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Sindh Mdcat Exclusive Course Chemistry Biochemistry — Solved Past Paper with Answers
All 20 MCQs from Sindh Mdcat Exclusive Course Chemistry Biochemistry, solved with the correct answer highlighted and a full explanation for every question. This is a free MDCAT Sindh / DUHS past paper — no signup, no ads. Practise it interactively in timed mode, drill more with free MDCAT MCQs, or browse all Sindh / DUHS papers.
Attempt this paper interactively →Q1. Which of the following bond is responsible for joining the amino acids in proteins?
- A. Peptide Bond✓
- B. Ionic Bond
- C. Metallic Bond
- D. Disulfide Bond
Explanation: Option A is correct, as within a protein chain, each amino acid is linked to each other by a peptide bond which is a covalent bond.Options B and D describe those bonds that hold proteins in their complex three dimensional structures, but the question asks for the bond that holds individual amino acids within proteins.Option C describes a metallic bond which is only found in metals, and not in proteins as it is molecular in nature.
Why the other options are wrong
- B. Options B and D describe those bonds that hold proteins in their complex three dimensional structures, but the question asks for the bond that holds individual amino acids within proteins.
- C. Option C describes a metallic bond which is only found in metals, and not in proteins as it is molecular in nature.
- D. Options B and D describe those bonds that hold proteins in their complex three dimensional structures, but the question asks for the bond that holds individual amino acids within proteins.
Q2. Enzymes that are functioning within the cell are called:
- A. Endoenzymes✓
- B. Exoenzymes
- C. Holoenzymes
- D. Both A & C
Explanation: Enzymes that function within the cell are called endoenzymes, out of the cell are called exoenzymes. Holoenzyme is the biochemically active part of an enzyme.
Why the other options are wrong
- B. Enzymes that function witout of the cell are called exoenzyme
- C. Holoenzyme is the biochemically active part of an enzyme.
- D. This option is incorrect.
Q3. Monosaccharides contain _ carbon atoms.
- A. 2-3
- B. 3-10✓
- C. 5-20
- D. 20-25
- E. Only 5
Explanation: Monosaccharides can contain carbon atoms ranging from 3-10.
Why the other options are wrong
- A. As explained above, this option is incorrect.
- C. As explained above, this option is incorrect.
- D. As explained above, this option is incorrect.
- E. As explained above, this option is incorrect.
Q4. Which of the following is correct about Ascorbic acid?
- A. Soluble in water
- B. Easily destroyed by oxidation
- C. Its deficiency causes anemia
- D. It helps in healing the wounds
- E. All of these answers✓
Explanation: All the options are correct.
Why the other options are wrong
- A. This statement is correct. Ascorbic acid, also known as vitamin C, is highly soluble in water. This property allows it to be easily absorbed and transported throughout the body.
- B. This statement is correct. Ascorbic acid is a highly sensitive molecule that can be easily destroyed by oxidation. Exposure to air, heat, light, and certain cooking methods can lead to the degradation of ascorbic acid, resulting in a loss of its nutritional value.
- C. Vitamin C deficiency has been associated with various forms of anemia, but it is still unclear whether vitamin C (ascorbic acid) is directly involved in hematopoiesis or if anemia arises indirectly through the interactions of vitamin C with folate and iron metabolism.
- D. This statement is correct. Ascorbic acid plays a vital role in collagen synthesis, a protein necessary for wound healing. It supports the formation of connective tissue, aids in tissue repair, and enhances the body's ability to heal wounds
Q5. The normal pH of blood is:
- A. 7.05
- B. 7.75
- C. 7.35✓
- D. 7.25
Explanation: Our blood has a normal pH range of 7.35 to 7.45. This means that blood is naturally slightly alkaline or basic.
Why the other options are wrong
- A. Incorrect option, as it is very acidic for the bloods normal pH.
- B. Incorrect option, as it is basic for the bloods normal pH.
- D. Incorrect option, as it is acidic for the bloods normal pH.
Q6. Amylopectin has _ linkage.
- A. α 1 - 4 glycosidic
- B. β 1 - 4 and 1 - 6 glycosidic
- C. α 1-4 and 1 - 6 glycosidic✓
- D. β 1 - 4 glycosidic
Explanation: Amylopectin has both α 1-4 and 1 - 6 glycosidic bond linkages.
