exam qns + ans Flashcards

1
Q

Suggest two advantages to eukaryotic cells of having membrane-bound organelles.

A

Compartmentalisation of cell contents allows for:

MP1: different conditions in different compartments. to maintain optimum conditions for each enzymatic reaction

MP2: to increase surface area for the embedding of enzymes and proteins into the membrane

MP3: isolates harmful substances from the rest of the cell

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2
Q

Describe two functions of Golgi body.

A

MP1: Chemical modification of proteins / lipids occurs in GA such as glycosylation / addition of
short carbohydrate chains to form glycoproteins / glycolipids;

MP2: Proteins are transported to the Golgi apparatus where they are temporarily stored and
concentrated;

MP3: The vesicles containing protein to be secreted and lysosomal enzymes bud off / pinch off the Golgi apparatus as secretory vesicles and lysosomes respectively;

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3
Q

Outline the structure and function of the Golgi body in cells. [6]

A
  • Golgi apparatus (GA) consists of one or several stacks of flattened, curved single membrane-bound sacs called cisternae;
  • New cisternae are formed at cis / receiving / forming face by fusing with transport vesicles from the rER and sER
  • Golgi vesicles bud off from shipping / trans / secreting face whereby its movement to their target
    sites is guided by microtubules / cytoskeleton;

+ any 3 functions (i.e. maximum 3 marks):

  • Chemical modification of proteins / lipids occurs in GA such as glycosylation / addition of short carbohydrate chains to form glycoproteins / glycolipids;
  • Proteins are transported to the Golgi apparatus where they are temporarily stored and concentrated;
  • The vesicles containing protein to be secreted and lysosomal enzymes bud off / pinch off the Golgi apparatus as secretory vesicles and lysosomes respectively;
  • Synthesis and secretion of polysaccharides such as pectin to form new plant cell wall;
  • Transport of lipids via Golgi vesicles to cell surface membrane to enter the lymphatic system;
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4
Q

Describe the properties of the phospholipid bilayer and the aquaporin channels in relation to the movement of water across the cell surface membrane.

A
  • Water is a polar molecule, therefore it is unable to pass through hydrophobic core of phospholipid bilayer;
  • As phospholipid bilayer is fluid and water is a small molecule, some water molecules will pass through small temporary gaps in the membrane via osmosis;
  • Aquaporins have hydrophilic pores which allows water to pass through the phospholipid bilayer;
  • Aquaporins have hydrophobic regions / amino acids which form hydrophobic interactions with hydrophobic fatty acid tails of the phospholipid bilayer, allowing it to be embedded in the membrane;
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5
Q

Explain how proteins are held in the membrane.

A
  • Proteins have hydrophobic regions which form hydrophobic interactions with hydrophobic fatty acid tails of the phospholipid bilayer
  • The hydrophilic regions form hydrogen bonds with the aqueous medium and hydrophilic phosphate heads of the phospholipid molecules.
  • They are attached to cytoskeleton on the cytoplasmic side;
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6
Q

Explain how endocytosis allows foreign material such as aluminium silicate to be taken up by cells.

A
  • Phagocytosis involves the extension of pseudopodia to engulf the foreign material;
  • As the cell surface membrane pinches off, a phagocytic vesicle is formed enclosing the foreign material
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7
Q

Describe the role of proteins, glycolipids and cholesterol in the structure of the cell surface membrane. [8]

A
  • Channel proteins aid in facilitated diffusion as they have hydrophilic pores which allow hydrophilic molecules such as charged ions and polar molecules to pass through the hydrophobic fatty acid tails of the phospholipid bilayer;
  • Carrier proteins aid in facilitated diffusion / active transport as they have specific binding sites and undergo change in conformation when a solute binds to it;
  • Some proteins function as enzymes such as adenylate/adenylyl cyclase; Receptor proteins have sites of attachments which allow for cell signalling between cells;
  • Glycoproteins are involved in cell-cell recognition where cells are able to recognize and bind to one another;
  • Glycoproteins function in intercellular junctions for cell-cell adhesion which allows cells to bind with each other to form a tissue;
  • Formed by addition of short carbohydrate chains to phospholipids/lipids on outer surface of cell membrane;
  • For cell-cell recognition
  • For cell-cell adhesion
  • They act as receptor sites for signal molecules to bind which allows for cell signalling;
  • For electrical insulation (e.g. myelin sheath which insulates axons);
  • For protection (e.g. glycocalyx which protects the cell surface membrane);
  • Regulate fluidity of cell membrane;
  • Maintains mechanical stability of cell membrane;
  • Prevents leakage of small polar molecules;
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8
Q

