Topic 4: Biodiversity and Natural Resources

Question 1

How does a plant cell compare to an animal cell?

Answer 1

Plant cells contain most of the organelles that animal cells do, plus a few extras that animal cells don’t have.

Question 2

Ultrastructure/components of a plant cell

Answer 2

• Cell Wall
• Middle Lamella
• Plasmodesmata
• Pits
• Chloroplast
• Amyloplast
• Vacuole and Tonoplast

Question 3

Cell wall

Answer 3

Description:
A rigid structure that surrounds plant cells. It’s made mainly of the carbohydrate cellulose.

Function:
Supports plant cells.

Question 4

Middle lamella

Answer 4

Description
The outermost layer of the cell.

Function:
The layer acts as an adhesive, sticking adjacent plant cells together. It gives plant stability

Question 5

Plasmodesmata

Answer 5

Description:
Channels in the cell walls that link adjacent cells together.

Function:
Allow transport of substances and communication between cells.

Question 6

Pits

Answer 6

Description:
Regions of the cell wall where the wall is very thin. They’re arranged in pairs - the pit in one cell is lined up with the pit in the adjacent cell.

Function:
Allow transport of substances between cells.

Question 7

chloroplasts

Answer 7

Description:
Small, flattened structure surrounded by a double membrane, and also has membranes inside called thylakoid membranes - which are embedded with chlorophyll pigment. These membranes are stacked up in some parts of the chloroplast to form grana. Grana are linked together by lamellae - thin, flat pieces of thylakoid membrane.

Function:
The site where photosynthesis takes place. Some parts of photosynthesis happen in the grana, and other parts happen in the stroma (a thick fluid found in the chloroplasts).

Question 8

Amyloplast

Answer 8

Description:
A small organelle enclosed by a membrane. They contain starch granules.

Function:
Storage of starch grains. They also convert starch back to glucose for release when the plant requires it.

Question 9

Vacuole and Tonoplast

Answer 9

Description:
The vacuole is a compartment surrounded by a membrane called the *tonoplast**.

Function:
The vacuole contains cell sap, which is made up of water, enzymes, minerals and waste products. Vacuoles keep the cells turgid - this stops plants wilting. They’re also involved in the breakdown and isolation of unwanted chemicals in the cell. The tonoplast controls what enters and leaves the vacuole.

Question 10

4.9) Starch structure and function

Answer 10

starch = the main energy source in plants, broken down to release glucose.
Starch is insoluble in water so doesn’t cause water to enter cell by osmosis
starch is a mixture of:
Amylose - long, unbranched chain of a-glucose. coiled structure = compact & good for storage. Joined with 1,4 glycosidic bonds
Amylopectin - long branched chain of a-glucose, allows enzymes quick access to glycosidic bonds for quick energy release. Joined by 1,4 & 1,6 glycosidic bonds

Question 11

4.9) Cellulose structure and function

Answer 11

Function: strengthen plant cell walls.
Structure:
- Straight chains of b-glucose joined by 1,4 glycosidic bonds
- consecutive b-glucose molecules are rotated 180* to each other. Due to the inversion many hydrogen bonds form between cellulose chains, linking them together to form microfibrils - strong threads that provide support for cells.
cellulose microfibrils are arranged in a net-like/cross-hatch structure to provide strength.

Question 12

4.11) Describe the structure, function, and position in the stem of sclerenchyma fibres

Answer 12

Function - provide support
Position in stem - on the outside
Structure - bundles of dead cells, long and vertical with a hollow lumen. Contain end walls. Cell walls are thickened with lignin. No pits. more cellulose than other plant cells.

Question 13

4.11) Describe the structure, function, and position in the stem of xylem vessels

Answer 13

Function - transport water and mineral ions up the plant & Provide support.
Structure - dead cells joined end to end to form long, tube-like structures found in bundles. Have a hollow lumen, no cytoplasm and no end walls for easy transport.
Walls are thickened with lignin for support. Lignin is impermeable to water so the cell is dead.
pits allow water + mineral ions to move in and out.
Position in stem - On the inside

Question 14

4.11) Describe the structure, function, and position in the stem of Phloem vessels

Answer 14

Function - translocation of organic solutes in both directions.
Structure - made up of sieve tube elements, joined end to end to form sieve tube. Sieve tube elements are living cells that contain a thin layer of cytoplasm but no nucleus and few organelles. A companion cell is connected to each sieve tube and carry out the living functions for the sieve cells. At either end of the sieve tube elements are sieve plates, containing pores that allow solutes to pass through .
Position in stem - in the middle of the xylem and the sclerenchyma.

