Topic 4: Biodiversity and Natural Resources

Question 1

4.2 i) Outline what biodiversity is and what it includes?

Answer 1

Biodiversity is the variety of living organisms in an environment. It includes:
- Species diversity:
Species richness - the number of different species in a given habitat.
Species evenness - how close in numbers each species is in an environment.
- Genetic diversity: the variation of alleles within a species (or within a population of a species).

Question 2

4.2 i) Explain the term ‘endemism’

Answer 2

Endemism is when a species is unique to a single place (it isn’t found anywhere else in the world).

Question 3

4.2 ii) How can biodiversity be measured within a habitat using species richness?

Answer 3

Species richness is the number of different species in an area. Counting the number of different species is too time-consuming. Instead, a sample of the population is taken and estimates about the whole habitat are based on the sample:
Choose a small area to sample within the habitat being studied. To avoid bias, the sample should be chosen randomly (i.e. using a random number generator to select coordinates). Count the number of individuals of each species in the sample area (e.g. using a quadrat). Repeat the process to gain a better indication of the whole habitat.

Question 4

4.2 ii) How can biodiversity be measured within a species using genetic diversity?

Answer 4

You can measure the genetic diversity within a species using the heterozygosity index (heterozygotes have two different alleles at a particular locus).
H = no. of heterozygotes / no. of individuals in the population
A higher proportion of heterozygotes means that the population has a high genetic diversity.

Question 5

4.2 iii) How can biodiversity be compared in different habitats?

Answer 5

Biodiversity can be compared in different habitats using a formula to calculate an index of diversity (D) - this formula takes account of both species richness and evenness:
D = N(N - 1) / Σn(n - 1)
Where:
N = The total no. of organisms (of all species)
n = The total no. of organism in one species
Σ = The sum of
The higher the number, the more diverse the area is.
A highly diverse habitat would have high species richness and high species evenness. In comparison, a community with an identical number of species, but which is dominates by one of them, is considered to be less diverse. Species evenness often has a greater impact than species richness in terms of biodiversity.

Question 6

4.3 Outline and explain the concept of a ‘niche’.

Answer 6

A niche can be defined as the way an organism exploits its environment. The niche a species occupies within an habitat includes its interactions with other living organisms, and its interactions with non-living organisms. Every species occupies its own unique niche. If two species live in the same habitat and occupy the same niche, they will compete directly with each other - the better adapted organism will out-compete the other and exclude it from the habitat.

Question 7

4.3 Outline and discuss three examples of adaption of organisms to their environment.

Answer 7

Behavioural adaptations are any actions by organisms that help them to survive or reproduce.
Physiological adaptations are internal processes (often systematic responses to an external stimuli) that help organism survive or reproduce.
Anatomical adaptations are structural features of an organisms body that help it survive and reproduce.

Question 8

4.4 Explain how natural selection can lead to adaptation and evolution.

Answer 8

A selection pressure can affect an organisms chance of survival and reproduction. Genetic variations within a species, a result of random mutations, crossing over and independent assortment (during meiosis), can result in individuals who have advantageous genes, and therefore an advantageous phenotype. These individuals have a selective advantage, and are more likely to survive, reproduce and pass on their advantageous alleles to their offspring. Over time, the number of individuals with the advantageous alleles increases. Evolution is therefore a change in allele frequency over time.

Question 9

4.4 List three factors which affect the ability of a population to adapt to new conditions.

Answer 9

1. The strength of the selection pressure.
2. The size of the gene pool.
3. The reproductive rate of the organism.

Question 10

4.5 i) Outline how the Hardy-Weinberg equation can be used to predict allele frequency.

