Compare the location and functions of embryonic and tissue (adult) stem cells. (7)

1. Stem cells are undifferentiated cells capable of repeated division to both more stem cells and cells that will later differentiate to form specialised cells.

2. Embryonic stem cells (ESCs) are found in the inner cell mass of the blastocyst.

3. ESCs are capable of repeated division…

4. …to form more ESCs and other cell types.

5. ESCs are: capable of forming all the other cell types of the body / pluripotent

6. Tissue stem cells (TSCs) are found throughout the juvenile and adult body.

7. TSCs are capable of repeated division to form more TSCs and other cell types.

8. TSCs: can only form cells of the organ to which they belong / are multipotent.

9. e.g. bone marrow tissue cells can only give rise to bone marrow cells, red blood cells, platelets, phagocytes and lymphocytes.

Give an account, with examples, of the different body tissue types and their functions. (8)

1. Body tissue cells derive from somatic stem cells …

2. … by repeated mitosis.

3. Epithelial tissue - covers the organ surfaces.

4. Protection - skin / secretion - intestinal glands / absorption - villi.

5. Connective tissue - gives shape to organs and supports them.

6. Protection - skull bones / structural framework - ribs / storage of energy - adipose tissue / connecting body organs - blood / connecting epithelial to muscle tissue - cartilage (in tendons)

7. Muscle tissue - which causes locomotion or movement within organs.

8. Skeletal muscle - locomotion / smooth muscle - in arterioles control of access to capillary bed / cardiac muscle - contraction of the heart.

9. Nervous tissue - which transmits messages between the central nervous system and the rest of the body (and within the central nervous system).

10. Neurons - conduct impulses / glial cells - maintain a constant environment for neurons.

Describe the way in which the use of stem cells has contributed to:

A. corneal transplants; (3)

B. skin grafts. (3)

A. Corneal transplants:

1. Traditionally, the person would have to wait for donated corneal tissues to become available.

2. With the use of stem cells isolated from the patient's healthy eye, new tissue can be cultured…

3. …which can be transplanted back into the damaged eye,

4. and will repair the damaged corneal epithelium.

B. Skin grafts:

1. Traditionally, a section of the patient's healthy skin is cultured over 3 weeks.

2. During this time, dehydration and infection are risks for the patient.

3. Skin stem cells from the patient can be cultured to produce new epidermal cells much more quickly.

4. This reduces the time during which the patient is at risk.

A. Describe the location and structure of DNA. (8)

B. Give an account of the replication of DNA. (7)

A. Describe the location and structure of DNA.(maximum of 8 marks):

1. DNA is located on chromosomes in the nucleus.

2. A DNA molecule consists of two strands wound in a double helix.

3. Each strand consists of subunits called nucleotides.

4. A nucleotide consists of a deoxyribose sugar molecule, a phosphate group and an organic / nitrogenous base.

5. The deoxyribose and phosphate are linked to their neighbours to form a sugar-phosphate backbone.

6. There are four organic bases: adenine (A), thymine (T), guanine (G) and cytosine (C).

7. Bases are linked in complementary pairs A-T and G-C.

8. Bases link the two DNA strands by hydrogen bonds.

9. The two DNA strands have an antiparallel structure / explanation.

10. Deoxyribose is found at the 3′ end of each strand and phosphate is found at the 5′ end.

B. Give an account of the replication of DNA.(maximum of 7 marks):

1. DNA is unwound and unzipped…

2. …by the enzyme helicase…

3. …to form two template strands.

4. DNA polymerase is the enzyme which adds nucleotides to the new DNA strand.

5. DNA polymerase needs a primer to start replication.

6. DNA polymerase can only add complementary nucleotides to the deoxyribose / 3′ end of the DNA strand.

7. This results in one strand / the leading strand being continuously replicated,

8. and the other strand / the lagging strand being replicated in fragments,

9. which are joined together (by the enzyme ligase).

Give an account of the structure and function of proteins. (8)

A. Protein structure (maximum of 4 marks):

1. Proteins consist of amino acids joined together (in chains).

2. The amino acids are joined by strong peptide bonds…

3. …to produce the primary structure of the polypeptide.

4. Further bonding between amino acids…

5. …produces the secondary and tertiary structures.

6. Folding of the polypeptide chains forms the three-dimensional shape of the protein.

B. Protein function (maximum of 4 marks):

1. Some proteins are enzymes + named example (e.g. the digestive enzyme amylase).

2. Some proteins are hormones + named example (e.g. insulin).

3. Some proteins are antibodies which help the body fight infections.

4. Some proteins transport substances + named example (e.g. haemoglobin).

5. Some proteins provide structure + named example (e.g. collagen).

Give an account of gene expression under the following headings.

A. Transcription (6)

B. Post-translational modification (PTM) (4)

A. Transcription (maximum of 6 marks):

1. Transcription is the formation of a mRNA molecule on a DNA template.

2. DNA unwinds, and strands separate by the action of helicase.

3. RNA nucleotides attach to exposed bases of DNA.

4. DNA A pairs with RNA U, DNA T - RNA A, DNA G - RNA C and DNAC - RNA G.

5. RNA polymerase joins nucleotides (to each other).

6. This produces the primary transcript.

7. Which contains introns and exons.

8. Exons are protein coding sections of mRNA and introns are non-coding sections.

B. Post-translational modification (PTM) (maximum of 4 marks):

1. It takes place after the polypeptide has been completed on the ribosome.

2. It allows several proteins to be formed from one gene.

3. It may involve enzyme cutting and combining of polypeptide chains.

4. e.g. insulin from pro-insulin.