Question 15

4.10) Strength of Plant Fibres

Answer 15

• The cell wall contains cellulose microfibrils in a net-like/cross-hatch arrangement to provide strength.
• When some structural plant cells (like sclerenchyma and xylem) have finished growing they produce as secondary cell wall between the normal cell wall and the cell membrane, the secondary cell wall is usually thicker and has more lignin.

Because plants fibres are strong they can be exploited by humans to make things like rope or fabric.

Question 16

4.15) What is meant by the term ‘Sustainability’

Answer 16

Sustainability referrers to the using resources in such a way that the requirements of the current generation are met without depleting these resources for future generations. To make products sustainability requires the use of renewable resources.

Question 17

4.15) How can Plant fibres contribute to sustainability?

Answer 17

Plants fibres can be used to make ropes and fabrics instead of plastic, which is made from oil.
This is more sustainable because:
Less fossil fuels are used up
crops can be regrown (renewable)
plant fibres are biodegradable (minimises environmental pollution)
Easier to grow and process, and cheaper.

Question 18

4.15) How can starch contribute to sustainability?

Answer 18

Bioplastics are plastics made from starch instead of oil and Bioethanol is a fuel that is made from starch instead of oil.
Making plastics and fuels from starch is more sustainable because:
Uses less fossil fuels (a non-renewable resources)
Crops can be regrown (renewable)

Question 19

Core practical 6: Identifying structures in a stem

1) How is magnification calculated?
2) What is the function of a stage micrometre
3) Why are lines on either side of the specimen on the slide drawn using a wax pencil?

Answer 19

1) magnification = size of image / size of object

2) A stage micrometre can be used to calibrate an eyepiece graticule to make measurements.

3) To prevent the dye from spreading

Question 20

4.13) William Withering’s Digitalis soup experiment

Answer 20

in which he isolated the active ingredients from foxglove and then tested different doses on patients and recorded the findings, led to the development of contemporary drug testing protocols.

Question 21

4.13) The three stages of contemporary drug testing

Answer 21

• Phase 1: A range of doses of the drug is tested on a very small group of healthy people to see if it is safe to use (looking for side-effects_
• Phase 2: The drug is tested on a small group of patients with the condition to see if it has any effect on the condition
• Phase 3: A large group of patients is given the drug to assess the effectiveness of it, as well as record type and frequency of any side-effects. This phase involves a double blind trial, meaning neither the doctors nor the patients know who is given the actual drug and who is given a placebo (a chemically inactive substance that resembles the drug being tested to determine psychological impact or the strength og the placebo effect).

Question 22

4.2 i) Biodiversity

Answer 22

​Biodiversity is the variety of living organisms in an area. It includes:
- Species diversity: The number of different species and the abundance of each species in an area.
- Genetic diversity: The variation of alleles within a species (or population of species).

Question 23

4.2 i) Define endemism

Answer 23

The state of a species being unique to a particular geographical location and not found anywhere else.

Question 24

4.2 ii) How to calculate genetic diversity (biodiversity within a species)

Answer 24

heterozygosity index (H):

H = number of heterozygotes / number of individuals in the population

Question 25

4.2 ii) The fruit fly has many different alleles which code for eye colour. In a particular population of 456 fruit flies, 276 were found to be heterozygous at the locus for eye colour. Calculate the heterozygosity index for the flies at the locus for eye colour.

Answer 25

H = number of heterozygotes / number of individuals in the population
276/456
= 0.61

Question 26

4.2 iii) How to calculate species diversity in order to compare biodiversity in different habitats

Answer 26

Count the number of different species (the species richness) and the number (abundance) of individuals in each species. Then use an index of diversity to calculate the species diversity. The higher the number the more diverse the area is.

Question 27

4.2 iii) There are 3 different species of flower in this field - a red species, a white and a blue. There are 11 organisms altogether. There are 3 of the red species, 5 of the white and 3 of the blue. Work out the species diversity of this field.

