotype).
A non-grey female is crossed with a heterozygous male.
Draw a genetic diagram to show the expected ratio of phenotypes in the offspring.

Answer 10

Parental phenotypes: Non-grey, Grey
Parental genotypes: gg, Gg
Parental gametes: g and g, G and g
Offspring genotypes and phenotypes:
Gg = Grey
gg = Non-grey
Ratio: 1:1

Question 11

Monohybrid cross with multiple alleles

In one snail species, shell colour is controlled by a gene with 3 alleles. The shell may be brown, pink or yellow. The allele for brown (C^B) is dominant to the other two alleles. The allele for pink (C^P) is dominant to the allele for yellow (C^Y).
A cross between two pink-shelled snails produced only pink-shelled and yellow-shelled snails.
Use a genetic diagram to explain why.

Answer 11

Parental phenotypes: Pink-shelled, Pink-shelled
Parental genotypes: CᴾCʸ, CᴾCʸ
Parental gametes: Cᴾ and Cʸ, Cᴾ and Cʸ
Offspring genotypes and phenotypes:
CᴾCᴾ, CᴾCʸ = Pink-shelled
CʸCʸ = Yellow-shelled
Ratio: 3:1

Question 12

Monohybrid cross with codominance and multiple alleles

The inheritance of the ABO blood groups in humans is controlled by three alleles of a single gene, Iᴬ, Iᴮ and Iᴼ. The alleles Iᴬ and Iᴮ are codominant, and the allele Iᴼ is recessive to Iᴬ and recessive to Iᴮ.
Two heterozygous parents plan to have a child. One has blood group A and the other has blood group B.
Use a genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in their offspring.

Answer 12

Parental phenotypes: Blood group A, Blood group B
Parental genotypes: IᴬIᴼ, IᴮIᴼ
Parental gametes: Iᴬ and Iᴼ, Iᴮ and Iᴼ
Offspring genotypes and phenotypes:
IᴬIᴮ = AB
IᴬIᴼ = A
IᴮIᴼ = B
IᴼIᴼ= O
Ratio: 1:1:1:1

Question 13

Explain the evidence from a pedigree diagram which would show that the allele for [named phenotype] is dominant

Answer 13

• [Named phenotype] parents [n & n] have child [n] WITHOUT [named phenotype]
• So both parents [n & n] must be heterozygous / carriers of recessive allele
- If it were recessive, all offspring would have [named phenotype]

Question 14

Explain the evidence from a pedigree diagram which would show that the allele for [named phenotype] is recessive

Answer 14

• Parents [n & n] WITHOUT [named phenotype] have child [n] WITH [named phenotype]
• So both parents [n & n] must be heterozygous / carriers of recessive allele

Question 15

Dihybrid cross (basic)

In fruit flies, the allele for grey body (G) is dominant to the allele for ebony body (g) and the allele for normal wings (N) is dominant to the allele for vestigial wings (n). These genes are not linked.
Vestigial-winged flies, heterozygous for grey body colour, were crossed with ebony-bodied flies, heterozygous for normal wings.
Complete the genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.

Answer 15

Parental phenotypes: Grey body & vestigial wings, Ebony body & normal wings
Parental genotypes: Ggnn, ggNn
Parental gametes: Gn and gn, gN and gn
Offspring genotypes and phenotypes:
GgNn = Grey, normal
Ggnn = Grey, vestigial
ggNn = Ebony, normal
ggnn = Ebony, vestigial
Ratio: 1:1:1:1

Question 16

Dihybrid cross with codominance

In a species of flowering plant, the T allele for tallness is dominant to the t allele for dwarfness. In the same species, two alleles Cᴿ (red) and Cᵂ (white) code for the colour of flowers.
When homozygous red-flowered plants were crossed with homozygous white-flowered plants, all the offspring had pink flowers. A dwarf, pink-flowered plant was crossed with a heterozygous tall, white-flowered plant.
Complete the genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.