Why the other options are wrong
- A. As amylopectin has branching, it has α 1 - 6 glycosidic linkage along with α 1 - 4 glycosidic linkages.
- B. Amylopectin has chain and branching of α glucoses and not β glucoses . Hence, this option is wrong.
- D. Amylopectin has chain and branching of α glucoses and not β glucoses . Hence, this option is wrong.
Q7. Which one is not a monosaccharide?
- A. Galactose
- B. Fructose
- C. Mannose
- D. Lactose✓
Explanation: Lactose is not a monosaccharide but a disaccharide,as shown below:
Why the other options are wrong
- A. Galactose is a monosaccharide with 6 carbons.
- B. Fructose is a 6-C monosaccharide.
- C. Mannose is a 6-C monosaccharide.
Q8. Based on the physico-chemical properties, proteins may be classified into the following types:
- A. Simple proteins
- B. Compound proteins
- C. Derived proteins
- D. All of the above✓
Explanation: Proteins can be classified as: (a) Simple proteins. (b) Conjugated proteins. (c) Derived proteins.
Why the other options are wrong
- A. Simple proteins. On hydrolysis they yield only the amino acids and occasional small carbohydrate compounds. Examples are: albumins, globulins, glutelins, albuminoids, histones and protamines.
- B. These are simple proteins combined with some non-protein material in the body. Examples are: nucleoproteins, glycoproteins, phosphoproteins, hemoglobins and lecithoproteins.
- C. Derived proteins. These are proteins derived from simple or conjugated proteins by physical or chemical means. Examples are: denatured proteins and peptides.
Q9. L - Asparaginase enzyme has been used for the treatment of:
- A. Jaundice
- B. Blood cancer✓
- C. Rickets
- D. Heart disease
Explanation: L-asparaginase is an enzyme that has been used in the treatment of certain types of cancer, particularly acute lymphoblastic leukemia. L-asparaginase works by depleting levels of the amino acid asparagine in the blood. Cancer cells, unlike normal cells, are unable to synthesize their own asparagine and rely on external sources. By depleting asparagine, L-asparaginase interferes with the growth and survival of cancer cells, ultimately leading to their death.
Why the other options are wrong
- A. Jaundice can be a symptom of various underlying medical conditions, such as liver diseases, hepatitis, or obstruction of the bile ducts. When the liver is unable to process bilirubin properly or when there is an increased breakdown of red blood cells, bilirubin levels in the body can rise, leading to jaundice.
- C. Rickets is a condition that results from a deficiency of vitamin D, calcium, or phosphate, which can lead to weakened or soft bones and other symptoms.
- D. Heart disease is a broad term that encompasses a range of conditions affecting the heart. The most common type of heart disease is coronary artery disease (CAD), which can lead to heart attacks. CAD occurs when the blood vessels supplying the heart muscle (coronary arteries) become narrowed or blocked by a buildup of plaque (atherosclerosis). This can lead to chest pain (angina) or a heart attack.
Q10. Sucrose is a non reducing sugar because?
- A. It does not exist in open chain form
- B. Carbonyl groups are not free to reduce fehling's solution
- C. They have at least one free carbonyl group
- D. They form 1-4 glycosidic linkage
- E. Both a and b✓
Explanation: A sugar that serves as a reducing agent due to its free aldehyde or ketone functional groups in its molecular structure. Sucrose is a non-reducing sugar because the two monosaccharide units are held together by a glycosidic linkage between C1 of α-glucose and C2 of β-fructose. Sucrose has no aldehyde or ketone group to reduce Fehling's solution , so both options A and B are true. Hence the correct option is E
Why the other options are wrong
- A. Sucrose is a disaccharide composed of glucose and fructose units linked together by a glycosidic bond. In its cyclic form, both glucose and fructose are in a closed-ring structure, and the carbonyl groups are not free to undergo oxidation reactions. Hence, sucrose does not have a free aldehyde or ketone group (carbonyl group) that is required to participate in reducing reactions.
- B. It contains copper ions, which are reduced by the aldehyde or ketone groups of reducing sugars, causing the solution to change color from blue to red or yellow, indicating the formation of a copper oxide precipitate. Since sucrose lacks a free carbonyl group in its cyclic form, it cannot reduce Fehling's solution.