D=N(N-1)/’sum of’ n(n-1)

Answer 27

N = total number of all organisms of all species
n = total number of organisms of one species

D= 11(11-1) / 3(3-1) + 5(5-1) + 3(3-1)
D = 110 / 6 + 20 + 6
D = 3.44

Question 28

4.3) What is meant by the term ‘niche’

Answer 28

A niche is the role of a species within its habitat, this includes its interactions with other organisms and its interactions with the non-living environment. Every species has its own unique niche and species which share the same niche will compete with each other; the better-adapted species will outcompete the other forcing it to alter the niche it occupies or it could die.

Question 29

4.3) How can organisms be adapted to their niche?

Answer 29

Adaptions are features that increase an organism’s chance of survival and reproduction

• Behavioural adaptions are ways an organism acts that increase its chance of survival e.g ‘paying dead’
• Physiological adaptions are processes inside an organism’s body that increases its chance of survival. E.g. tanning of skin when exposed to the sun over long periods,
• Anatomical (structural) adaptions are structural features of an organism’s body that increase its chance of survival i.e. it’s body shape .

Question 30

4.12) what is the importance of water in plants?

Answer 30

Water is required in plants for photosynthesis, maintaining structural rigidity, transport if substances, and thermoregulation.

Question 31

4.12) What is the importance of magnesium ions in plants?

Answer 31

Magnesium ions are important as they are involved in chlorophyll production. They also activate some plant enzymes.

Question 32

4.12) What is the importance of nitrate ions in plants?

Answer 32

Nitrate ions supply nitrogen for making DNA, RNA, proteins, and chlorophyll.

Question 33

4.12) What is the importance of calcium ions in plants?

Answer 33

Calcium ions are a component of the plant cell wall - they form calcium pectate. They’re also essential for plant growth.

Question 34

4.1) What has happened to the variety of life over time?

Answer 34

Over time the variety of life on Earth has become extensive but is now being threatened by human activity such as deforestation.

Question 35

4.4) Understand how natural selection can lead to adaption and evolution

Answer 35

1) A variety of phenotypes exist within a population due to mutation.
2) An environmental change occurs and as a result of that the selection pressure changes.
3) Some individuals possess advantageous alleles which give them a selective advantage and allow them to survive and reproduce.
4) The advantageous alleles are passed on to their offspring (this process is natural selection)
5) Over time, the frequency of alleles in a population changes (this is evolution).

Question 36

4.5 i) How can the Hardy-weinberg equation be used to see whether a change in allele frequency is occuring in a population over time?

Answer 36

1) The Hardy-Weinberg principle predicts that the frequencies of alleles in a population won’t change from one generation to the next.
2) But this prediction is only true under certain conditions: a large population where there is no immigration, emigration, mutations or natural selection and there is random mating.
3) If the allele frequencies do change between generations in a large population then immigration, emigration, natural selection or mutations have happened.

Question 37

4.5 i) The hardy Weinberg equation
a) what does p+q=1 stand for?
b) What does p^2+2pq+q^2=1 stand for?

Answer 37

a) p+q=1 Is used to predict the allele frequency in a population where:
p = the frequency of dominant alleles
q = the frequency of recessive alleles

b) p^2+2pq+q^2=1 is used to predict the genotype and phenotype frequency in a population where:
p^2 = the frequency of the homozygous dominant genotype (AA)
2pq = the frequency of the heterozygous genotype (Aa)
q^2 = the frequency of the homozygous recessive genotype

Question 38

4.5 i) The frequency of cystic fibrosis (genotype ff) in the UK is currently approximately 1 birth in every 2500. From this information can you estimate the % of people in the UK that are cystic fibrosis carriers (Ff)
Use the following equations:
p+q=1
p^2+2pq+q^2=1

Answer 38

1) Calculate q:
Freq. of CF (ff) is 1 in 2500
ff = q^2 = 1/2500 = 0.0004
so, q = square root of 0.0004 = 0.02

2) Calculate p:
Using p+q=1
p=1-q
p=1-0.02 = 0.98

3) Calculate 2pq:
2pq = 2 x 0.98 x 0.02 = 0.039

The frequency of genotype Ff is 0.039, so the % of the UK population that are carriers is 3.9%.

Question 39

4.5 ii) What happens if two populations become reproductively isolated?

