Questions + Answers Flashcards

1
Q

How to calibrate the eyepiece graticule?

A

Use a special slide called the stage micrometer. Slide has a scaled etched onto it.
Line up the scales on the eyepiece graticule and the stage micrometer.

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2
Q

Difference between active and passive immunity characteristics.

A

Active immunity requires exposure to an antigen, passive immunity does not. Active immunity takes a while for protection to develop, passive immunity has immediate protection. Active immunity produces memory cells, passive immunity doesn’t. Active immunity protection is long-term because the antibody is produced in response to complementary antigen being present, passive immunity is short term because the antibodies given are broken down.

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3
Q

How is glucose absorbed by co-transport in the ileum?

A

Sodium ions are actively transported out of the ileum epithelial cells, into the blood by the sodium-potassium pump. This creates a concentration gradient where there is now a higher concentration of sodium ions in the lumen of the ileum than inside the cell. Sodium ions then diffuse from the lumen of the ileum into the epithelial cell, down the concentration gradient. They do this via sodium-glucose transporter proteins. The co-transporter carries glucose into the cell with sodium. This increases the concentration of glucose inside the cell. Glucose diffuses out of the cell, into the blood, down it’s concentration gradient through a protein channel by facilitated diffusion.

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4
Q

Describe how monoclonal antibodies can be used to target a drug to cancer cells. [4 marks]

A

Monoclonal antibodies are made against antigens specific to cancer cells. An anti-cancer drug is attached to the antibodies. The antibodies bind to the antigens on cancer cells because their binding sites have a complementary shape. This delivers the anti-cancer drug to the cells.

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5
Q

What are monoclonal antibodies?

A

Monoclonal antibodies are antibodies produced from a single group of genetically identical B-cells (plasma cells). They’re all identical in structure. Used to target specific substances or cells.

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6
Q

Why do you need to squash the tissue when preparing a slide of plant root tip cells?

A

Place a cover slip over the cells and push down firmly to squash the tissues.
This makes the tissue thinner and allow light to pass through it.

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7
Q

Why shouldn’t you smear the cover slip sideways?

A

Could damage the chromosomes.

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8
Q

Explain two ways in which the structure of fish gills is adapted for efficient gas exchange. (2)

A

Many lamellae give it a large surface area.

Thin surface means short diffusion pathway.

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9
Q

Explain how the counter current mechanism in fish gills ensures the maximum amount of the oxygen passes into the bold flowing through the gills. (3)

A

Blood and water flow in opposite directions.
Blood is always passing water with a higher concentration of oxygen.
The diffusion gradient is maintained throughout the whole length of the gill.

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10
Q

Describe the mass flow hypothesis for the mechanism of translocation in plants. (4)

A

Mass flow hypothesis involves sugars in the source being actively transported into the phloem by companion cells. This lowers the water potential of the sieve cell and water enters by osmosis. An increase in pressure causes mass movement towards the sink. Sugars are converted in the sink for respiration and for storage.

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11
Q

Messenger RNA (mRNA) is used translation to form polypeptides.
Describe how mRNA is produced in the nucleus. (Transcription).
(6)

A

DNA helicase causes the hydrogen bonds between the bases to break. The DNA double helix unwinds, where one strand acts as a template. The RNA nucleotides are attracted to the exposed bases. RNA nucleotides bind with their complementary pairs. RNA polymerase moves along the template strand causing RNA nucleotides to join to each other. Pre-mRNA is formed. Splicing occurs where introns are removed from the pre-mRNA to make mRNA.

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12
Q

Describe the structure of proteins. (5)

A

A protein is a polymer of amino acids. They are joined by peptide bonds. They are formed by condensation reactions. The primary structure of proteins are the sequence of amino acids in the polypeptide chain. The secondary structure is the way the chain of amino acids in the polypeptide is folded, due to hydrogen bonding. The tertiary structure is the way the whole molecule is folded, due to hydrogen, ionic and disulfide bonds. The quaternary structure is two or more polypeptide chains linked together.

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13
Q

Describe how proteins are digested in the human gut. (4)

A

The peptide bonds in the protein are hydrolysed. Endopeptidases hydrolyse internal peptide bonds, making polypeptides into smaller peptide chains. Exopeptidase’s remove terminal amino acids.
Dipeptidases hydrolyse dipeptides into amino acids.

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14
Q

Describe how you would use a biochemical test to show that a reducing sugar is present. (2)

A

Add Benedict’s reagent and heat.

If the solution turns orange then a reducing sugar is present.

