Read each question and click on the ONE answer option you believe is correct. The explanation will appear after you click any option.
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In a phospholipid molecule within a bilayer, identify the component that links the phosphate head group to the fatty acid tails.
Click the CORRECT answer:
- Cholesterol
- Glycerol
- An amino acid
- A carbohydrate chain
- The phosphate group itself
Reasoning for Correct Answer #2: In typical phospholipids found in cell membranes, a glycerol molecule serves as the central backbone. Two of glycerol’s hydroxyl groups are esterified to fatty acid tails (hydrophobic), and the third hydroxyl group is esterified to a phosphate group (hydrophilic head). The phosphate group may be further linked to other small polar molecules (like choline or ethanolamine). Therefore, glycerol is the molecule that connects the fatty acid tails to the phosphate head group.
Cholesterol molecules contain a small polar hydroxyl (-OH) group and a larger non-polar ring structure and tail. Explain the typical orientation of cholesterol when embedded within a cell membrane’s phospholipid bilayer.
Click the CORRECT answer:
- The non-polar section aligns with the phosphate heads, while the hydroxyl group embeds in the fatty acid tail region.
- Cholesterol molecules float freely within the aqueous environment on either side of the membrane.
- The polar hydroxyl group interacts with the aqueous environment and phosphate heads, while the non-polar section lies among the fatty acid tails.
- Cholesterol spans the entire membrane with the hydroxyl group on one side and the non-polar tail on the other.
- The entire cholesterol molecule embeds deeply within the fatty acid tail region, avoiding the phosphate heads.
Reasoning for Correct Answer #3: Cholesterol is amphipathic, meaning it has both hydrophilic (polar) and hydrophobic (non-polar) parts. This dictates its orientation in the bilayer. The small, polar hydroxyl (-OH) group is hydrophilic and positions itself near the polar phosphate heads of the phospholipids, interacting with them and the surrounding aqueous environment. The rest of the molecule (the steroid rings and hydrocarbon tail) is bulky and hydrophobic, so it embeds within the non-polar fatty acid tail region of the bilayer.
State one role of cholesterol in the phospholipid bilayers of animal cell membranes.
Click the CORRECT answer:
- Acts as the primary energy storage molecule within the membrane.
- Forms channels for ion transport across the membrane.
- Maintains or regulates membrane fluidity (buffering against temperature changes).
- Serves as the main structural component forming the bilayer itself.
- Acts as an enzyme to break down membrane lipids.
Reasoning for Correct Answer #3: A key function of cholesterol in animal cell membranes is to regulate fluidity. It acts as a ‘fluidity buffer’. At higher temperatures, it restricts phospholipid movement, making the membrane less fluid and more stable. At lower temperatures, it disrupts the tight packing of phospholipid tails, preventing solidification and maintaining fluidity. It does not form the primary structure (that’s phospholipids), store energy, form channels, or act as an enzyme in this context.
Explain why sodium ions (Na⁺) cannot cross phospholipid bilayers by simple diffusion.
Click the CORRECT answer:
- They are too large to fit between the phospholipid molecules.
- They are non-polar and thus repelled by the phosphate heads.
- They require specific channel proteins which are always closed to sodium.
- They are lipid-soluble and readily pass through the membrane core.
- They are charged ions and are repelled by the hydrophobic/non-polar core of the bilayer.
Reasoning for Correct Answer #5: Simple diffusion across a membrane relies on the ability of a substance to pass directly through the lipid bilayer. The core of the bilayer is made of hydrophobic (non-polar) fatty acid tails. Sodium ions (Na⁺) are charged particles (ions) and are therefore hydrophilic (water-soluble). Charged and polar substances are effectively repelled by the hydrophobic core of the membrane and cannot easily pass through it. Their transport requires the assistance of membrane proteins (channels or transporters).
Compare facilitated diffusion and active transport by stating one similarity and two differences.
Click the CORRECT answer:
- Similarity: Both move substances down the concentration gradient. Difference 1: Active transport uses proteins, facilitated diffusion does not. Difference 2: Active transport requires ATP, facilitated diffusion does not.
- Similarity: Both require membrane transport proteins. Difference 1: Facilitated diffusion moves substances down the gradient, active transport moves them against. Difference 2: Active transport requires ATP energy, facilitated diffusion does not.
- Similarity: Both require ATP energy. Difference 1: Facilitated diffusion moves substances against the gradient, active transport moves them down. Difference 2: Both use only channel proteins.
- Similarity: Both move substances against the concentration gradient. Difference 1: Both use membrane proteins. Difference 2: Facilitated diffusion requires ATP, active transport does not.
- Similarity: Both are passive processes. Difference 1: Facilitated diffusion uses proteins, active transport does not. Difference 2: Active transport moves substances against the gradient.