Answer 39

If two populations become reproductively isolated, new species will be formed due to accumulation of different genetic information in populations over time due to different environments and selection pressures. This is speciation and may be either allopatric (in which groups of organisms are geographically isolated) or sympatric (in which they are isolated by other means, within the same area, such as random mutations).

Question 40

4.5 ii) What can lead to reproductive isolation of a species?

Answer 40

Changes in the alleles and phenotypes due to random mutations, which prevent them from successfully breeding together. These changes include:
Seasonal changes (i.e. individuals in the same population develop different flowering or mating seasons).
Mechanical changes (i.e. changes in genitalia prevent successful mating)
Behavioural changes (i.e. changes in courtship rituals that aren’t attractive to the main population)

Geographical isolation:
floods, volcanic eruptions and earthquakes can create physical barriers that isolate some individual from the main population. Adaption and evolution occurs due to change in conditions.

Question 41

4.6 i) What is meant by classification?

Answer 41

Classification is a means of organising the variety of life based on relationships between organisms using differences and similarities in phenotypes and in genotypes, and is built around the species concept.

Question 42

4.6 i) What are the classes that organisms are grouped into?

Answer 42

Domain
Kingdom
Phylum
Class
Order
Family
Genus
Species

Question 43

4.6 i) What are the five kingdoms organisms can be placed into?

Answer 43

Prokaryotae
Protoctista
Fungi
Plantae
Animalia

Question 44

4.6 i) List the features of prokaryotaes and give one example

Answer 44

Features: prokaryotes, unicellular, no nucleus, less than 5um
Example: bacteria

Question 45

4.6 i) List the features of Protoctista and give one example

Answer 45

Features: Eukaryotic cells, usually live in water, single-celled or simple multicellular organisms.
Examples: algae, protozoa

Question 46

4.6 i) List the features of plantae and give one example

Answer 46

Features: eukaryotic, multicellular, cellulose cell walls, can photosynthesis, contain chlorophyll, autotrophic (produce their own food)
Examples: mosses, ferns, flowering plants

Question 46

4.6 i) List the features of fungi and give one example

Answer 46

Features: Eukaryotic, chitin cell wall, saprotrophic (absorb substances from dead or decaying organisms)
Examples: moulds, yeasts, mushrooms

Question 47

4.6 i) List the features of Animalia and give one example

Answer 47

Features: eukaryotic, multicellular, no cell walls, heterotrophic (consume plants and animals).
Examples: molluscs, insects, fish, reptiles, birds, mammals

Question 48

4.6 i) What is the binomial system?

Answer 48

Binomials are the unique scientific names given to organisms (in Latin), the first word is the genus name and the second word is the species name - e.g. humans are homo sapiens.
Binomials allows for species to be universally identified as it is the same across the entire globe.

Question 49

What is taxonomy?

Answer 49

Taxonomy is the science of classification. It involves naming organisms and organising them into groups based on their similarities and differences. This makes it easier to identify and to study them.

Question 50

4.16) Evaluate the methods used by zoos

Answer 50

Advantages:
- Invaluable resource for scientific research (Scientists are able to closely study the genetics, behaviours and habitat needs of captive animals).
- This provides greater understanding of species’ needs and aids conservation efforts in the wild.
- Scientists can carry out studies that would be difficult to do in wild populations.
- Zoos contribute to educating the public about endangered species and conservation efforts.

Disadvantages:
- Captive breeding of small species population can reduce genetic diversity however this is prevented by stud books.
- Certain animals will not breed in captivity.
- Not all zoos can provide adequate habitats for animals with specific needs.
- Zoo animals may not behave in the same way they would in the wild, so scientist data from zoo-based studies may not be reliable .
- Many ppl question the ethics of keeping animals in captivity.

Question 51

4.16) Evaluate the methods used by seed banks

Answer 51

Advantages:
- It’s cheaper to store seeds than to store fully grown plants.
- Larger number of seeds can be sored than fully grown plants.
- Seeds can be stored anywhere as long as it’s cool and dry.
- Seeds are less likely to be damaged by disease, natural disaster or vandalism than plants.
- Scientific Research - scientists can study how plant species can be successfully grown from seeds, which is useful for reintroducing them back into the wild.
- SB’s can be used to grow endangered plants for use in medical research, as new crops or for new materials. This mean we don’t have to remove endangered plants from the wild.
- SB’s contribute to education, helping to raise public awareness and interest in conserving biodiversity.