Explain how compounds are taken up by cells. [8]

A
  • Passive transport is the movement of substances down a concentration gradient without the use of ATP and types of such transport are diffusion, facilitated diffusion and osmosis.
  • Small hydrophobic non-polar molecules such as O2 move across the phospholipid bilayer by diffusion;
  • Charged ions like Na+ and K+ move across the phospholipid bilayer by facilitated diffusion, through channel proteins, which form hydrophilic pores in the membrane;
  • (Large) polar molecules like glucose also move across the phospholipid bilayer via facilitated diffusion, through carrier proteins, which have specific binding sites and undergo change in conformation when a solute binds to it;
  • Water molecules move across the phospholipid bilayer via osmosis, through channel proteins known as aquaporins, or through temporary gaps between phospholipids in the membrane;
  • Osmosis is the net movement of water molecules down a water potential gradient from a region of less negative water potential to a region of more negative water potential across a partially/selectively permeable membrane until a dynamic equilibrium is reached;
  • Some compounds may be taken up by cells through active transport, which is the movement of substances against a concentration gradient using energy in the form of ATP and pumps;
  • Some compounds are taken up by cells through endocytosis, where large molecules are taken up using energy in the form of ATP;
  • Cell surface membrane pinches/buds off to form a vesicle containing the substance
  • Pinocytosis is an unspecific process which involves the uptake of droplets of extracellular fluid via tiny vesicles via the invagination of the cell surface membrane; (against a concentration gradient)
  • Phagocytosis involves the uptake of large, solid materials via the extension of pseudopodia, which are then enclosed in phagocytic vesicles (phagosomes) and digested by hydrolytic enzymes in lysosomes;
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9
Q

explain how the binding of insulin to the insulin tyrosine kinase receptor triggers a response in the target cell that reduces the blood glucose concentration

A
  1. insulin binds to the ligand binding site of the receptor-tyrosine kinase (RTK) that has a complementary shape to insulin
  2. the binding triggers dimerisation of RTK proteins and a three-dimensional conformational change that activates the tyrosine kinase activity in the intracellular receptor tails
  3. the receptor tail carries out cross-phosphorylation that adds phosphate from ATP to the tyrosine amino acid residue on the tail of the other RTK protein
  4. this fully activated RTK phosphorylates other relay proteins that bind to the phosphorylated tyrosine on the RTK
  5. Signal amplification occurs when these relay proteins activate other downstream proteins and kinases in a phosphorylation cascade
  6. Various cellular responses take place eg. increase in GLUT4 on the plasma membrane for glucose uptake from the blood/ increase glycogenesis by glycogen synthesis/ increased rate of glycolysis
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10
Q
A
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11
Q

Describe the molecular structure and functions of haemoglobin. [15]

A
  1. 4 polypeptide chains; 2 alpha chains, 2 beta chains;
  2. each polypeptide chain is repeatedly coiled to form α-helix.
  3. Further folding of secondary structures (α-helices) to form
  4. Globular protein;
  5. Specific 3 dimensional tertiary structure, which is the compact spherical shape.
  6. The quaternary structure of haemoglobin is held by hydrogen bonds, ionic bonds and
    hydrophobic interaction;
  7. Hydrophobic pocket of each subunit contains one haem group
  8. Haem group is a porphyrin ring with an iron Fe2+ in the center;
  9. Each haem group is able to bind reversibly to one oxygen molecule to transport oxygen
  10. Each haemoglobin molecule has four haem groups and can carry 4 oxygen molecules to
    form oxyhaemoglobin.
  11. Reference to quaternary structure being held by weak bonds,
  12. allowing cooperative binding where;
  13. binding oxygen to one subunit causes a conformation change making easier for another
    subunit to bind to oxygen;
  14. The hydrophobic amino acids in the interior of the protein and hydrophilic amino acids
    found at the exterior surface
  15. Soluble
  16. Involved in metabolic reaction i.e. Transport oxygen
  17. Compact, Many haemoglobin molecules can be packed and dissolved into cytoplasm of
    red blood cell; Rej Hb dissolves in blood
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12
Q

Explain how the use of iPSCs may overcome some of the ethical concerns of using other types of stem cells in medical research and treatment.