Answer 10

When a gene has two alleles, you can calculate the frequency of one of the alleles given the frequency of the other allele:
p + q = 1
p = the frequency of the dominant allele
q = the frequency of the recessive allele
You can calculate the frequency of one genotype, given the frequencies of the others:
p2 + 2pq + q2 = 1
p2 = the frequency of the homozygous dominant genotype
2pq = the frequency of the heterozygous genotype
q2 = = the frequency of the homozygous recessive genotype

e.g. Given the proportion of people with a heterozygous recessive genotype (qq) in a population, first convert this to a decimal, and then square root it to gain the frequency of the recessive allele (q) within a population. Then calculate the frequency of the dominant allele (p) using 1 - q = p. Using the values for p and q, you can then calculate the 2pq and p2.

Question 11

4.5 i) Explain how and why the Hardy-Weinberg equation and principles can be used to see whether a change in allele frequency is occurring in a species over time.

Answer 11

The allele frequency (how often an allele occurs) in a population can be calculated using the Hardy-Weinberg equation.
The Hardy-Weinberg principle predicts that the frequency of alleles in a population won’t change from one generation to the next - this is on the basis that no immigration, emigration, mutations or natural selection occurs. There also needs to be random mating - where all possible genotypes can breed with each other.
If the frequency of an allele has changed between generations, the Hardy-Weinberg principle doesn’t apply - one of the above factors must have been affecting allele frequency.

Question 12

4.5 ii) Outline how geographical isolation can lead to speciation.

Answer 12

Speciation is the development of a new species. A population can become reproductively isolated due to geographical isolation. Geographical isolation takes place when a physical barrier divides a population of a species. Environmental conditions on either side of the barrier will be slightly different, providing different selection pressures. Genetic variations within each gene pool, a result of random mutations, crossing over and independent assortment (during meiosis), can result in individuals who have advantageous genes, and therefore an advantageous phenotype. These individuals have a selective advantage, and are more likely to survive, reproduce and pass on their advantageous alleles to their offspring. Over time, the number of individuals with the advantageous alleles will increase in each population. Due to the accumulation of different genetic information in each population, they might eventually evolve into two separate species. A species is defined as a group of similar organisms that can reproduce to give fertile offspring

Question 13

4.6 i) Explain what classification entails.

Answer 13

Classification is a means of organising the variety of life based on relationships between organisms using differences and similarities in phenotypes and genotypes. It is based around the species concept, where a species is defined as a group of similar organisms that can reproduce to give fertile offspring.
There are eight taxonomic groups:
Domain, Kingdom, Phylum, Class, Order, Family, Genus, Species.
As you move down the hierarchy, there are more groups at each level but fewer organisms in each group. Since members of a taxon share common features (in phenotype and genotype), it is likely that they also share a common evolutionary ancestor.

Question 14

4.6 i) Outline the common features of each of the five kingdoms.

Answer 14

1) Animalia: euakaryotic, multicellular, no cell wall, heterotrophic
2) Plantae: eukaryotic, multicellular, cell wall made from cellulose, contains chlorophyll and can photosynthesise, autotrophic
3) Fungi: eukaryotic, multicellular, chitin cell wall, heterotrophic - more specifically saprotrophic (absorbs nutrients from dead or decaying matter)
4) Prokaryotae: prokaryotic, unicelluar, no nucleus
5) Protoctista: eukaryotic, single-celled / simple multicellular organisms, usually live in water, heterotrophic / autotrophic

Question 15

4.6 ii) Explain the process that leads to new taxonomic groups.

Answer 15

Scientists can share new discoveries through meetings and scientific journals (this allows to comment on and assess the investigation). New data is critically evaluated by the scientific community (to check that experiments are valid and conclusions fair) e.g. the research is peer-reviewed (to detect invalid claims and support valid ones) before being accepted.

Question 16

4.6 ii) Outline the concept of the three domains of life.

Answer 16

Phylogeny is the study of evolutionary history of groups of organisms. Molecular phylogeny involves the examination of molecules (such as DNA and proteins) to see which species are related and how closely related these organisms are.
In the older, five kingdom system of classification, all organisms were placed into one of five kingdoms. In the new, three domain system, all organisms are placed into one of three domains. Organism that were in the kingdom Prokaryotae (unicellular organisms without a nucleus) are separated into two domains - the Archaea and Bacteria. The Prokaryotae were classified into two domains because molecular phylogeny suggested that Archaea and Bacteria are more distantly related than originally thought. Organisms from the other four kingdoms (organisms with cells that contain a nucleus) are placed in the third kingdom - Eukaryota.