Answer 16

Parental phenotypes: Dwarf & pink-flowered, Tall & white-flowered
Parental genotypes: ttCᴿCᵂ, TtCᵂCᵂ
Parental gametes: tCᴿ and tCᵂ, TCᵂ and tCᵂ
Offspring genotypes and phenotypes:
TtCᴿCᵂ = Tall pink
ttCᴿCᵂ = Dwarf pink
TtCᵂCᵂ = Tall white
ttCᵂCᵂ = Dwarf white
Ratio: 1:1:1:1

Question 17

What is a sex-linked gene?

Answer 17

A gene with a locus on a sex-chromosome (normally X)

Question 18

Explain why males are more likely to express a recessive X-linked allele

Answer 18

• Females (XX) have 2 alleles → only express recessive allele if homozygous recessive / can be carriers
• Males (XY) have 1 allele (inherited from mother) → recessive allele always expressed

Question 19

Monohybrid cross with sex-linkage

A single gene that’s carried on the X chromosome controls the presence of hair on the skin of cattle. Its dominant allele causes hair to be present and its recessive allele causes hairlessness.
A male and a female with hair have two female offspring with hair.
What is the probability that their next calf born is hairless?

Answer 19

Parental phenotypes: Female with hair, Male with hair
Parental genotypes: XᴴXʰ, XᴴY
Parental gametes: Xᴴ and Xʰ, Xᴴ and Y
Offspring genotypes and phenotypes:
XᴴXᴴ = Female with hair
XᴴXʰ = Female with hair
XᴴY = Male with hair
XʰY = Male hairless
Probability: 0.25

Question 20

Monohybrid cross with sex-linkage and codominance

A gene on the X chromosome controls fur colour in cats. The allele G codes for ginger fur and the allele B codes for black fur.
These alleles are codominant.
Heterozygous females have patches of both so their phenotype is described as tortoiseshell.
A tortoiseshell female was crossed with a black male.
Use a genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.

Answer 20

Parental phenotypes: Tortoiseshell female, Black male
Parental genotypes: XᴳXᴮ, XᴮY
Parental gametes: Xᴳ and Xᴮ, Xᴮ and Y
Offspring genotypes and phenotypes:
XᴳXᴮ = Tortoiseshell female
XᴮXᴮ = Black female
XᴳY = Ginger male
XᴮY = Black male
Ratio: 1:1:1:1

Question 21

Dihybrid cross with sex linkage

In fruit flies, a gene for body colour has a dominant allele for grey body (G) and a recessive allele for black body (g). A gene for eye colour has a dominant allele for red eyes (R) and a recessive allele for white eyes (r) and is located on the X chromosome.
A heterozygous grey-bodied, white-eyed female fly was crossed with a black-bodied, red-eyed male fly. Complete a genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring from this cross.

Answer 21

Parental phenotypes: Grey-bodied & white-eyed female, Black-bodied & red-eyed male
Parental genotypes: GgXʳXʳ, ggXᴿY
Parental gametes: GXʳ and gXʳ, gXᴿ and gY
Offspring genotypes and phenotypes:
GgXᴿXʳ = Grey-bodied red-eyed female
ggXᴿXʳ = Black-bodied red-eyed female
GgXʳY = Grey-bodied white-eyed male
ggXʳY = Black-bodied white-eyed male
Ratio: 1:1:1:1

Question 22

Explain the evidence from a pedigree diagram which would show that the allele for [named phenotype] on the X-chromosome is recessive

Answer 22

• Mother [n] WITHOUT [named phenotype] has child [n] WITH [named phenotype]
• So mother [n] must be heterozygous / carrier of recessive allele

Question 23

Explain the evidence from a pedigree diagram which would suggest that [named recessive phenotype] is caused by a gene on the X chromosome

Answer 23

Only males tend to have [named recessive phenotype].