Reasoning for Correct Answer #2:
Similarity: Both facilitated diffusion and active transport utilize specific membrane transport proteins (channels or carriers) to mediate the movement of substances across the membrane.
Difference 1 (Direction): Facilitated diffusion moves substances *down* their concentration or electrochemical gradient (from high to low). Active transport moves substances *against* their gradient (from low to high).
Difference 2 (Energy): Facilitated diffusion is a passive process and does not require the cell to expend metabolic energy (ATP). Active transport is an active process that requires energy, usually directly or indirectly from ATP hydrolysis.
Option 2 correctly states these key points.
Prostaglandins are lipids involved in inflammation, formed partly from phospholipids in the Smooth Endoplasmic Reticulum (SER) involving the enzyme COX. Suggest an advantage for this reaction pathway occurring in the SER rather than the cytoplasm.
Click the CORRECT answer:
- The SER provides direct access to the nucleus for gene regulation of the pathway.
- The SER membrane contains the necessary phospholipid substrates (like arachidonic acid released from membrane phospholipids).
- The cytoplasm contains specific inhibitors that would block the COX enzyme if it were located there.
- The SER has a significantly lower pH which is optimal for COX enzyme activity compared to the neutral cytoplasm.
- Prostaglandins are proteins and therefore must be synthesized on ribosomes associated with the SER membrane.
Reasoning for Correct Answer #2: The Smooth Endoplasmic Reticulum (SER) is a major site of lipid synthesis and metabolism. Prostaglandins are derived from fatty acids (like arachidonic acid) which are released from membrane phospholipids. Locating the synthesis pathway (including the COX enzyme) within or associated with the SER membrane ensures close proximity to the necessary lipid substrates readily available within the membrane itself. Option 5 is incorrect as prostaglandins are lipids, not proteins. Options 1, 3, and 4 are less likely or unsupported advantages compared to substrate availability.
Aspirin reduces the activity of the enzyme COX by modifying an R-group of one of its amino acids. Suggest how this chemical modification reduces COX’s catalytic activity.
Click the CORRECT answer:
- It increases the enzyme’s affinity for the substrate, preventing product release.
- It causes the enzyme to denature completely by breaking all its peptide bonds.
- It changes the enzyme’s primary structure by inserting several new amino acids into the polypeptide chain.
- It alters the active site shape or chemistry, preventing proper substrate binding or the catalytic reaction.
- It provides additional energy to the enzyme, speeding up the reaction uncontrollably.
Reasoning for Correct Answer #4: Enzyme activity depends on the precise three-dimensional structure of its active site, which specifically binds the substrate and facilitates the chemical reaction. Aspirin covalently modifies a specific amino acid residue (serine) within or near the active site of COX enzymes. This chemical modification alters the shape, charge distribution, or steric properties of the active site, thereby preventing the natural substrate (arachidonic acid) from binding correctly or hindering the catalytic steps necessary for prostaglandin synthesis. This leads to reduced or inhibited enzyme activity. It doesn’t typically cause complete denaturation or change the primary sequence.
Outline the process of cell signalling involving prostaglandins that leads to a response, such as inflammation.
Click the CORRECT answer:
- Prostaglandins bind to specific intracellular receptors located in the cytoplasm or nucleus, directly altering gene transcription.
- Prostaglandins are released from cells, diffuse locally, and bind to specific G-protein coupled receptors on the surface of target cells, initiating intracellular signal transduction pathways that lead to the inflammatory response.
- Target cells involved in inflammation engulf prostaglandins via phagocytosis, which triggers the release of inflammatory mediators.
- Prostaglandins directly enter target cells through lipid channels and activate cytoplasmic protein kinases by direct phosphorylation.
- Prostaglandins integrate themselves into the target cell’s plasma membrane, significantly changing its fluidity and directly causing inflammatory changes.
Reasoning for Correct Answer #2: Prostaglandins function as local hormones or paracrine signalling molecules. They are synthesized and released by cells, then act on nearby target cells by binding to specific receptors, typically G-protein coupled receptors (GPCRs), located on the target cell’s surface membrane. This ligand-receptor binding activates intracellular signalling cascades (signal transduction), involving second messengers like cAMP or changes in ion concentrations, which ultimately alter cell behaviour to produce the inflammatory response (e.g., vasodilation, pain signalling, immune cell recruitment). They generally do not act on intracellular receptors like steroid hormones (#1).
Identify the white blood cell described as: A large cell with a bean-shaped nucleus that can develop into a macrophage.
Click the CORRECT answer:
- Lymphocyte
- Neutrophil
- Eosinophil
- Monocyte
- Basophil
Reasoning for Correct Answer #4: Monocytes are the largest type of circulating white blood cells. They are characterized by their abundant cytoplasm (often greyish-blue) and a large nucleus that is typically indented, kidney-shaped, or bean-shaped, rather than segmented (like neutrophils) or perfectly round (like lymphocytes). A key feature of monocytes is their ability to migrate from the bloodstream into tissues, where they differentiate into long-lived, highly phagocytic macrophages.