Question 17

4.7 Describe the structure an function of the cell wall.

Answer 17

The cell wall is a rigid structure that surrounds cell walls, made mainly of the carbohydrate cellulose. It supports plant cells.

Question 18

4.7 Describe the structure an function of the middle lamella.

Answer 18

The middle lamella is the outermost layer of the cell. It acts as an adhesive, sticking adjacent plant cells together, and providing stability.

Question 19

4.7 Describe the structure an function of plasmodesmata.

Answer 19

Plasmodesmata (singular: plasmodesma) are channels in the cell walls that link adjacent cells together. They allow transport of substances and communication between cells.

Question 20

4.7 Describe the structure an function of pits.

Answer 20

Pits are regions of the cell wall where the wall is very thing. They’re arranged in pairs so that the pit in one cell is lined up with the pit in the adjacent cell. They allow transport of substances between cells.

Question 21

4.7 Describe the structure an function of an amyloplast.

Answer 21

An amyloplast is a small organelle enclosed by a membrane, containing starch granules. They provide storage for starch grains, and convert starch back to glucose for release when the plan requires it.

Question 22

4.7 Describe the structure an function of a vacuole.

Answer 22

The vacuole contains the cell sap, which is made up of water, enzymes, minerals and waste products. Vacuoles keep the cells turgid (preventing them from wilting) and are also involved in the breakdown and isolation of unwanted chemicals in the cell. It is surrounded by a membrane called the tonoplast, which controls what enters and leaves the vacuole.

Question 23

4.9 Explain the structure and function of starch.

Answer 23

Starch acts as an energy storage molecule for plants. It is a mixture of two polysaccharides:
1. Amylose is a long, unbranched chain make from alpha-glucose monomers, joined with 1-4 glycosidic bonds. The position and angles of these bonds cause a coiled structure, making it compact and therefore good as a storage molecule.
2. Amylopectin is a long, branched chain made from alpha-glucose molecules, joined with both 1-4 and 1-6 glycosidic bonds. These side branches enable to enzymes to access and break down the glycosidic bonds easily, releasing glucose quickly when needed.
Starch is also insoluble in water, preventing it from effecting the concentration of water in the cell and the osmotic balance.

Question 24

4.9 Explain the structure and function of cellulose.

Answer 24

Cellulose is formed from long, unbranched chains of β-glucose, joined by 1,4-glycosidic bonds. The glycosidic bonds are straight, so the polysaccharide remain as straight chains.
Hydrogen bonds form between the β-glucose molecules in neighbouring cellulose chains, forming cellulose microfibrils.
In the cell wall, layers of cellulose microfibrils are laid down at different angles (in a net-like arrangement). This ensures that the cell wall remains strong and supportive.

Question 25

4.10 Why and how are plant fibres exploited by humans.

Answer 25

Xylem vessels and sclerenchyma fibres form plant fibres. It is their strength that makes plant fibres useful - this is a result of two things:

1) Layers of cellulose microfibrils are laid down at different angles (in a net-like arrangement). The strength of the microfibrils (due to the numerous hydrogen bonds) and their arrangement in the cell wall gives plant fibres their strength.
2) When structural plant cells (like sclerenchyma and xylem) have finished growing, they produce a secondary cell wall between the normal cell wall and the cell membrane. The secondary cell wall is thicker than the normal cell wall and usually has more of a woody substance called lignin. The growth of a secondary cell wall is called secondary thickening, and the lignin gives plant fibres must greater tensile strength.

