Biochem FEEDs (2023)

Question 1

The enzyme HGPRT is found to be deficient. This will result in:
a. Increased synthesis of Krebs cycle intermediates
b. Activation of xanthine oxidase
c. Increased purine synthesis
d. Hyperuricemia

Answer 1

d. Hyperuricemia

Rationale: HGPRT (hypoxanthine-guanine phosphoribosyltransferase) is crucial for the salvage pathway of purines. A deficiency in HGPRT results in the accumulation of hypoxanthine and guanine, which are subsequently converted to uric acid, leading to hyperuricemia. This condition is associated with Lesch-Nyhan syndrome.

Question 2

Deoxyribonucleotides are formed by reduction of:
a. Ribonucleosides
b. Ribonucleoside monophosphates
c. Ribonucleoside diphosphates
d. Ribonucleoside triphosphates

Answer 2

c. Ribonucleoside diphosphates

Rationale: Deoxyribonucleotides are synthesized from ribonucleoside diphosphates by the enzyme ribonucleotide reductase, which reduces the 2’-hydroxyl group of the ribose ring to a hydrogen atom, producing deoxyribonucleotides.

Question 3

The action of allopurinol is best described as:
a. Competitive inhibition of xanthine oxidase
b. Inhibiting PRPP synthesis
c. Increasing the solubility of uric acid in plasma
d. Inhibiting leukocyte movement by affecting microtubule formation

Answer 3

a. Competitive inhibition of xanthine oxidase

Rationale: Allopurinol is a structural analog of hypoxanthine. It acts as a competitive inhibitor of xanthine oxidase, the enzyme responsible for converting hypoxanthine and xanthine to uric acid. This inhibition reduces uric acid production, helping to manage conditions like gout.

Question 4

Which of the following is not a cause of gout?
a. Inefficient purine base salvage reactions
b. Increased PRPP synthetase activity
c. Overproduction of pyrimidine bases
d. von Gierke’s disease

Answer 4

c. Overproduction of pyrimidine bases

Rationale: Gout is caused by the accumulation of uric acid, which is a product of purine metabolism, not pyrimidine metabolism. The overproduction of pyrimidine bases does not contribute to gout.

Question 5

Which of the following is absent in deoxyribonucleosides?
a. Phosphate group
b. 5-carbon sugar in pentose ring form
c. Purine or pyrimidine base
d. Glycosidic bond

Answer 5

a. Phosphate group

Rationale: Deoxyribonucleosides consist of a deoxyribose sugar attached to a purine or pyrimidine base via a glycosidic bond. They do not contain a phosphate group, which differentiates them from deoxyribonucleotides.

Question 6

True of nitrogenous bases:
a. In nucleotides, they are attached to the phosphate by a phosphodiester bond
b. A prime (‘) symbol is used to define the carbon atoms in its ring
c. The N-9 atom of purines is linked to carbon 1 of the sugar
d. Nitrogenous base pairs are linked together by N-glycosidic bonds

Answer 6

c. The N-9 atom of purines is linked to carbon 1 of the sugar

Rationale: In nucleotides, the N-9 nitrogen of purine bases (adenine and guanine) is linked to the 1’ carbon of the sugar (ribose or deoxyribose) via an N-glycosidic bond. The other options are incorrect because (a) nitrogenous bases are attached to the sugar, not the phosphate, by a glycosidic bond, (b) the prime (‘) symbol is used to define carbon atoms in the sugar, not in the nitrogenous base, and (d) base pairs are linked together by hydrogen bonds, not N-glycosidic bonds.

Question 7

PRPP acts as a substrate for:
a. Purine synthesis
b. Pyrimidine synthesis
c. Both Purine and pyrimidine synthesis
d. Neither purine nor pyrimidine synthesis

Answer 7

c. Both purine and pyrimidine synthesis

Rationale: PRPP (phosphoribosyl pyrophosphate) is a crucial substrate for the synthesis of both purine and pyrimidine nucleotides. It donates the ribose-phosphate group that is required for the formation of nucleotides.