Identify the white blood cell described as: A cell with a large spherical nucleus and little cytoplasm, responding to non-self antigens.
Click the CORRECT answer:
- Monocyte
- Lymphocyte
- Neutrophil
- Platelet
- Eosinophil
Reasoning for Correct Answer #2: Lymphocytes are the primary cells of the adaptive immune system, responsible for recognizing specific non-self antigens. Morphologically, they typically appear as cells with a large, dense, round or slightly indented nucleus that occupies most of the cell volume, leaving only a thin rim of light blue cytoplasm. Monocytes are larger with more cytoplasm and a bean-shaped nucleus. Neutrophils have multi-lobed nuclei. Platelets are cell fragments. Eosinophils have bi-lobed nuclei and prominent granules.
Identify the white blood cell described as: A cell with a lobed nucleus that is phagocytic.
Click the CORRECT answer:
- Lymphocyte
- Basophil
- Neutrophil
- Monocyte
- Red blood cell
Reasoning for Correct Answer #3: Neutrophils are characterized by their distinctive multi-lobed nucleus (usually 3-5 lobes connected by thin strands) and cytoplasm containing fine granules. They are highly phagocytic cells, acting as crucial first responders in the innate immune system, particularly against bacterial infections. Monocytes (#4) are also phagocytic but typically have a single, unlobed (though often indented) nucleus. Lymphocytes (#1) have large, round nuclei and are not primarily phagocytic (except for some specialized types). Basophils (#2) have lobed nuclei but are mainly involved in allergic reactions. Red blood cells (#5) lack nuclei and are involved in oxygen transport.
Camel red blood cells are elliptical, whereas human red blood cells are biconcave discs. Dromedary camels live in hot, dry deserts and their blood can become viscous due to dehydration. Suggest how the elliptical shape of camel red blood cells might be an adaptation to this environment.
Click the CORRECT answer:
- The elliptical shape maximizes surface area for oxygen uptake in dry air.
- The elliptical shape allows easier blood flow, especially when blood viscosity increases due to dehydration.
- The elliptical shape helps cells store more water internally.
- The elliptical shape prevents the cells from binding oxygen too tightly.
- The elliptical shape allows cells to divide more rapidly when needed.
Reasoning for Correct Answer #2: Camels are adapted to survive dehydration. When dehydrated, blood plasma volume decreases, increasing blood viscosity (thickness) and haematocrit (proportion of red blood cells). The flexible, elliptical shape of camel RBCs is thought to be advantageous under these conditions because it allows them to deform and align more easily to flow through narrow capillaries and resist osmotic stress better than biconcave discs, thus helping to maintain blood circulation even when the blood is highly viscous. The biconcave shape of human RBCs maximizes surface area for gas exchange, but the elliptical shape appears beneficial for flow under dehydration stress.
Oxygen dissociation curves show that llama haemoglobin has a higher oxygen saturation than human haemoglobin at the low oxygen partial pressure typical of high altitudes (e.g., 6.4 kPa). Explain how this demonstrates that llamas are better adapted to high altitudes.
Click the CORRECT answer:
- Llama haemoglobin releases oxygen more readily at low partial pressures, delivering it faster to tissues.
- Llama haemoglobin binds oxygen less tightly, allowing quicker loading in the lungs.
- Llama haemoglobin has a higher affinity for oxygen, allowing it to load more oxygen efficiently in the lungs where oxygen pressure is low.
- Human haemoglobin carries more oxygen per molecule at all partial pressures.
- The higher saturation means llama blood becomes fully saturated at lower altitudes than human blood.
Reasoning for Correct Answer #3: At high altitudes, the partial pressure of oxygen (PO₂) in the inhaled air and thus in the lungs is significantly lower than at sea level. An oxygen dissociation curve shifted to the left, showing higher saturation at a given low PO₂, indicates a higher affinity of haemoglobin for oxygen. Llama haemoglobin exhibits this left-shifted curve compared to human haemoglobin. This higher affinity is adaptive because it allows the llama’s blood to bind and become adequately saturated with oxygen even when the oxygen pressure in the lungs is low, ensuring sufficient oxygen uptake for transport to the tissues. Releasing oxygen readily (#1) or binding less tightly (#2) would be counterproductive for loading oxygen in low-PO₂ environments.
An increase in carbon dioxide concentration shifts the oxygen dissociation curve for human haemoglobin. Describe the direction of this shift and its significance.
Click the CORRECT answer:
- The curve shifts left, increasing oxygen affinity, helping loading in the lungs.
