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the one Flashcards

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1
Q

MCQ 1-5

A

1-C
2-C
3-C
4-A
5-D

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2
Q

MCQ 6-10

A

6-A (TEM>SEM)
7-B
8-C
9-D
10-D

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3
Q

MCQ 11-15

A

11-C
12-C
13-D
14-A
15-A

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4
Q

2 other roles of mitosis (2)

A

Repair by replacing damaged cells with identical healthy cells.
Tissue repair
Development of body plans
Clonal expansion

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5
Q

Why do plants reproduce asexually by mitosis instead of meiosis

A

Mitosis produces 2 genetically identical daughter cells, whereas meiosis produces 4 genetically different daughter cells.

Mitosis ensures all daughter plants receive advantageous/desirable alleles.

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6
Q

Why do bacteria undergo binary fission instead of mitosis (2)

A

Bacteria have simplified cytoskeleton with no centrioles

So bacteria unable to produce spindle fibers needed to pull chromosomes to opposite poles of the cell

Bacteria have no nucleus

Bacteria have circular DNA, not linear

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7
Q

Advantages and disadvantages of cloning in plants and animals (6)

Advantages:

A

Cloned plants can be produced non-seasonally / all year round so that supply is still reliable

Guarantees quality of crop and livestock yield / genetic modifications can be carried to the next generation

Can be used when sexual reproduction is not possible / rare species

More cloned plants can be produced more quickly than from germinating seeds.

Desirable alleles always passed on to offspring.

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8
Q

disadvantages of cloning in plants and animals (6)

A

Low hybrid vigour / increased chance of inbreeding depression

Increased chance of homozygous recessive genotypes could lead to genetic disorders / health issues (e.g. large offspring syndrome in cloned cows)

Cloned plants have low genetic diversity so more susceptible to the same environmental change / selection pressure

Success rate of SCNT is still very low

Ethical issues due to reduced lifespan of clones

Some people may be against the consumption of cloned crops/animals

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9
Q

Chromosome frog q

A

1024 (2)

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10
Q

Dependent and independent variables for hydrogen peroxide experiment (2)

A

Independent variable is concentration of hydrogen peroxide

Dependent variable is volume of O2 produced

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11
Q

Calculate the initial rate of activity at 2 a.u.

A

0.117 cm3 per second or round to 0.18 cm3 per second

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12
Q

How do you produce 8 a.u. Solution (2)

A

Serial dilutions with dilution factor of 0.4
Use a measuring cylinder to measure 4cm3 of 20 a.u. and 6cm3 of distilled water. Mix and stir to produce a homogenous solution.

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13
Q

Standard Deviation: (2)

A

2.3

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14
Q

Comment on the repeatability of the results (2)

A

the higher the standard deviation, the lower the repeatability.

At 2 a.u. the standard deviation is at most 1, so highly repeatable.

But at 20 a.u. the standard deviation increases up to 9.0, so has low repeatability at this concentration

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15
Q

Why was potato used instead of liver (2)

A

Potato has less catalase, so less O2 lost before delivery tube can be connected

Potatoes will give off the same volume of gas in a longer time, so the percentage uncertainty using potatoes is less than using liver

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16
Q

Suggest an improvement to method (1)

A

Take the cylinders from the same potato

17
Q

Suggest 2 reasons why volume of O2 measured may be inaccurate (2)

A

O2 is a soluble gas so will dissolve in the water of the water trough
Oxygen is also used up by respiration in potato cells.

18
Q

Explain the graph shape for 20 a.u. (3)

A

At 20 a.u the H2O2 substrate concentration is high. So high frequency of collisions between enzyme catalase and H2O2 substrate.

So high rate of formation of enzyme substrate complex when substrate enters complementary active site. So high initial rate of product formation

As reaction progresses, H2O2 used up, so substrate concentration decreases. Substrate concentration is limiting factor, and rate of reaction decreases.

More oxygen produced at 20 a.u because there are more moles of H2O2 at 20 a.u for same given volume.