Question 26

4.11 Outline the structures, position in the stem and function of xylem vessels

Answer 26

The function of xylem vessels is to transport water and mineral ions up the plant, as well as providing support. They’re long, tube-like structures, formed from dead cells. The end walls between the cells of the column are either lost or highly perforated (with the cytoplasm and cell organelles broken down) forming a hallow lumen. This makes an uninterrupted tube, allowing water and mineral ions to be transported around the plant. Secondary thickening results in a lignified second cell wall, increasing the tensile strength of the xylem vessel. The lignin also serves to waterproof the cell walls, reducing permeability and preventing water-loss. Water and mineral ions move into and out of the vessels through pits in the walls where there’s no lignin. In the stem, xylem vessels group together with phloem tissue to form vascular bundles (the xylem is closer to the centre of the stem than the phloem). The transport of water and mineral ions is a one-way process.

Question 27

4.11 Outline the structures, position in the stem and function of phloem tissue.

Answer 27

The main function of the phloem is translocation of organic solutes (e.g. sucrose) from where they’re made in the plant to where they’re needed. Sieve tube elements are living cells, joined together to form sieve tubes. The highly perforated end walls of each sieve tube element are called sieve plates - these pores in the plant cell walls facilitate transport of solutes between sieve tube elements. Sieve tube elements have no nucleus, no ribosomes, no vacuole and a very thing layer of cytoplasm. Each sieve tube element is normally associated with one or more nucleated companion cells, to which they are connected by plasmodesmata - these cells perform the metabolic functions that maintain the sieve tube. In the stem, xylem vessels group together with phloem tissue to form vascular bundles (the xylem is closer to the centre of the stem than the phloem). Translocation is a two-way process.

Question 28

4.11 Outline the structures, position in the stem and function of sclerenchyma fibres.

Answer 28

The function of sclerenchyma fibres is to provide support. They’re long, tube-like structures, formed from dead cells. Secondary thickening results in a lignified second cell wall, increasing the tensile strength of the sclerenchyma. Unlike xylem vessels, they do in fact have end walls, and are not involved in transport. They also have more cellulose than other plant cells. Sclerenchyma fibres are usually associated with the vascular bundles in the stem (although they are the furthest from the centre of the stem).

Question 29

4.12 Outline the importance of water in plant cells.

Answer 29

The movement of water through the xylem provides a mass flow system for the transport of inorganic ions. Water’s dipole nature, and the hydrogen bonds between molecules allows it to be very cohesive, transporting substances in a continuous column up the xylem vessel.
Water is a solvent, allowing substances to dissolve in solution, and thereby enabling biological reactions to occur in solution. Ionic substances dissolve easily in water: the slightly positive end will be attracted to the negative ion and the slightly negative end attracted to a positive ion, resulting in water molecules surrounding each ion.
The specific heat capacity of water is very high because a large amount of energy is required to break the strong hydrogen bonds. Because water warms up and cools down slowly, organism can avoid rapid changes in their internal temperature. Water also help to regulate temperature by evaporating from leaves to help cool plants down.
Water also helps to maintain structural rigidity (by exerting pressure in cell vacuoles).

Question 30

4.12 Outline the importance of inorganic ions (nitrate, calcium ions and magnesium ions) to plants.

Answer 30

Nitrate ions are needed to make amino acids. They later go on to form proteins, such as enzymes. Nitrate ions are also found in chlorophyll, nucleic acids (e.g. DNA) and ATP. A lack of nitrogen ions can result in leaves becoming yellow.
Magnesium ions are needed for the production of chlorophyll. A lack of magnesium ions can result in leaves becoming yellow, and veins with reddish-brown tints.
Calcium ions are vital for the structure of the cell wall and the permeability of the cell membrane. A lack of calcium ions can result in stunted growth.

Question 31

4.13 Outline William Withering’s digitalis soup.

Answer 31

William Withering, a scientist in the 1700’s, discovered that an extract of foxglove could be used to treat dropsy (or oedema - this is swelling caused by the accumulation of tissue fluid, brought about by heart failure and a high blood pressure). Withering made a chance observation - a patient suffering from dropsy recovered after being treated with a traditional remedy containing foxglove. Foxglove, however, was known to be poisonous. So Withering started testing different concentrations of foxgloves on patients with dropsy known as his ‘digitalis soup’): while too much digitalis would be poisonous, too little was ineffective. It was through this trial and error method that he discovered the correct dose.