- The curve shifts right, decreasing oxygen affinity, facilitating oxygen release in tissues.
- The curve shifts left, decreasing oxygen affinity, hindering oxygen loading.
- The curve shifts right, increasing oxygen affinity, enhancing oxygen uptake by tissues.
- The curve shifts vertically upwards, increasing the total oxygen carrying capacity.
Reasoning for Correct Answer #2: An increase in carbon dioxide concentration in the blood leads to a decrease in pH (increased acidity due to formation of carbonic acid). Both increased CO₂ and decreased pH reduce haemoglobin’s affinity for oxygen. This decreased affinity is represented graphically as a shift of the oxygen dissociation curve to the right (the Bohr shift or Bohr effect). The physiological significance of this rightward shift is that haemoglobin releases oxygen more readily at the lower PO₂ and higher CO₂/lower pH conditions found in metabolically active tissues, thus facilitating efficient oxygen delivery where it is most needed.
Explain the importance of the Bohr shift (the effect of increased carbon dioxide/lower pH on oxygen dissociation) in metabolically active organs.
Click the CORRECT answer:
- It increases haemoglobin’s oxygen affinity, ensuring maximum oxygen is carried away from active tissues back to the lungs.
- It decreases haemoglobin’s oxygen affinity, promoting oxygen release to meet the high respiratory demand of active tissues.
- It helps haemoglobin bind carbon dioxide more effectively for transport away from active tissues.
- It triggers the production of more red blood cells specifically in active tissues to increase oxygen supply.
- It shifts the curve left, allowing haemoglobin to pick up waste products other than CO₂ more easily.
Reasoning for Correct Answer #2: Metabolically active tissues (like exercising muscle) produce large amounts of CO₂ and lactic acid (if anaerobic respiration occurs), leading to a localized increase in PCO₂ and decrease in pH. The Bohr shift describes the phenomenon where these conditions reduce haemoglobin’s affinity for oxygen (shifting the dissociation curve to the right). This is physiologically advantageous because it means haemoglobin unloads its oxygen more readily precisely in these active tissues, where the oxygen demand for cellular respiration is highest. It ensures efficient oxygen delivery to support the tissue’s metabolic needs.
In a typical monocotyledon root cross-section, identify the tissues typically found at these locations within the central vascular cylinder (stele): immediately inside the Casparian strip; the large-diameter water-conducting vessels; the sugar-conducting tissue often found between arms of the water-conducting tissue.
Click the CORRECT answer:
- Inside Casparian strip: Phloem; Water-conducting: Xylem; Sugar-conducting: Endodermis
- Inside Casparian strip: Endodermis; Water-conducting: Xylem; Sugar-conducting: Phloem
- Inside Casparian strip: Cortex; Water-conducting: Phloem; Sugar-conducting: Xylem
- Inside Casparian strip: Pericycle; Water-conducting: Phloem; Sugar-conducting: Xylem
- Inside Casparian strip: Epidermis; Water-conducting: Xylem; Sugar-conducting: Phloem
Reasoning for Correct Answer #2: The Casparian strip is within the walls of the endodermal cells. Therefore, the layer immediately inside the strip (functionally speaking, controlling passage into the stele) is the endodermis itself. The large-diameter water-conducting vessels are xylem elements. The sugar-conducting tissue is phloem. In monocot roots, xylem and phloem typically alternate in a ring around a central pith. Phloem is found between the xylem ‘arms’ or bundles. Thus, Endodermis (inside strip boundary), Xylem (water), Phloem (sugar) is the correct sequence of identification based on the functional descriptions. (Note: Pericycle is structurally just inside the endodermis, but endodermis defines the boundary).
Outline the role of the endodermis in a plant root.
Click the CORRECT answer:
- It increases the surface area for water absorption from the soil.
- It actively transports sugars from the phloem into the surrounding cortex cells for storage.
- It contains dividing cells that generate the pressure required for root growth through the soil.
- It forces water and dissolved minerals entering the vascular tissue (stele) to pass through the selectively permeable cell membranes (symplast pathway) due to the presence of the waterproof Casparian strip.
- It primarily functions as a storage tissue for starch reserves absorbed from the cortex.
Reasoning for Correct Answer #4: The key feature of the root endodermis is the Casparian strip, a band of suberin (a waterproof substance) impregnating the radial and transverse cell walls. This strip blocks the apoplast pathway (movement through cell walls and intercellular spaces). Consequently, any water and dissolved mineral ions moving inwards from the cortex must cross the selectively permeable plasma membrane of an endodermal cell to enter its cytoplasm (joining the symplast pathway) before they can pass into the vascular cylinder (xylem). This ensures selective uptake of minerals into the vascular tissue.
State an example of an organic compound translocated (transported long-distance) in the phloem tissue of a plant root.