Question 24

Explain the evidence from a pedigree diagram which would show that the gene for [named phenotype] is not on the X chromosome

Answer 24

If dominant
• [Named phenotype] father [n] has daughter [n] WITHOUT [named phenotype]
• Father [n] would pass on allele for [named phenotype] on X chromosome so daughter [n] would have [named phenotype]
OR
If recessive
• [Named phenotype] mother [n] has son [n] WITHOUT [named phenotype]
• Mother [n] would pass on allele for [named phenotype] on X chromosome so son [n] would have [named phenotype]

Question 25

Explain how autosomal linkage affects inheritance of alleles

Answer 25

• Two genes located on same autosome (non-sex chromosome) • So alleles on same chromosome inherited together - Stay together during independent segregation of homologous chromosomes during meiosis • But crossing over between homologous chromosomes can create new combinations of alleles - If the genes are closer together on an autosome, they are less likely to be split by crossing over

Question 26

Autosomal linkage In fruit flies, the genes for body colour and for wing development are not on the sex chromosomes. The allele for grey body colour, G, is dominant to the allele for black body colour, g. The allele for long wings, L, is dominant to the allele for short wings, l. A cross was carried out between flies with grey bodies & long wings (heterozygous for both genes) and flies with black bodies & short wings. The result of this cross was 225 offspring with a grey body & long wings and 220 with a black body & short wings. Explain these results.

Answer 26

• The two genes are linked / autosomal linkage • No crossing over occurs / genes are close together • So only GL and gl gametes produced / no Gl and gL gametes produced / no Ggll and ggLl offspring produced

Question 27

Autosomal linkage In tomato plants, the genes for height and for the type of leaf are on the same homologous pair of chromosomes. The allele T, for a tall plant, is dominant to the allele t, for a dwarf plant. The allele M, for normal leaves, is dominant to the allele m, for mottled leaves. - Tall plants and normal leaves 1850 offspring - Tall plants and mottled leaves 63 offspring - Dwarf plants and normal leaves 52 offspring - Dwarf plants and mottled leaves 579 offspring Crosses were carried out between plants heterozygous for both genes. The position of the two alleles for both genes was the same in each parent plant as shown in the diagram. (T on chromosome with M and t with m). Explain the results. Give the expected ratio of phenotypes if the genes for height and type of leaf were on different homologous pairs of chromosomes.

Answer 27

• Genes are linked (so mainly TM and tm gametes are produced) • Crossing over has occurred • So few Tm and tM gametes produced / fewer tall, mottled and dwarf, normal offspring produced • If not linked (use punnett square) - 9:3:3:1 (tall, normal: tall, mottled: dwarf normal: dwarf, mottled)

Question 28

What is epistasis?

Answer 28

Interaction of (products of) non-linked genes where one masks / suppresses the expression of the other.

Question 29

Dihybrid cross with epistasis The inheritance of fruit colour in summer squash plants is controlled by two genes that are not linked, A and B. Each gene has two alleles. The diagram shows the interaction of these two genes. White →(aa) Green →(B) Yellow Complete the genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross. (aabb & AaBb)

Answer 29

Parental phenotypes: white, yellow Parental genotypes: aabb, AaBb Parental gametes: ab, AB and ab and aB and Ab Offspring genotypes and phenotypes: AaBb = Yellow Aabb = Green aaBb = White aabb = White Ratio: 2:1:1

Question 30

Describe when a chi-squared (X²) test can be used

Answer 30

• When determining if observed results are significantly different from expected results (frequencies) - E.g. comparing the goodness of fit of observed phenotypic ratios with expected ratios • Data is categorical (can be divided into groups e.g. phenotypes)

Question 31

Suggest why in genetic crosses, the observed phenotypic ratios obtained in the offspring are often not the same as the expected ratios

Answer 31

• Fusion / fertilisation of gametes is random • Autosomal linkage / epistasis / sex-linkage • Small sample size → not representative of whole population • Some genotypes may be lethal (cause death)

Question 32

Describe how a chi-squared value can be calculated

Answer 32

X² = Σ((O-E)² / E) O = frequencies observed E = frequencies expected (multiply total n with each expected ratio as a fraction)

Question 33

Describe how a chi-squared value can be analysed

Answer 33

1. Number of degrees of freedom = number of categories - 1 (eg. 4 phenotypes = 3 degrees of freedom) 2. Determine critical value at p = 0.05 (5% probability) from a table 3. If X² value is [greater / less] than critical value at p < 0.05 • Difference [is / is not] significant so [reject / accept] null hypothesis • So there is [less / more] than 5% probability that difference is due to chance

Question 34

What is a population?