- C. They have at least one free carbonyl group: This statement is incorrect for sucrose. Unlike reducing sugars (e.g., glucose or fructose), sucrose does not have a free aldehyde or ketone group, as the carbonyl groups are involved in forming the glycosidic linkage between the glucose and fructose units.
- D. This statement is correct. Sucrose is formed by the condensation reaction between the anomeric carbon of glucose (C1) and the hydroxyl group at carbon 4 (C4) of fructose, resulting in the formation of a 1,4-glycosidic linkage.
Q11. A dipolar charged but overall electrically neutral ion is called:
- A. Zwitterion✓
- B. Unit electron
- C. Cation
- D. Anion
Explanation: A zwitterion is an ion that contains two functional groups. In simple terms, it is an ion possessing both positive and negative electrical charges. Therefore, zwitterions are mostly electrically neutral (the net formal charge is usually zero). So option A is the correct answer.
Why the other options are wrong
- B. A unit electron is single negative charge -1e.
- C. A cation is positive charge +1.
- D. An anion is unit negative charge -1. Since all these charges are single charges , so options B , C and D are incorrect.
Q12. Alanine, lysine and tryptophan are:
- A. Carbohydrates
- B. Lipids
- C. Enzymes
- D. Amino acids✓
Explanation: All the three named compounds(lysine, alanine and Tryptophan) are amino acids in nature. However, Alanine is neutral Alpha-amino acid. Lysine is a basic amino acid. Tryptophan is also a neutral amino acid.
Why the other options are wrong
- A. Alanine , lysine and tryptophan are amino acids not carbohydrates.
- B. Alanine , lysine and tryptophan are amino acids not lipids.
- C. Alanine , lysine and tryptophan are amino acids not enzymes.
Q13. The energy produced by 1g of carbohydrate is:
- A. 4.8 cal
- B. 4.2 kcal✓
- C. 9 joules
- D. 9.2 kcal
Explanation: Carbohydrates, proteins and fats in the foods provide energy for the body functions and physical activities. They, however, do not have same energy content. The following is the energy-content by respective nutrients:1 gram of carbohydrates- 4kcal1 gram of protein- 4 kcal1 gram of fats- 9 kcalSo, the correct answer is '4.2 kcal'
Why the other options are wrong
- A. According to the explanation given below b is the correct answer
- C. According to the explanation given below b is the correct answer
- D. According to the explanation given below b is the correct answer
Q14. Ferritin deposits iron in spleen. It is an example of:
- A. Structural protein
- B. Respiratory protein
- C. Storage protein✓
- D. Contractile protein
Explanation: Ferritin are actually the stores of iron that are utilized when the serum iron depletes.
Why the other options are wrong
- A. This option is incorrect
- B. This option is incorrect
- D. This option is incorrect
Q15. Starch is a polymer of:
- A. Fructose
- B. d-D- Glucose (d=alpha)✓
- C. B-D Glucose
- D. Sucrose
Explanation: Starch is constituted by α-D-glucose units, which can be organized to form two distinct molecules, amylose and amylopectin. Use the image below to help recall:
Why the other options are wrong
- A. Fructose is a monosaccharide, meaning it is a single sugar unit and not a polymer. It is a ketose sugar with a chemical formula of C6H12O6.
- C. B-D glucose is also a monosaccharide and a six-carbon sugar, but it is not used in the formation of starch. Instead, it is a component of cellulose, a structural polysaccharide found in plant cell walls.
- D. Sucrose is a disaccharide made up of glucose and fructose units linked by a glycosidic bond. It is not a polymer of glucose and is not used in the formation of starch.
Q16. Which of the following is NOT a fatty acid?
- A. Phenyl acetic acid✓
- B. Stearic acid
- C. Palmitic acid
- D. Oleic acid
Explanation: Fatty acids are a group of heterogeneous compounds that consist of a long aliphatic alkyl chain which may be saturated or unsaturated and a terminal carboxylic acid functional group.Phenyl acetic acid consists of an aromatic benzene ring bonded with the carboxylic acid group. Therefore, it does not qualify to be categorised as a fatty acid. Hence option A is correct.Option B,C and D are all examples of fatty acids and their structures conform to the definition of fatty acids as shown:
Why the other options are wrong
Q17. According to Watson and Crick’s model of DNA, the DNA molecule consists of a double helix. What type of forces are responsible to keep two strands of DNA together?