Click the CORRECT answer:
- Starch
- Cellulose
- Sucrose
- Lignin
- Mineral ions (e.g., nitrate)
Reasoning for Correct Answer #3: The primary function of phloem is the long-distance transport (translocation) of sugars produced during photosynthesis from source tissues (usually leaves) to sink tissues (like roots, fruits, growing points). The main sugar transported in the phloem sap of most plants is the disaccharide sucrose. Starch (#1) is a storage polysaccharide, not typically transported long distances. Cellulose (#2) and lignin (#4) are structural components of cell walls. Mineral ions (#5) are mainly transported upwards from the roots in the xylem.
What is the name of the narrow channels that pass through plant cell walls, connecting the cytoplasm of adjacent cells and allowing communication and transport?
Click the CORRECT answer:
- Plasmodesmata
- Stomata
- Lenticels
- Pits
- Middle lamella
Reasoning for Correct Answer #1: Plasmodesmata are membrane-lined channels that span the cell walls connecting the cytoplasm of adjacent plant cells. They create a continuous cytoplasmic network throughout the plant tissue, known as the symplast, allowing for direct cell-to-cell transport of water, small solutes, signalling molecules, and even some macromolecules. Stomata (#2) are pores on leaf surfaces for gas exchange. Lenticels (#3) are pores on stems for gas exchange. Pits (#4) are thin areas in secondary walls, often facilitating water movement between xylem cells. The middle lamella (#5) is the outermost layer cementing adjacent cell walls together.
Complete the following statements about common polysaccharides: Amylose consists of α-glucose monomers joined by ______ glycosidic bonds. Cellulose consists of β-glucose monomers joined by ______ glycosidic bonds. Glycogen functions as energy storage primarily in ______.
Click the CORRECT answer:
- Amylose: 1,6 bonds; Cellulose: 1,4 bonds; Glycogen function: Plants
- Amylose: 1,4 bonds; Cellulose: 1,6 bonds; Glycogen function: Animals
- Amylose: 1,4 bonds; Cellulose: 1,4 bonds; Glycogen function: Animals
- Amylose: 1,4 bonds; Cellulose: 1,4 bonds; Glycogen function: Bacteria
- Amylose: 1,6 bonds; Cellulose: 1,6 bonds; Glycogen function: Plants
Reasoning for Correct Answer #3:
Amylose: A component of starch, it’s a linear chain of α-glucose joined by α-1,4 glycosidic bonds.
Cellulose: A structural polysaccharide in plants, it’s a linear chain of β-glucose joined by β-1,4 glycosidic bonds.
Glycogen: The main glucose storage polysaccharide in animals (and fungi), it consists of α-glucose joined by α-1,4 bonds, with frequent α-1,6 branch points.
Therefore, the correct completions are: Amylose (1,4), Cellulose (1,4), Glycogen (Animals).
A DNA template strand has the sequence CAC TAC TCC AAC. What is the sequence of the primary transcript (mRNA) produced from this template?
Click the CORRECT answer:
- G UG AUG AGG UUG
- C AC UAC UCC AAC
- G TG ATG AGG TTG
- G UG UAC UCC UUG
- U GU AUG AGG AAC
Reasoning for Correct Answer #1: Transcription synthesizes mRNA using one DNA strand as a template. The base pairing rules are: DNA C pairs with RNA G, DNA G pairs with RNA C, DNA T pairs with RNA A, and DNA A pairs with RNA U (Uracil replaces Thymine in RNA).
Template DNA: 5′- CAC TAC TCC AAC -3′ (Assuming standard 5′ to 3′ reading, though template strand direction isn’t specified, the pairing works either way)
Let’s apply pairing rules:
C -> G
A -> U
C -> G
T -> A
A -> U
C -> G
T -> A
C -> G
C -> G
A -> U
A -> U
C -> G
Resulting mRNA: 3′- GUG AUG AGG UUG -5′. Conventionally, mRNA sequences are written 5′ to 3′. While the pairing gives the sequence, option #1 presents it directly as GUG AUG AGG UUG. Let’s recheck assuming the template is read 3′ to 5′:
Template DNA: 3′- CAC TAC TCC AAC -5′
mRNA synthesized: 5′- GUG AUG AGG UUG -3′
Option 1 matches this 5′-3′ sequence.