Answer 34

A group of organisms of the same species in one area at one time that can interbreed

Question 35

What is a gene pool?

Answer 35

All the alleles of all the genes in a population at any one time

Question 36

What is allele frequency?

Answer 36

Proportion of an allele of a gene in a gene pool (decimal or percentage)

Question 37

What does the Hardy-Weinberg principle state and what are the conditions under which the principle applies?

Answer 37

• Allele frequencies will not change from generation to generation, given: - Population is large - No immigration / emigration (to introduce / remove alleles) - No mutations (to create new alleles) - No selection for / against particular alleles - Mating is random

Question 38

What is the Hardy-Weinberg equation?

Answer 38

p² + 2pq + q² = 1 This can be used simultaneously with: p + q = 1 p = frequency of one dominant allele of the gene q = frequency of the other recessive allele of the gene p² = frequency of homozygous dominant genotype 2pq = frequency of heterozygous genotype q² = frequency of homozygous recessive genotype Note - if alleles are codominant, either can be assigned p and q.

Question 39

Using the Hardy-Weinberg equation (basic) The gene for the Rhesus blood group has two alleles. The allele for Rhesus positive, R, is dominant to that for Rhesus negative, r. 16% of the population of Europe is Rhesus negative. Use the Hardy-Weinberg equation to calculate the percentage of this population that you would expect to be heterozygous for the Rhesus gene.

Answer 39

Find q² = 16% = 0.16 Find q = √0.16 = 0.4 Find p = 1 - 0.4 = 0.6 Find 2pq = 2 × 0.4 × 0.6 = 0.48 Convert = 48%

Question 40

Using the Hardy-Weinberg equation with codominance In one species of flowering plant, two alleles Cᴿ (red) and Cᵂ (white) code for flower colour. When both are present, flowers are pink. A population contained 9% red-flowered plants. Use the Hardy-Weinberg equation to calculate the percentage of pink-flowered plants in this population.

Answer 40

Find q² = 9% = 0.09 Find q = √0.09 = 0.3 Find p = 1 - 0.3 = 0.7 Find 2pq = 2 × 0.3 × 0.7 = 0.42 Convert = 42%

Question 41

Using the Hardy-Weinberg equation (more challenging) In fruit flies, a gene for body colour has a dominant allele G (grey) for and a recessive allele g (black). A population contained 64% grey-bodied flies. Use the Hardy-Weinberg equation to calculate the percentage of flies heterozygous for this gene.

Answer 41

Find q² = p² + 2pq = 64% = 0.64 so q² = 1 - 0.64 = 0.36 Find q = √0.36 = 0.6 Find p = 1 - 0.6 = 0.4 Find 2pq = 2 × 0.6 × 0.4 = 0.48 Convert = 48%

Question 42

Explain why individuals within a population of a species may show a wide range of variation in phenotype

Answer 42

• Genetic factors - Mutations = primary source of genetic variation - Crossing over between homologous chromosomes during meiosis - Independent segregation of homologous chromosomes during meiosis - Random fertilisation of gametes during sexual reproduction • Environmental factors (depends on context - eg. food availability, light intensity)

Question 43

What is evolution?

Answer 43

• Change in allele frequency over time / many generations in a population • Occurring through the process of natural selection

Question 44

Describe factors that may drive natural selection

Answer 44

• Predation, disease and competition for the means of survival • These result in differential survival and reproduction, ie. natural selection

Question 45

Explain the principles of natural selection in the evolution of populations

Answer 45

1. Mutations - Random gene mutations can result in [named] new alleles of a gene 2. Advantage - Due to [named] selection pressure, the new allele might benefit its possessor lexplain whyl → organism has a selective advantage 3. Reproductive success - Possessors are more likely to survive and have increased reproductive success 4. Inheritance - Advantageous allele is inherited by members of the next generation (offspring) 5. Allele frequency - Over many generations, [named] allele increases in frequency in the gene pool