Disadvantages:
- Testing the seeds for viability can be expensive and time-consuming.
- May be difficult to collect seeds from some plants as they may grow in remote locations.
– Only studying plants from seeds in a SB limits the data to small interbred populations. So info may not be representative of wild plants.

Question 52

4.16) Evaluate the process of reintroducing organism from zoos and seedbanks back into the wild

Answer 52

Benefits:
- Will help prevent extinction in the wild.
- Organisms that rely on these plants and animals for food or habitat may also benefit from their presence.
- This contributes towards restoring lost or degradable habitats.

Negative effects:
- These organisms may carry new diseases that will harm other organisms living in that habitat.
- Reintroduced animals may lack the ability to find food or communicate effectively with members of their own species.

Question 53

Studbooks

Answer 53

Studbooks are used by zoos and captive breeding programmes to help maintain genetic diversity.
The studbook for an individual species shows the history and location of all the captive animals of that species in the places that are co-operating in an overall breeding plan. These are used by zoos to prevent interbreeding an reducing the gene pool of an organism.

Question 54

Core Practical 8: Tensile strength of plant fibres

Answer 54

Variables:
IV: Diameter of the fibres
DV: Mass applied until breakage
CV’s: species of plant (NZ flax), length of fibres, Age/source of fibre, temp and humidity.

Safety:
Dropping masses on feet - use box or soft item
Scalpels - sharp
glasses - to avoid pointed fibre, poking eye

Methods:
1) Micrometer used to accurately measure diameter.
2) Keep length of fibre the same
3) Secure one end of the fibre to the desk, using a clamp stand. Hang masses on the other end.
4) Increase mass by 100g each time until the fibre snaps.
5) Repeat 3x with same fibre and calc. mean.
6) Work out tensile strength = force/cross-sectional area.
7) Repeat with more fibres of different diameters.
8) Draw a scatter graph and look for a correlation.

Question 55

Core Practical 9: Antimicrobial properties of plants

Answer 55

Variables:
IV: Type of plant extract
DV: Area of Zone of inhibition
CV’s: Size of paper disk, concentration of extract, temp (25dc), humidity, time left to grow, sterile conditions.

Safety:
- E.coli - safety when handling life organisms, use aseptic techniques.
- Ethanol - highly flammable
- keep agar plates sealed

Method:
Control = Plain paper disk
1) Crush each plant extract, having weighed the same quantities (garlic, chili, mint etc), and add the same volume of ethanol.
2) Using aseptic techniques (sterilising desk using ethanol, wash hands, set up bunson to flame forceps, avoiding opening petri dish, and autoclave equipment before and after use.
3) Add extract to each sterile paper disk, and place on agar plate
4) Leave at 25dc for 24-48 hrs, to grow.
5) Measure the diameter of the zone of inhibitions + calc. the area for each one
6) Repeats + calc. means.
7) Draw bar charts with SE bars.

Question 56

Core Practical 6: Looking for plant stems

Answer 56

a) Use a rhubarb - preparing longitudinal section
- stained with toluidine blue (TBO), which stains lignin blue/green.
- Observe under the microscope (spiral/longitudinal rings of lignin)

b) X-sections of stems:
- Using razor blade, cut very thin sections of stem (geranium).
- Place sections on a slide, add TBO stain, add a cover slip, and observe under a light microscope (should be able to identify the sclerenchyma, phloem and xylem).

Question 57

Core Practical 7: Mineral deficiencies in Plants

Answer 57

Variables:
- Iv: mineral deficiencies
- DV: change in mass
- CV’s: Time left to grow, temp, light intensity, vol. of solution, age of seedlings.

Method:
Control = solution with all the minerals
1) Setting up seedlings:
- Take mass of each seedling.
- Place seedling into a boiling tube wrapped in cotton wool to support them.
2) Place in each solution
- wrap tests tube in foil to prevent algae growth.
3) Leave to grow for 2-4 weeks (in same conditions).
4) Measure the mass + calc. % change in mass
5) Repeat 3x with each solution, calc. the means.
6) Draw bar graph to compare.