The DNA template strand sequence encoding four amino acids (aa1 to aa4) begins CAC TAC TCC AAC. Using a standard genetic code table where DNA triplets CAC, CAG, CAT, CAA all code for Val; TAC codes for Met; TCC, TCA, TCG, TCT, AGC, AGT all code for Ser; and AAC, AAT code for Asn, identify amino acids aa1, aa2, aa3, and aa4. (Simplified code provided for this question)
Click the CORRECT answer:
- aa1: Val, aa2: Met, aa3: Ser, aa4: Ser
- aa1: Val, aa2: Met, aa3: Ser, aa4: Asn
- aa1: Gln, aa2: Tyr, aa3: Arg, aa4: Leu
- aa1: Val, aa2: Stop, aa3: Ser, aa4: Asn
- aa1: His, aa2: Tyr, aa3: Pro, aa4: Asn
Reasoning for Correct Answer #2: The question provides the DNA *template* strand and a simplified code relating DNA template triplets directly to amino acids.
Triplet 1 on template: CAC -> codes for Val (aa1 = Val)
Triplet 2 on template: TAC -> codes for Met (aa2 = Met)
Triplet 3 on template: TCC -> codes for Ser (aa3 = Ser)
Triplet 4 on template: AAC -> codes for Asn (aa4 = Asn)
Therefore, the amino acid sequence is Val-Met-Ser-Asn.
Consider the DNA template sequence starting CAC TAC… If the first C (at position 1) were substituted, explain the likely effect on the protein, given that DNA triplets CAC, CAA, CAG, CAT all code for Valine (Val).
Click the CORRECT answer:
- A frameshift mutation would occur, changing all subsequent amino acids.
- A nonsense mutation would occur, causing premature termination.
- No effect would occur, as the first base is irrelevant to coding.
- No effect would occur, because the resulting codon would still code for Valine due to degeneracy.
- A different amino acid would be coded, likely altering protein function significantly.
Reasoning for Correct Answer #4: The original prompt likely intended to ask about substituting the *third* base (C at position 3) of the CAC triplet, as explained previously. If we assume that was the intent, substituting the C at position 3 could result in CAA, CAG, or CAT. The question explicitly states that *all* these triplets, along with the original CAC, code for Valine. Therefore, this substitution is a silent mutation; it changes the DNA and corresponding mRNA codon, but due to the degeneracy of the genetic code, the same amino acid (Valine) is incorporated. Thus, there would be no effect on the protein sequence. (If the question *strictly* meant position 1, substituting C could lead to AAC, GAC, TAC etc., which *would* change the amino acid, making #5 correct in that strict interpretation, but inconsistent with the provided context about degeneracy for Val). Based on context, #4 is the intended answer logic.
Consider the DNA template sequence starting CAC TAC… Explain the effect of deleting the nucleotide at position 3 (the second C).
Click the CORRECT answer:
- The protein sequence would remain unchanged due to the genetic code’s degeneracy.
- Only the first amino acid (coded by the first triplet) would be changed.
- It causes a frameshift mutation, altering the reading frame and likely the entire downstream amino acid sequence.
- It introduces a silent mutation, substituting one amino acid for a similar one.
- The DNA polymerase would skip that position, leaving a gap but maintaining the reading frame.
Reasoning for Correct Answer #3: Deleting a single nucleotide (or any number not divisible by three) within the coding sequence of a gene causes a frameshift mutation. The genetic code is read in non-overlapping triplets (codons). Removing one base shifts the grouping of all subsequent bases into different codons. Original template: CAC | TAC | TCC | AAC … -> codes Val-Met-Ser-Asn… After deleting C at pos 3: CAT | ACT | CCA | AC… -> codes Val-?-?-?… The reading frame is shifted, resulting in different codons being read from that point onwards, usually leading to a completely altered amino acid sequence and often a premature stop codon, producing a non-functional protein.
During which stage(s) of the mitotic cell cycle does nuclear DNA replication occur?
Click the CORRECT answer:
- Prophase only
- S phase only
- Anaphase and Telophase
- Interphase (specifically S phase)
- G1 phase and G2 phase
Reasoning for Correct Answer #4: The cell cycle is broadly divided into Interphase and M phase (mitosis + cytokinesis). Interphase is further divided into G1 (Gap 1), S (Synthesis), and G2 (Gap 2). Nuclear DNA replication, the process where the cell duplicates its chromosomes, occurs specifically during the S phase of Interphase. G1 and G2 are growth phases, and M phase involves the segregation of the duplicated chromosomes.
Outline how DNA is replicated inside the nucleus prior to cell division.
Click the CORRECT answer:
- The enzyme helicase unwinds the DNA, DNA polymerase synthesizes new strands continuously using free RNA nucleotides, and ligase joins fragments.
- DNA polymerase unwinds the DNA, helicase synthesizes new strands using one old strand as a template, and ligase proofreads the sequence.
- Helicase unwinds the DNA double helix, DNA polymerase adds complementary DNA nucleotides to both template strands (synthesizing one continuously – leading strand, and the other discontinuously in fragments – lagging strand), and DNA ligase joins the Okazaki fragments on the lagging strand.
- The entire DNA molecule is copied base-by-base by ribosomes using mRNA as a template, followed by proofreading by ligase.