A

*Using iPSCs avoids the exploitation of pluripotent stem cells from blastocyst / embryonic stem cells
*Thus there is no destruction of human embryos nor denial ot the right of life
*Nor the ethical problem on treatment of unused embryos
*There si no exploitation of women such as lowering of her future fertility from the extraction procedures involved ni ova donation

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13
Q

source of zygotic stem cells and embryonic stem cells

A

ZSC: single cell formed from the fusion between a sperm and an egg cell

ESC: cells obtained from the inner cell mass blastocyst formed a week after fertilisation

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14
Q

Describe the features of blood stem cells and explain their normal functions. [8]

A
  1. Is a type of adult stem cell found in bone marrow of long bones; 2. capable of dividing and renewing themselves for long periods; 3. unspecialized, does not have tissue specific structures for them to perform specialized functions; 4. can give rise to specialized cell types via differentiation ; Ref to differential gene expression 5. Blood stem cells (e.g. lymphoid and myeloid stem cells) are multipotent so are not totipotent or pluripotent 6. They have the ability to differentiate into a limited range of cell type, usually of a closely related family of cells; 7. Myeloid stem cells give rise to red blood cells, granulocytes, platelets; 8. Lymphoid stem cells give rise to T and B lymphocytes; 9. primary function is to maintain the steady state of functioning of tissue by generating replacement ® repair for cells lost through disease, tissue injury or normal wear-and-tear;
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15
Q

Explain how glucagon helps to maintain blood glucose level. [3]

A

When blood glucose levels decrease below the setpoint of 90mg/100ml of blood, it is
detected by the α cells in the islets of Langerhans which secrete glucagon directly into
the bloodstream.
Glucagon will be transported via the circulatory system and binds to receptors found on
cell surface membrane of target tissues / liver cells / effectors.
This triggers cell signalling and liver cells respond by stimulating glycogenolysis and
gluconeogenesis. Negative feedback is achieved to increase blood glucose
concentration back to the normal levels.

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16
Q

With reference to Fig. 6.2, describe how the binding of GLP-1 to its receptor results in cell
proliferation. [4]

A

Binding of GLP-1 to extracellular side of G protein-coupled receptor (GPCR) activates
the receptor and causing it to change its conformation.
The cytoplasmic side of the receptor then binds an inactive G protein, causing G protein
to exchange its bound GDP for GTP.
The G protein is activated and dissociates from the receptor. Activated G protein binds
and activates adenylate cylase, which catalyse the conversion of large number of ATP to
cAMP.
cAMP, a second messenger, binds and activates a large number of protein kinase A
(PKA). Protein kinase A activates CREB protein by phosphorylation.
Activated CREB moves into the nucleus and acts a transcription factor increasing the rate
of transcription of IRS-2 gene.

17
Q

Explain how the structure of the TGFβR allows it to carry out its function as a receptor. [2]

A

TGFβR has an extracellular binding site is complementary to ligand / TGF;
TGFβR is a transmembrane protein embedded on cell surface membrane allowing it
to receive signal and transmit to the cell’s interior.
The intracellular tails of TGFβR allows for dimerisation of receptors / The
intracellular tail of TGFβR functions as threonine kinase.

18
Q

suggest why anaerobic respiration cannot be used continuously in the long run

A
  1. An AR produced net 2 ATP per glucose molecule during substrate level phosphorylation during glycolysis, incompletely oxidising glucose. compared to 32 ATP per glucose molecule in AR
  2. pryuvate is reduced to lactic acid which lowers the pH of muscle cells and can denature cellular enzymes or proteins in muscle cells
19
Q

Explain the term epistasis in this context (dominant epistasis) [3]

A

Genes A and B code for functional proteins that are involved in the same metabolic pathway to synthesise green pigment in organism

Protein A inhibits an enzyme that catalyses the formation of green pigment which becomes a substrate of the enzyme B to form yellow pigment

the presence of one dominant allele at gene locus A will mask the phenotypic expression of gene B

20
Q

Explain how a plant breeder could produce a pure breeding tomato plant with purple stems and cut leaves from the plants described in the passage above. [3]

A

A purple stem and cut leaves tomato plant (AaDd) is selected and allowed to self-pollinate / self -fertilise ; A test cross of all purple stem and cut leaves from the offspring with a double homozygous recessive plant for both characteristic i.e. green stem and uncut leaves / aadd was carried out ; 1/16 of the offspring would give rise to all purple stem and cut leaves offspring in the test cross indicating that the plant is pure breeding (AADD); [3]