Question 8

True of dihydrofolate reductase:
a. It is inhibited by methotrexate
b. It catalyzes an ATP-requiring reaction
c. Its absence results in an accumulation of thymidine nucleotides
d. All of the above are true

Answer 8

a. It is inhibited by methotrexate

Rationale: Dihydrofolate reductase is an enzyme that catalyzes the reduction of dihydrofolate to tetrahydrofolate, a cofactor required for the synthesis of thymidine nucleotides. Methotrexate is a competitive inhibitor of this enzyme. The other options are incorrect as the reaction catalyzed by dihydrofolate reductase does not require ATP, and its absence results in a deficiency of thymidine nucleotides, not an accumulation.

Question 9

Why are humans not affected by sulfonamides and trimethoprim?
a. Mammalian cells do not synthesize folate
b. Folate is not essential for DNA synthesis in humans
c. Humans can synthesize folate from PABA unlike microorganisms
d. Humans have enzymes that can digest sulfonamides and trimethoprim and render them inactive

Answer 9

a. Mammalian cells do not synthesize folate

Rationale: Sulfonamides and trimethoprim target bacterial folate synthesis. Humans obtain folate through their diet and do not synthesize it, rendering these drugs ineffective against human cells. Folate is essential for DNA synthesis in humans (b), humans cannot synthesize folate from PABA (c), and humans do not have enzymes that digest sulfonamides and trimethoprim (d).

Question 10

A 50-pair DNA chain has 30 bases made up of thymine residues. How many guanine bases would you expect to see in this chain?
a. 10
b. 20
c. 30
d. 40

Answer 10

b. 20

Rationale: In a 50-pair DNA chain, there are 100 bases in total. Given that there are 30 thymine residues, there must be 30 adenine residues (as thymine pairs with adenine). This leaves 40 bases that must be cytosine and guanine. Since cytosine and guanine pair together in equal numbers, there must be 20 cytosine and 20 guanine bases.

Question 11

Which of the following is an intracellular messenger?
a. cGMP
b. 6-mercaptopurine
c. ATP
d. UDP-glucose

Answer 11

a. cGMP

Rationale: cGMP (cyclic guanosine monophosphate) is a second messenger important in many biological processes. It is used for signal transduction in cells, particularly in processes such as vasodilation. The other options, such as 6-mercaptopurine, ATP, and UDP-glucose, have roles in metabolism and other cellular processes but are not typically classified as intracellular messengers.

Question 12

What is the rate-limiting enzyme of pyrimidine synthesis?
a. PRPP synthetase
b. Carbamoyl phosphatase
c. Aspartate transcarbamoylase
d. Adenosine deaminase

Answer 12

b. Carbamoyl phosphatase

Rationale: The correct enzyme is actually carbamoyl phosphate synthetase II (CPS II), which catalyzes the formation of carbamoyl phosphate from glutamine and bicarbonate. This enzyme is the rate-limiting step in pyrimidine synthesis. The other enzymes listed are involved in different pathways.

Question 13

The origin of replication is an area in the DNA strand where replication begins. Which statement regarding the origin of replication is true?
a. Prokaryotes have multiple origins of replication in one DNA strand
b. When the 2 strands separate from the origin of replication, a single replication fork is formed
c. The origin of replication is made up mostly of guanine-cytosine base pairs
d. The separation or melting of the strands is an ATP-dependent process

Answer 13

d. The separation or melting of the strands is an ATP-dependent process

Rationale: The process of strand separation at the origin of replication requires energy, which is provided by ATP hydrolysis. Prokaryotes typically have a single origin of replication per DNA molecule (a), two replication forks form at the origin (b), and the origin of replication is often rich in adenine-thymine base pairs, not guanine-cytosine (c), because A-T pairs are easier to separate.