19
Q

Suggest another means of immobilizing enzyme (1)

A

Membrane separation by microencapsulation
Enclose in a partially permeable membrane

20
Q

Suggest why activity of immobilized enzymes is reduced (2)

A

Shape of active site changed, so no longer complementary to shape of substrate molecule

Enzyme not able to mix with the substrate molecules fully, so lower rate of successful collision between enzyme and substrate

Delay with entrapment as reactants must diffuse into / products must diffuse out of the matrix

21
Q

Why is immobilized yeast more expensive than immobilized invertase? (1)

A

Yeast is living so may need correct conditions and nutrients to stay alive

immobilised yeast would still have to undergo downstream processing afterwards

22
Q

Lactose experiment (6)

A

practically carried out,
control variables are clearly identified,
processing of results is explained

common answers included Benedict’s test, colorimetry, repeating the experiment to calculate means, and using a t-test to compare them.

23
Q

Action of mucous membranes: (2)

A

Goblet cells secrete sticky mucus which trap virus

Ciliated epithelial cells waft mucus entrapped pathogens up and out of airways.

24
Q

Antibody vs Fab Fragment - 2 similarities and 1 difference (3)

A

Sim 1: Both have 2 variable region binding site to bind to separate 2 pathogenic antigens.
Sim 2: Both have disulphide bridges / hinge region joining the polypeptide chain subunits together.

Dif 1: Fab has 4 polypeptide chains of the same length, whereas antibody has 2 long heavy chains and 2 light short chains / Fab has no heavy chains

25
Q

Development of antibiotic resistance (3)

A

Random mutations results in genetic variation between bacteria

Antibiotics act as selection pressure.

Bacteria without alleles for antibiotic resistance die

Bacteria with alleles for antibiotic resistance survive and reproduce

This is natural directional selection

Repeats over many generations

26
Q

Synthetic biology (2)

A

Use of genetically engineered organisms to produce useful drugs/medicines

Bacteria genetically engineered with human gene for insulin to produce insulin,

Suitable for vegans/causes fewer allergic reactions.

27
Q

What type of immunity is this (1)

A

Artificial passive immunity

28
Q

Immunity Graph Letter fill (5)

A

Clonal Selection - A only

Clonal Expansion - B and D only

Antigen presentation - A C only

Highest conc of Memory Cells (E or F - don’t know.

It needs to be after the second exposure to pathogen, but is it E or F.
Remember y axis was antibody concentration, not cell concentration. You would think memory cell concentration is highest after the pathogen has been killed - i.e. once the antibody concentration has decreased back down).

Highest T Helper Cell activity - B and D only

29
Q

Describe change in deer population since 1993 (3)

A

General Decrease
Remains stable in last couple of years
Some fluctuations
+bonus marks for using data

30
Q

% change of wolf population

A

350% (2)

31
Q

Evaluate evidence from the graph supporting the scientists conclusion that grey wolves caused increase in beaver population(4)

A

Graph measures beavers as colonies, and not as beaver population
Causation =/= correlation
Decrease in wolf population from 2003 to 2005, but beaver population did not decrease in this time period.

However

Beaver population increased after reintroduction of wolves (use supporting data from graph).
Wolves kill more deer. Both deer and beavers are herbivores. So less interspecific competition between beavers and deer. So beavers have more plants available to eat.

32
Q

Why are beavers labelled as colonies

A

Too many to count as individuals idk?
To avoid recounting the same beavers?

33
Q

Additional piece of evidence?

A

Number of deer grazing next to the river over time
Health of trees close to the river/ number of trees close to river.

34
Q

How is reintroduction of wolf an example of conservation

A

Active human management is taken to restore/increase the biodiversity.
The beaver population increased following re-introduction, which increased species richness and evenness.

35
Q

6 marker gap fill:

A

Ammonification -
Nitrifying
Nitrosomonas
Nitrobacter
Amino acids/proteins/nucleic acids
Nitrogen fixing

36
Q

Effect of low O2 on plant growth (2)

A

Insufficient oxygen for respiration in decomposers/nitrifying bacteria
Activity of anaerobic pseudomonas increased
So less nitrates in soil
So less amino acids can be formed for protein synthesis

37
Q

Glucose structure (3)

A

Ch2OH on C5
Oxygen needed to be filled in
OH on bottom and H on top of C1

38
Q

Why can glucose be taken up by bacteria (3)

A

Glucose has polar -OH hydroxyl group

So can form hydrogen bonds with water molecules

So glucose is soluble in water

Can pass through channel proteins in the plasma membrane of bacteria

biology past paper s (11 decks)

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