Question 32

4.13 Outline modern drug testing.

Answer 32

In pre-clinical testing, the drug is first tried against isolated cells and tissue cultures, later being tested on live animals (this is to assess safety and determine whether the drug is effective against the target disease). Clinical trials then take place in three phases:
Phase 1: This involves testing a new drug on a small group of healthy individuals. This is to determine a safe dosage, to evaluate any side effects, and ensure that the drug is suitably absorbed, distributed, metabolised and excreted by the body.
Phase 2: The drug is then tested on a small group of patients to determine how effective it is.
Phase 3: During this phase, the drug is compared to existing treatments. It is tested on a large sample of patients for reliable results.
- Placebos and double blind trials are used to make results of clinical trials more valid:
The patients are split into two groups, with one given the drug and another given the placebo (an inactive substance). Patients often show signs of improvement because they believe that they’re receiving treatment (the placebo effect). The group with the placebo is used as a control to asses how effect the drug really is.
In double blind trials, neither the patients nor the doctors know which group has been given the placebo and which group has been given the drug. This reduces bias and ensured validity.

Question 33

4.14 Outline the conditions required for bacterial growth.

Answer 33

Bacteria need sufficient nutrients so they can respire and grow. If they rely on aerobic respiration, they need a sufficient supply of oxygen as well. The optimum temperature and pH is necessary for enzymes catalysing metabolic reactions.

Question 34

4.15 Explain how the use of plant fibres and starch may contribute to sustainability, including plant-based products to replace oil-based plastics.

Answer 34

It is more sustainable to replace oil-based products with plant-base products. Fossil fuels are non-renewable, so forming oil-based products is not a sustainable practice. Making products from plant fibres (e.g. ropes, fabrics and plastics) or starch (e.g. vehicle fuels and bioplastics) is more sustainable than making them from oil - less fossil fuel is used up, and crops can be regrown to maintain the supply for future generations.
Products made from plant fibres tend to be biodegradable, unlike oil-based ones (which can’t be broken down and remain in the environment for years).
It is also easier and cheaper to grow and process plants that extract and process oil.

Question 35

4.16 Evaluate the methods used by zoos in their conservation of endangered species.

Answer 35

Captive breeding programmes involve breeding animals in controlled environments. This helps to increase their numbers, and selective breeding on site between more distantly related organisms can increase genetic diversity, and prevent inbreeding depression. However, animals can have problems breeding outside their natural habitat, which is hard to recreate in a zoo. It is often considered cruel to keep animals in captivity.
Reintroduction programmes involve reintroducing animals in to the wild (their natural habitat). This can increase the biodiversity of an ecosystem, and restore habitats that have been lost. This can help to conserve other animals that depend on predation, for example. However, reintroduction can being new diseases to a habitat, harming organisms that live there and aren’t adapted to survive. Animals previously in captivity may not behave as they would if they’re been raised in the wild.
Zoos also help to educate people about conserving biodiversity by raising public awareness and interest. Scientific research can increase knowledge about the behavious, physiology and nutritional needs of animals (some of which can’t be carried out in the wild).

Question 36

4.16 Evaluate the methods used by seed banks in their conservation of endangered species.

Answer 36

Seed banks store seeds from different species of endangered plants. If the plant becomes extinct in the wild, the seed can be used to grow new plants.
They can help to conserve genetic diversity by storing a range of seeds from plants with different characteristics. Cool, dry conditions are needed for storage: this preserves the seeds for a long time, preventing decay and disease. Seeds are tested for viability, where they are planted, grown and new seeds harvested for storage. It’s cheaper to store seeds than to store fully grown plants, and larger number of seeds can be stores as less space is needed. Plants would also need the conditions from their original habitat, as well as maintaining and looking after them. However, testing seeds for viability can be expensive and time-consuming.