- A. Van der Waals forces
- B. Ionic bonding
- C. Hydrogen bonding✓
- D. Dipole-induced dipole forces
Explanation: The bases in each complementary base pair have hydrogen bonds between them, which keeps the two strands together and stable. 'A' and 'T' have 2 hydrogen bonds between them, while 'G' and 'C' have three hydrogen bonds between them. The answer is C, as the other options do not mention hydrogen bonding.
Why the other options are wrong
- A. Van der Waals forces are weak intermolecular forces that occur between atoms or molecules. These forces arise from temporary fluctuations in electron distribution, leading to the creation of temporary dipoles. There are three types of Van der Waals forces: dispersion forces, dipole-dipole interactions, and hydrogen bonding. Dispersion forces are present in all molecules and arise from the momentary imbalances in electron distribution. Dipole-dipole interactions occur between polar molecules with permanent dipoles. Hydrogen bonding is a specific type of dipole-dipole interaction that occurs when a hydrogen atom is bonded to an electronegative atom, such as oxygen, nitrogen, or fluorine.
- B. Ionic bonding is a type of chemical bonding that occurs between atoms with significantly different electronegativities. It involves the transfer of electrons from one atom to another, resulting in the formation of ions. One atom, typically a metal, loses electrons to become a positively charged ion (cation), while the other atom, usually a nonmetal, gains those electrons to become a negatively charged ion (anion).
- D. Dipole-induced dipole forces, also known as induced dipole forces or London dispersion forces, are a type of intermolecular force. They occur between molecules with a permanent dipole and molecules without a permanent dipole.When a polar molecule approaches a nonpolar molecule, the electric field of the polar molecule induces a temporary dipole in the nonpolar molecule. This induced dipole creates an attractive force between the two molecules.
Q18. Fats are a type of lipid called glycerides. They are esters of long chain carboxylic acids:
- A. Propene-1, 2, 3-triol
- B. Propane-1, 2, 3-triol✓
- C. Propene-1, 2, 3-diol
- D. Propane-1, 2, 3-diol
Explanation: The correct term for the compound is "Propane-1, 2, 3-triol."Fats, also known as triglycerides, are esters formed from glycerol (propane-1,2,3-triol) and three fatty acid molecules. Glycerol is a triol, meaning it has three hydroxyl (OH) groups. When esterified with fatty acids, these hydroxyl groups react with the carboxyl groups (-COOH) of the fatty acids, forming ester bonds. Therefore, the compound "Propane-1, 2, 3-triol" is the correct chemical name for glycerol, which is a component of fats.
Why the other options are wrong
- A. This option is incorrect.
- C. This option is incorrect.
- D. This option is incorrect.
Q19. Which one of the following bases is NOT present in RNA?
- A. Cytosine
- B. Adenine
- C. Thyamine✓
- D. Guanine
Explanation: The base "Thymine" (T) is not present in RNA. Instead, RNA contains the base "Uracil" (U). Thyamine is a nucleotide base found in DNA but not in RNA. In DNA, thymine pairs with adenine through two hydrogen bonds, forming a T-A base pair.
Why the other options are wrong
- A. Cytosine is one of the four nucleotide bases found in RNA and DNA. In RNA, cytosine pairs with guanine through three hydrogen bonds, forming a C-G base pair.
- B. Adenine is one of the four nucleotide bases found in both RNA and DNA. In RNA, adenine pairs with uracil through two hydrogen bonds, forming an A-U base pair. In DNA, adenine pairs with thymine, forming an A-T base pair.
- D. Guanine is one of the four nucleotide bases found in RNA and DNA. It pairs with cytosine through three hydrogen bonds in DNA, forming a G-C base pair.
Q20. The amino acids which largely exist in dipolar ionic form are:
- A. Acidic amino acids
- B. Basic amino acids
- C. Beta amino acids
- D. Alpha amino acids✓
Explanation: α-amino acids exist as dipolar ions or zwitter ions.
Why the other options are wrong
- A. As per the explanation, this option is incorrect.
- B. As per the explanation, this option is incorrect.
- C. As per the explanation, this option is incorrect.
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