5. It may involve adding of phosphate or carbohydrate groups.

6. Addition of phosphate enables enzymes / receptors to be switched 'on' and 'off'.

Give an account of gene mutation. (9)

1. A gene mutation is a change in the base type or sequence in a gene.

2. In a substitution (mutation) one base is replaced by another.

3. If substitution produces a new stop codon it is a nonsense mutation.

4. If substitution affects introns and exons it is a splice-site mutation.

5. If substitution changes one mRNA codon it is a missense mutation.

6. In a deletion (mutation) a base is removed.

7. In an insertion (mutation) a base is added.

8. A change to a single nucleotide / base is a point mutation.

9. Insertion and deletion (mutations) can potentially cause frameshift mutation.

10. A frameshift mutation alters all the triplets following it.

11. If any of these mutations occurs in a protein-coding gene, then the protein produced may be altered (or not produced at all).

Give an account of the amplification of DNA sequences by the Polymerase Chain Reaction (PCR). (8)

1. The components of the PCR (template DNA, primers, nucleotides and the Taq DNA polymerase) are mixed together.

2. Heating to 90 - 95°C…

3. …separates (denatures) the double-stranded DNA (containing the target DNA sequence).

4. Annealing at 54°C…

5. …allows the primers to bind (by forming hydrogen bonds) to their complementary sequences on the separated strands

6. Primers are short DNA sequences (each about 20 bases long),

7. that bind at either side of the target sequence (one on each of the complementary strands of the target DNA).

8. Extension at 72°C.

9. Taq polymerase now adds nucleotides to the 3′ ends of the primers and extends them into new complementary strands.

10. Repeated cycles multiply the target DNA exponentially…

11. …since each new double strand separates to become two templates for further synthesis.

Describe the control of the action of enzymes which are continuously present in the cell. (10)

1. Enzymes that are in constant use are continuously present.

2. The presence of the substrate may initiate a reaction.

3. The build-up of the product may reverse the direction of a reaction.

4. The active site is the part of the enzyme molecule to which the substrate (temporarily) binds.

5. Competitive inhibitors resemble the substrate and compete with it for the active site.

6. Increased substrate concentration reduces the effect of competitive inhibition.

7. Competitive inhibition is reversible.

8. Non-competitive inhibitors bind to allosteric sites…

9. …and alter the shape of the sites, preventing the substrate from binding to them.

10. Enzyme activators bind to allosteric sites and alter the shape of the active site to fit the substrate.

11. Feedback inhibition involves non-competitive inhibition.

12. Products from reactions late in metabolic pathway act as non-competitive inhibitors to enzymes earlier in the pathway.

State what is meant by the term 'feedback inhibition', give an example of it, and explain its advantage to the cell. (7)

1. Feedback inhibition in a metabolic pathway occurs when an enzyme's activity is reduced…

2. …by an increase in the concentration of a product from a later step in the pathway.

3. In the energy investment stage of glycolysis…

4. …phosphofructokinase is inhibited by ATP…

5. …which is released by glycolysis and the cytochrome system.

6. If the ATP concentration rises in the cell, phosphofructokinase will become less active…

7. …and the production of ATP will be reduced, or vice versa.

8. The production of ATP will be matched to the cell's need for energy…

9. …and resources will be conserved.

Give an account of the enzymes and coenzymes involved in the citric acid cycle. (6)

1. Each step in the citric acid cycle is mediated by a different enzyme.

2. Dehydrogenase enzymes remove hydrogen and electrons.

3. Coenzymes NAD / NAD+ and FAD accept hydrogen and electrons.

4. NAD / NAD+ accepts hydrogen to form NADH.

5. FAD accepts hydrogen to form FADH2.

6. NADH / FADH2 / reduced coenzymes carry hydrogen and electrons to the electron transport chain…

7. …which results in the generation of ATP.

Give an account of the electron transport chain. (8)

1. NADH and FADH2 carry H+ ions and high energy electrons…

2. …from glycolysis and the citric acid cycle.

3. At the inner mitochondrial membrane, NADH and FADH2 pass the hydrogen ions and high energy electrons…

4. …to carriers in the electron transport chain.

5. The electron transport chain is a collection of proteins attached to the inner mitochondrial membrane.

6. As the high energy electrons released from NADH and FADH2 pass along the chain, …

7. …they release their energy in stages.

8. This energy is used to pump H+ ions into the intermembrane space.

9. The final electron acceptor is oxygen.

10. The (negatively charged) oxygen combines with two H+ ions to form water.

Give an account of the role of creatine phosphate in muscles. (8)

1. Muscle cells do not contain any reserves of ATP which will fuel more than 2s of activity.

2. Creatine phosphate is an energy reserve found in cells with high and variable energy demands…

3. …such as muscle / brain.

4. Creatine phosphate is used by fast twitch muscle fibres.

5. Creatine phosphate breaks down anaerobically…

6. …to release energy and phosphate…

7. …which is used to convert ADP and Pi to ATP at a fast rate.

8. There is only sufficient creatine phosphate to fuel strenuous muscle activity for a maximum of 10s.

9. When energy demands are low, creatine phosphate reserves are restored…

10. …using ATP provided by aerobic respiration.

	Created by Lucy Guthrie over 7 years ago

Extended response questions and mark schemes adapted from SCHOLAR website. Covers all four units: - Human Cells - Physiology and Health - Neurobiology and Communication - Immunology and Public Health

Higher Human Biology: Extended Response Questions/Markschemes