Question 14

True of the lagging strand:
a. Chain elongation occurs in the 3’→ 5’ direction
b. It is the strand that is being copied in the direction of the advancing replication fork
c. It is synthesized continuously as the replication fork advances
d. The enzyme responsible for joining its strands is DNA ligase

Answer 14

d. The enzyme responsible for joining its strands is DNA ligase

Rationale: The lagging strand is synthesized discontinuously in short fragments called Okazaki fragments. These fragments are later joined together by DNA ligase. Chain elongation on the lagging strand occurs in the 5’→ 3’ direction (a is incorrect). The lagging strand is synthesized in the direction opposite to the replication fork (b and c are incorrect).

Question 15

This catalyzes DNA chain elongation:
a. DNA Polymerase I
b. DNA Polymerase III
c. Helicase
d. DNA Topoisomerase

Answer 15

b. DNA Polymerase III

Rationale: DNA Polymerase III is the primary enzyme responsible for DNA chain elongation during replication in prokaryotes. DNA Polymerase I is mainly involved in removing RNA primers and filling in the gaps, Helicase unwinds the DNA helix, and DNA Topoisomerase relieves the torsional strain ahead of the replication fork.

Question 16

This type of DNA repair resolves pyrimidine dimers:
a. Nucleotide excision repair
b. Mismatch repair
c. Base excision repair
d. Double strand break repair

Answer 16

a. Nucleotide excision repair

Rationale: Nucleotide excision repair (NER) is the primary mechanism for repairing bulky lesions such as pyrimidine dimers caused by UV radiation. This process removes a short single-stranded segment containing the lesion and uses the complementary strand as a template for repair.

Question 17

This is non-densely packed and transcriptionally active:
a. Euchromatin
b. Intron
c. Solenoid
d. Metaphase chromatid

Answer 17

a. Euchromatin

Rationale: Euchromatin is less densely packed than heterochromatin and is generally transcriptionally active, allowing for gene expression. The other options, such as introns, solenoids, and metaphase chromatids, do not describe a state of chromatin that is transcriptionally active.

Question 18

Which of the following is responsible for the unwinding of double-stranded DNA?
a. DNA Helicase
b. DNA Ligase
c. DNA Topoisomerase
d. DNA Primase

Answer 18

a. DNA Helicase

Rationale: DNA helicase unwinds the double-stranded DNA during replication, creating the replication fork. DNA ligase joins Okazaki fragments on the lagging strand, DNA topoisomerase relieves torsional strain, and DNA primase synthesizes RNA primers.

Question 19

How many percent of human DNA is made up of repetitive sequences?
a. 10%
b. 30%
c. 50%
d. 70%

Answer 19

b. 30%
Rationale: More than 50% of DNA in eukaryotic organisms is unique or non-repetitive sequences, while approximately 30% consists of repetitive sequences, such as centromeres, telomeres, and other repetitive elements.

Question 20

In epigenetics, which of the following will give rise to a transcriptionally active chromosome?
a. DNA methylation
b. Histone acetylation
c. snRNA pairing
d. Telomerase signaling

Answer 20

b. Histone acetylation

Rationale: Histone acetylation is an epigenetic modification that relaxes chromatin structure, making it more accessible to transcription factors and thereby promoting gene expression. DNA methylation typically silences gene expression, snRNA pairing and telomerase signaling are not directly related to transcriptional activation.

Question 21

A person ate mushrooms picked in a wooded area. Shortly thereafter, he was rushed to the hospital, where he died. He had no previous medical problems. Which stage of transcription was probably affected?
a. Initiation
b. Elongation
c. Termination
d. Post-transcriptional modification

Answer 21

a. Initiation

Rationale: Certain toxic mushrooms contain alpha-amanitin, a potent inhibitor of RNA polymerase II, which is essential for the initiation of transcription in eukaryotic cells. Inhibition of RNA polymerase II prevents the synthesis of mRNA, which is critical for protein production and cell survival.