- Topoisomerase cuts the DNA, DNA polymerase replicates it, and helicase rewinds the new molecules.
Reasoning for Correct Answer #3: DNA replication follows a semi-conservative model involving several key enzymes. Helicase unwinds the parental DNA double helix, creating replication forks. DNA polymerase synthesizes new strands by adding complementary DNA nucleotides (A-T, G-C) using the parental strands as templates. Due to the antiparallel nature of DNA and the 5′-to-3′ directionality of DNA polymerase, one new strand (leading strand) is synthesized continuously, while the other (lagging strand) is synthesized discontinuously in short pieces called Okazaki fragments. Finally, DNA ligase joins these Okazaki fragments together to create a continuous lagging strand.
Adenosine triphosphate (ATP) consists of an adenine base, a sugar molecule, and three phosphate groups. Identify the sugar molecule found in ATP.
Click the CORRECT answer:
- Deoxyribose
- Glucose
- Ribose
- Sucrose
- Fructose
Reasoning for Correct Answer #3: ATP (Adenosine Triphosphate) is a ribonucleotide. The ‘Adenosine’ part refers to the nucleoside formed by the nitrogenous base Adenine linked to a five-carbon sugar. In ATP, as in RNA, this sugar is Ribose. Deoxyribose (#1) is the sugar found in DNA (forming deoxyadenosine triphosphate, dATP, before incorporation). Glucose (#2), Sucrose (#4), and Fructose (#5) are common carbohydrates involved in energy metabolism but are not the sugar component directly within the ATP molecule.
An electron micrograph shows a bacterium with a curved rod shape and a visible flagellum, producing a 3D-like image of its surface. Which type of electron microscope was likely used?
Click the CORRECT answer:
- Transmission Electron Microscope (TEM) – Negative Stain
- Scanning Electron Microscope (SEM)
- Light Microscope – Phase Contrast
- Transmission Electron Microscope (TEM) – Thin Section
- Light Microscope – Fluorescence
Reasoning for Correct Answer #2: Scanning Electron Microscopy (SEM) is specifically designed to image the surface of specimens. It works by scanning a focused beam of electrons across the surface and detecting secondary or backscattered electrons emitted from the specimen. This process generates detailed images of the surface topography, often giving a striking 3D-like appearance, which matches the description of the bacterium image showing surface features like shape and flagellum. Transmission Electron Microscopy (TEM) passes electrons *through* a thin sample to reveal internal structures (#1, #4). Light microscopy (#3, #5) has much lower resolution and magnification capabilities.
Name the species of prokaryote, typically curved or comma-shaped, that causes cholera.
Click the CORRECT answer:
- Escherichia coli
- Mycobacterium tuberculosis
- Staphylococcus aureus
- Vibrio cholerae
- Plasmodium falciparum
Reasoning for Correct Answer #4: Cholera is an acute diarrhoeal illness caused by infection of the intestine with the bacterium Vibrio cholerae. These bacteria are characteristically Gram-negative and have a curved or comma-shaped rod morphology. E. coli (#1) and Staphylococcus aureus (#3) are other bacteria causing different types of infections. Mycobacterium tuberculosis (#2) causes tuberculosis. Plasmodium falciparum (#5) is a eukaryotic protist that causes malaria.
Identify the factual error in the following description of a typical prokaryotic cell: “Prokaryotic cells have cell surface membranes, 70S ribosomes, and a cellulose cell wall. Their DNA is circular and located free in the cytoplasm in a region called the nucleoid.”
Click the CORRECT answer:
- DNA is linear, not circular.
- Ribosomes are 80S, not 70S.
- The cell wall is made of cellulose.
- DNA is located within a nucleus.
- They lack a cell surface membrane.
Reasoning for Correct Answer #3: The description contains several correct features of typical prokaryotes: cell surface membrane present (#5 incorrect), 70S ribosomes (#2 incorrect), circular DNA (#1 incorrect), DNA in nucleoid region (not nucleus, #4 incorrect). The factual error lies in the composition of the cell wall. While prokaryotes typically have a cell wall, it is composed primarily of peptidoglycan (in bacteria) or other polymers (in archaea), not cellulose. Cellulose is the main structural component of plant cell walls.
In diagrams showing HIV attaching to a T-helper cell, what is the name of the protein structure immediately surrounding the viral RNA and enzymes, located inside the viral envelope?
Click the CORRECT answer:
- Envelope glycoprotein
- Reverse transcriptase
- Capsid
- Matrix protein
- Integrase
Reasoning for Correct Answer #3: The core structure of the Human Immunodeficiency Virus (HIV) particle, containing the viral genetic material (two strands of RNA) and essential enzymes (like reverse transcriptase #2, integrase #5, protease), is enclosed within a protein shell called the capsid (specifically, a conical capsid in HIV). This capsid is located inside the viral envelope. Between the capsid and the envelope is the matrix protein layer (#4). The envelope glycoproteins (#1) are embedded in the outer lipid envelope and mediate attachment and entry into host cells.