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15
Q

A high concentration of galactose slows down the breakdown of lactose by lactase.
Use your knowledge of competitive inhibition to suggest why. (2)

A

The inhibitor (galactose) is complementary to the active site. It binds with the active site. Means fewer or no enzyme substrate complexes.

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16
Q

Explain how the inner membrane is adopted to it’s function in mitochondria. (2)

A

The inner membrane contains cristae which provide a large surface area for enzymes in respiration to attach to.

17
Q

Give one feature of a prokaryotic cell that is not found in a eukaryotic cell. (1)

A

70s ribosomes

18
Q

Describe how a sample consisting of only chloroplasts could be obtained from homogenised plant tissue. (3)

A

Use differential centrifugation.
Discard the low-spin pellet.
The supernatant is spun at a higher speed until the pellet with chloroplasts is found.

19
Q

Describe how the structures of starch and cellulose molecules are related to their functions. (5)

A

Starch is insoluble so it is osmotically inactive. Starch is branched so glucose is easily released for respiration. Starch is a large molecule so it cannot leave the cell. Starch has a helical shape so it is compact.
Cellulose is long and has unbranched chains of B glucose. Cellulose is joined by hydrogen bonding. Form micro fibrils. Provide rigidity and strength.

20
Q

How are epithelial cells adapted for absorption?

A

They have microvilli for a large surface area. They have many mitochondria to release energy. They have carrier proteins for active transport and facilitated diffusion. Sodium ions and glucose are co-transported. Membrane bound enzymes digest dissacharides to produce glucose.

21
Q

Explain the advantage of the lugworm having haemoglobin with a dissociation curve to the left. (2)

A

It has a higher affinity for oxygen so it loads more oxygen. At a lower partial pressure of oxygen.

22
Q

Describe how tissue fluid returns to the circulatory system. (3)

A

The hydrostatic pressure is lower in the capillaries and higher in the tissues. The lower water potential in the capillaries causes water to move into them by osmosis. This is due to protein in the blood. The remainder of the tissue fluid is carried back by the lymphatic system.

23
Q

Describe how you would use the emulsion test to show that a seed contains lipids. (3)

A

Crush the seeds and shake with ethanol. Then add water and shake again. If the seed contains lipids then an emulsion will be seen, as the solution goes white.

24
Q

Describe now the structure of a phospholipid molecule is different from that of a triglyceride. (2)

A

A phospholipid has a phosphate group. A phospholipid has two fatty acids instead of three.

25
Q

What is meant by an unsaturated fatty acid. (2)

A

It is unsaturated because it has double bonds between the carbon atoms. Also it is not fully saturated with hydrogen.

26
Q

Describe how bacteria are destroyed by phagocytes. (3)

A

The phagocyte engulfs the bacteria to form a phagosome. The lysosome empties it’s contents into the phagosome. Enzymes are then released which digest the bacteria.

27
Q

Give two structures a bacterial cell may have that a white blood cell does not have. (2)

A

Capsule

Flagellum

28
Q

Explain why an enzyme only catalyses one reaction. (2)

A

The enzyme has an active site. Only the substrate fits to the active site.

29
Q

Give three ways in which courtship behaviour increases the probability of successful mating. (3)

A

It means members of the same species recognise each other.
It is an indication of sexual maturity.
It stimulates the release of gametes.

30
Q

Maltose if hydrolysed by the enzyme maltase.

Explain why maltase catalyses only this reaction. (3)

A

The active site of maltase has a specific shape. Only the maltose substrate is the right shape to form enzyme-substrate complexes.

31
Q

Explain what is meant by a phylogenetic group. (1)

A

They show evolutionary relationships within groups.

32
Q

Explain what is meant by a species. (2)

A

A species includes organisms that can reproduce to create fertile offspring.

33
Q

Explain how a mutation can result in the production of a non-functional protein receptor. (4)

A

There is a change in the DNA base. Therefore the amino acid sequence is changed. The position of hydrogen, ionic and disulphide bonds are altered due to the mutation. There is a change in the tertiary structure of the receptor.

34
Q

People with the CCR5 mutation show a greater resistance to developing AIDS. Explain why. (2)

A

The receptor is not complementary .

Therefore no replication of the virus.

35
Q

Name two structures present in plant cells that are not present in animal cells. (1)

A

Cell wall

Chloroplasts

36
Q

Differences between mitosis and meiosis. (4)

A

In mitosis the number of chromosomes stays the same but in meiosis the number of chromosomes halves. Mitosis produces diploid cells whilst meiosis produces haploid cells. In mitosis the cells produced are identical but in meiosis they are not.