Question 46

Explain the effects of stabilising selection

Answer 46

• Organisms with alleles coding for average / modal variations of a trait have a selective advantage (eg. babies with an average weight) • So frequency of alleles coding for average variations of a trait increase and those coding for extreme variations of a trait decrease • So range / standard deviation is reduced

Question 47

Explain the effects of directional selection

Answer 47

• Organisms with alleles coding for one extreme variation of a trait have a selective advantage (eg. bacteria with high resistance to an antibiotic) • So frequency of alleles coding for this extreme variation of the trait increase and those coding for the other extreme variation of the trait decrease

Question 48

Explain the effects of disruptive selection

Answer 48

• Organisms with alleles coding for either extreme variation of a trait have a selective advantage • So frequency of alleles coding for both extreme variations of the trait increase and those coding for the average variation of the trait decrease • This can lead to speciation

Question 49

Describe speciation (how new species arise from existing species)

Answer 49

1. Reproductive separation of two populations (of the same species) 2. This can result in accumulation of differences in their gene pools 3. New species arise when these genetic differences lead to an inability of members of the populations to interbreed and produce fertile offspring

Question 50

Describe allopatric speciation

Answer 50

1. Population is split due to geographical isolation (eg. new river formed) 2. This leads to reproductive isolation, separating gene pools by preventing interbreeding / gene flow between populations 3. Random mutations cause genetic variation within each population 4. Different selection pressures / environments act on each population 5. So different advantageous alleles are selected for / passed on in each population 6. So allele frequencies within each gene pool change over many generations 7. Eventually different populations cannot interbreed to produce fertile offspring

Question 51

Describe sympatric speciation

Answer 51

1. Population is not geographically isolated 2. Mutations lead to reproductive isolation, separating gene pools by preventing interbreeding / gene flow within one population, eg. • Gamete incompatibility • Different breeding seasons (eg. different flowering times) • Different courtship behaviour preventing mating • Body shape / size changes preventing mating 3. Different selection pressures act on each population 4. So different advantageous alleles are selected for / passed on in each population 5. So allele frequencies within each gene pool change over many generations 6. Eventually different populations cannot interbreed to produce fertile offspring

Question 52

Explain genetic drift and its importance in small populations

Answer 52

• Genetic drift = a mechanism of evolution in which allele frequencies in a population change over generations due to chance • As some alleles are passed onto offspring more / less often by chance - Regardless of selection pressures and whether alleles give a selective advantage • So strongest effects in small populations as gene pool is small and chance has a greater influence - Eg. when a population is sharply reduced in size (bottleneck effect) - Eg. when a small, new colony forms from a main population (founder effect) • This can reduce genetic diversity - some alleles can become fixed or lost entirely

Question 53

What is a community?

Answer 53

All the populations of different species living in the same place (habitat) at the same time.

Question 54

What is an ecosystem?

Answer 54

A community and the non-living (abiotic) components of its environment. - Ecosystems can range in size from very small to very large. - They are dynamic systems (populations rise / fall over time).

Question 55

What is a niche?

Answer 55

• The specific role of a species within its habitat, eg. what it eats, where and when it feeds • Governed by its adaptation to both abiotic (non-living) and biotic (living) conditions

Question 56

Explain the advantage of species occupying different niches

Answer 56

• Less competition for food / resources • If two species tried to occupy the same niche, one would outcompete the other

Question 57

What is carrying capacity?

Answer 57

The maximum (stable) population size of a species that an ecosystem can support.

Question 58

List the factors that influence carrying capacity

Answer 58

• Abiotic factors - Eg. light intensity, temperature, soil ph & mineral content, humidity • Interactions between organisms - Interspecific competition - between organisms of different species - Intraspecific competition - between organisms of the same species - Predation (predators kill and eat other animals, called prey)

Question 59

Explain how abiotic factors may affect population size / carrying capacity

Answer 59

• If conditions favourable, organisms more likely to survive & reproduce → increasing carrying capacity • Eg. increasing light intensity increases rate of photosynthesis in plants - This increases carrying capacity of a variety of plant species - So increases the number and variety of habitats, niches and food sources for animals - So increasing carrying capacity of a variety of animal species