Explain how the destruction of T-helper cells by HIV affects the immune system’s ability to resist pathogens.
Click the CORRECT answer:
- It leads to over-activation of B cells, causing widespread autoimmune diseases.
- It enhances the activity of cytotoxic T cells, resulting in an overly aggressive cell-mediated response.
- It primarily affects the innate immune system (e.g., phagocytes), leaving the adaptive immune response intact.
- It causes the remaining immune cells to directly destroy pathogens more effectively, compensating for the loss.
- It severely impairs both humoral (antibody-mediated) and cell-mediated immunity by reducing the activation signals (cytokines and co-stimulation) needed for effective B cell and cytotoxic T cell responses.
Reasoning for Correct Answer #5: T-helper lymphocytes (CD4+ T cells) are central regulators of the adaptive immune response. They are essential for activating B cells to produce antibodies (humoral immunity) and for activating cytotoxic T cells (CD8+ T cells) to kill infected cells (cell-mediated immunity). They do this through direct cell contact and the secretion of crucial signaling molecules called cytokines. HIV preferentially infects and destroys T-helper cells. Their depletion leads to a severe deficiency in both branches of adaptive immunity, making the individual highly susceptible to opportunistic infections and certain cancers, which defines Acquired Immunodeficiency Syndrome (AIDS).
Some individuals are resistant to HIV infection due to a mutation in the gene for the CCR5 protein, which normally acts as a co-receptor for HIV entry into T-helper cells. Suggest how this mutation provides protection.
Click the CORRECT answer:
- The mutation causes T-helper cells to actively destroy the HIV particle upon contact.
- The mutation prevents the T-helper cell from expressing any surface proteins, effectively hiding it from HIV.
- The mutation increases the production of antibodies specifically against the CCR5 protein, blocking HIV binding.
- The mutation alters the shape or prevents the surface expression of the CCR5 protein so that HIV cannot effectively bind to it, thereby blocking viral entry into the cell.
- The mutation allows the CCR5 protein to bind HIV normally but introduces a block that prevents the virus from replicating once inside the cell.
Reasoning for Correct Answer #4: Most strains of HIV require two receptors on the surface of T-helper cells to gain entry: the primary CD4 receptor and a co-receptor, usually CCR5 or CXCR4. The common resistance-conferring mutation, known as CCR5-delta32, is a deletion in the CCR5 gene. Individuals homozygous for this mutation produce a truncated, non-functional CCR5 protein that is not expressed on the cell surface. Without the CCR5 co-receptor present, CCR5-tropic strains of HIV (the most common type for initial infection) cannot bind effectively and complete the fusion process necessary to enter the T-helper cell, thus providing strong resistance to infection.
Outline the key steps in producing monoclonal antibodies against the CCR5 protein using the hybridoma method.
Click the CORRECT answer:
- Inject CCR5 into myeloma cells -> Fuse myeloma cells with T-helper cells -> Select fused cells -> Culture.
- Inject CCR5 protein into a mouse -> Extract antibody-producing plasma cells (B cells) from spleen -> Fuse plasma cells with immortal myeloma cells -> Select and clone hybridoma cells producing anti-CCR5 antibodies -> Culture clones.
- Culture T-helper cells -> Extract CCR5 protein from them -> Inject protein directly into isolated plasma cells -> Culture these plasma cells.
- Inject mouse with pre-made anti-CCR5 antibodies -> Extract myeloma cells -> Fuse myeloma cells with spleen cells -> Select hybridomas.
- Fuse mouse spleen cells randomly with myeloma cells -> Inject the resulting mixture of fused cells with CCR5 protein -> Select cells that start producing anti-CCR5 antibodies.
Reasoning for Correct Answer #2: The standard hybridoma technique involves:
1. Immunization: Injecting an animal (e.g., a mouse) with the antigen of interest (CCR5 protein) to elicit an immune response.
2. Isolation of B cells: Harvesting B lymphocytes (specifically, antibody-producing plasma cells) from the immunized animal’s spleen or lymph nodes.
3. Fusion: Fusing these B cells with immortal myeloma cells (cancerous plasma cells) using agents like PEG to create hybrid cells (hybridomas).
4. Selection: Growing the fused cells in a selective medium (like HAT medium) that allows only the hybridomas to survive (unfused myeloma cells die, unfused B cells have limited lifespan).
5. Screening and Cloning: Identifying and isolating hybridoma clones that produce the desired monoclonal antibody (anti-CCR5) and culturing them indefinitely.
Option 2 accurately outlines these key steps.