Question 22

This is required for transcription initiation:
a. Sigma factor
b. Rho factor
c. Spliceosome
d. RNA polymerase

Answer 22

d. RNA polymerase

Rationale: RNA polymerase is the enzyme responsible for synthesizing RNA from a DNA template. In prokaryotes, the sigma factor (a) is also essential for initiation, but RNA polymerase itself is required for the process in both prokaryotes and eukaryotes. The other options, Rho factor (b) and spliceosome (c), are involved in termination and RNA processing, respectively.

Question 23

What is the metabolic fate of exons?
a. They are excised upon processing of heterogenous nuclear RNA (hnRNA) to messenger RNA (mRNA)
b. They are retained upon processing of ribosomal RNA and are converted to miRNA and siRNA
c. They are translated into mRNA sequences and participate in the synthesis of proteins
d. They are added to mRNA during the splicing reactions by the spliceosome to prevent mRNA digestion by exonucleases

Answer 23

c. They are translated into mRNA sequences and participate in the synthesis of proteins

Rationale: Exons are the coding sequences of a gene that are retained in the final mRNA product after RNA splicing. They are translated into protein sequences during translation. The other options are incorrect as exons are not excised (a), converted into miRNA or siRNA (b), or added to mRNA to prevent digestion (d).

Question 24

Which of the following is a component of the 5’ cap?
a. 7-methylguanosine
b. Ribonuclease P
c. The sequence AAUAAA in the intron
d. Small nuclear ribonucleic acid (snRNA)

Answer 24

a. 7-methylguanosine

Rationale: The 5’ cap of eukaryotic mRNA consists of a 7-methylguanosine residue linked to the first nucleotide of the mRNA via a 5’-5’ triphosphate bridge. This modification is crucial for mRNA stability, nuclear export, and translation initiation. The other options are not components of the 5’ cap.

Question 25

What is the difference between DNA synthesis and RNA synthesis? a. The whole DNA strand is transcribed to form RNA but not in DNA synthesis b. The base pairing sequence is different for Guanine in DNA and RNA c. RNA polymerase does not require a primer, unlike DNA synthesis d. ATP is required as an energy source for DNA synthesis but not for RNA synthesis

Answer 25

c. RNA polymerase does not require a primer, unlike DNA synthesis Rationale: RNA polymerase can initiate RNA synthesis de novo without the need for a primer, whereas DNA polymerase requires a short RNA primer synthesized by primase to initiate DNA synthesis. This is a key difference between the two processes. The other options are incorrect because only specific regions of DNA are transcribed into RNA (a), base pairing is the same for guanine in both DNA and RNA (b), and ATP is required as an energy source for both DNA and RNA synthesis (d).

Question 26

The DNA template segment has this sequence: 3’ AAGCTTCCGC 5’. What would be the RNA product? a. 5’ UUCGAAGGCG 3’ b. 3’ UUCGAAGGCG 5’ c. 5’ GCGGAAGCUU 3’ d. 3’ TTCGTTGGCG 5’

Answer 26

a. 5’ UUCGAAGGCG 3’ Rationale: RNA is synthesized in the 5' to 3' direction using the DNA template strand, which is read in the 3' to 5' direction. The complementary RNA sequence to the DNA template strand (3’ AAGCTTCCGC 5’) would be 5’ UUCGAAGGCG 3’.

Question 27

In Rho-independent RNA transcription, the termination sequence is signaled by: a. The presence of the sigma factor which causes RNA polymerase to detach from the DNA template strand b. The rho factor itself c. A hairpin turn complementary to a region of the DNA template near the termination region (inverted repeats) d. Topoisomerases inhibiting the formation of negative supercoils during the process of elongation

Answer 27

c. A hairpin turn complementary to a region of the DNA template near the termination region (inverted repeats) Rationale: Rho-independent termination involves the formation of a hairpin structure in the RNA transcript followed by a series of uracil residues. This structure destabilizes the RNA-DNA hybrid, causing the RNA polymerase to dissociate from the DNA template.