Question 60

Explain how interspecific competition may affect population size

Answer 60

• Reduces [named resource] available to both species, limiting their chances of survival & reproduction - So reduces population size of both species • If one species is better adapted, it will outcompete the other - So population size of less well adapted species declines, potentially leading to extinction

Question 61

Explain how intraspecific competition may affect population size

Answer 61

• As population size increases, resource availability per organism decreases, so competition increases - So chances of survival & reproduction decrease → population size decreases • As population size decreases, resource availability per organism increases, so competition decreases - So chances of survival & reproduction increase → population size increases

Question 62

Explain the changes which occur in populations of predators & prey

Answer 62

Populations fluctuate in cycles, the predator population peaking after the prey (lag time): 1. Prey population increases so predators have more food • So more predators survive & reproduce 2. Predator population increases so more prey killed & eaten • So less prey survive & reproduce 3. Prey population decreases so predators have less food • So less predators survive & reproduce 4. Predator population decreases so less prey killed & eaten • So more prey survive & reproduce (cycle repeats)

Question 63

Describe how the size of a population of slow-moving or non-motile organisms can be estimated

Answer 63

1. Divide area into a grid / squares eg. place 2 tape measures at right angles 2. Generate a pair of coordinates using a random number generator (eg. on a calculator) 3. Place a quadrat here and count number / frequency of [named species] 4. Repeat a large number of times (10 or more) and calculate a mean per quadrat 5. Population size = (total area of habitat / quadrat area) x mean per quadrat

Question 64

Describe how the mark-release-recapture method can be used to estimate the size of a population of motile organisms

Answer 64

• Capture sample of species, mark and release • Ensure marking is not harmful / does not affect survival • Allow time for organisms to randomly distribute before collecting second sample • Population = (number in sample 1 x number in sample 2) / number marked in sample 2 Note - marking doesn't have to be physical. It could be recording the base sequence, for example. Recapturing an organism with an identical base sequence would show the organism has been caught ('marked') before.

Question 65

A student collected 17 lizards and marked them before releasing them back into the same area. Later, she collected 20 lizards, 10 of which were marked. Calculate the number of lizards in this area.

Answer 65

(17 × 20) / 10 = 34

Question 66

What assumptions does the mark-release-recapture method make?

Answer 66

1. Sufficient time for marked individuals to mix / distribute evenly within the population 2. Marking not removed and doesn't affect chances of survival / predation 3. Limited / no immigration / emigration 4. No / few births / deaths / breeding / change in population size (or birth & death rate are equal)

Question 67

Suggest why the mark-release-recapture method can produce unreliable results in very large areas

Answer 67

• Unlikely that organisms will distribute randomly / evenly • Less chance of recapturing organisms (that were marked initially)

Question 68

Describe and explain how primary succession occurs

Answer 68

Succession = change in a community over time due to change in abiotic factors / species 1. Colonisation by pioneer species (first to colonise) 2. Pioneer species (and other species at each stage in succession) change abiotic conditions • Eg. they die and decompose, forming soil which retains water (humus / organic matter) 3. So environment becomes less hostile / more suitable for other species with different adaptations AND less suitable for previous species, so better adapted species outcompete previous species 4. As succession goes on, biodiversity increases 5. Climax community reached - final stable community (no further succession)

Question 69

Describe features of a climax community

Answer 69

• Same species present / stable community over a long time • Abiotic factors (fairly) constant over time • Populations (fairly) stable (around carrying capacity)

Question 70

Explain how conservation of habitats involves management of succession

Answer 70

• Further succession can be prevented to stop a climax community forming - By removing or preventing growth of species associated with later stages eg. by allowing grazing • This preserves an ecosystem at a certain point / in its current stage of succession (plagioclimax) • So early species are not outcompeted by later species and habitats / niches are not lost

Question 71

Describe the conflict between human needs and conservation as well as the importance of managing this

Answer 71

• Human demand for natural resources (eg. timber) is leading to habitat destruction / biodiversity loss • Conservation is needed to protect habitats / niches / species / biodiversity • Management of this conflict maintains the sustainability of natural resources - Meeting current needs without compromising the ability of future generations to meet theirs