Question 28

The post-transcriptional processing of rRNA involves which of the following events? a. Hydrolytic cleavage of portions of the primary transcript b. Addition of a methylguanosine cap c. Addition of a Poly-A-tail d. Insertion of a CCA nucleotide to the 3’ terminus

Answer 28

a. Hydrolytic cleavage of portions of the primary transcript Rationale: The primary transcript of rRNA undergoes hydrolytic cleavage to produce the mature rRNA species. The other options, such as the addition of a methylguanosine cap, a Poly-A tail, or a CCA nucleotide, are related to the processing of other types of RNA.

Question 29

When the structure of the TATA box is altered, which phase of transcription is affected? a. Initiation b. Elongation c. Termination d. Splicing

Answer 29

a. Initiation Rationale: The TATA box is a promoter element crucial for the initiation of transcription. Alterations in its structure can affect the binding of transcription factors and RNA polymerase, thereby impacting the initiation phase of transcription.

Question 30

Which drug inhibits the elongation phase of transcription? a. AZT b. ddI c. Rifampin d. Actinomycin D

Answer 30

d. Actinomycin D Rationale: Actinomycin D inhibits the elongation phase of transcription by intercalating into DNA and preventing the movement of RNA polymerase along the DNA template. The other drugs, such as AZT, ddI, and Rifampin, have different mechanisms of action related to DNA synthesis or other processes.

Question 31

Inadequate intake of iodine may cause the following, EXCEPT: a. Goiter b. Hypothyroidism c. Cretinism d. Wilson’s disease

Answer 31

d. Wilson’s disease Rationale: Wilson's disease is a genetic disorder related to copper metabolism, not iodine deficiency. Iodine deficiency can lead to goiter (a), hypothyroidism (b), and cretinism (c).

Question 32

Principal cation in the intracellular fluid compartment: a. Potassium b. Sodium c. Copper d. Phosphorus

Answer 32

a. Potassium Rationale: Potassium is the principal cation in the intracellular fluid compartment. Sodium (b) is the main extracellular cation, and copper (c) and phosphorus (d) have other roles in the body but are not the primary intracellular cations.

Question 33

In children, a severe deficiency of vitamin D causes: a. Skin cancer b. Osteomalacia c. Rickets d. Night blindness

Answer 33

c. Rickets Rationale: Rickets is a disease in children caused by a severe deficiency of vitamin D, leading to impaired bone mineralization. Osteomalacia (b) is the adult equivalent, and vitamin D deficiency is not linked to skin cancer (a) or night blindness (d).

Question 34

Which of the following vitamins would most likely be deficient in a person with scurvy? a. Thiamine b. Cobalamin c. Pantothenic acid d. Vitamin C

Answer 34

d. Vitamin C (Ascorbic Acid) Rationale: Scurvy is caused by a deficiency of vitamin C, which is essential for collagen synthesis. The other vitamins, such as thiamine (a), cobalamin (b), and pantothenic acid (c), are not related to scurvy.

Question 35

True of vitamin D, EXCEPT: a. D2 is the form which is vegetable in origin b. D3 is the form which is animal in origin c. Rickets manifests as brittle bones due to demineralization d. Induces intestinal and renal absorption of phosphates

Answer 35

c. Rickets manifests as brittle bones due to demineralization Rationale: While rickets does involve bone demineralization, it typically results in soft, weak bones rather than brittle bones. Brittle bones are more characteristic of osteoporosis. Vitamin D2 (a) is of vegetable origin, vitamin D3 (b) is of animal origin, and vitamin D induces the absorption of phosphates (d).

Question 36

Pernicious anemia is the classic consequence of: a. Thiamine deficiency b. Cobalamin deficiency c. Pantothenic acid deficiency d. Folic acid deficiency

Answer 36

b. Cobalamin deficiency Rationale: Pernicious anemia is caused by a deficiency in vitamin B12 (cobalamin), often due to a lack of intrinsic factor needed for its absorption. This deficiency leads to impaired red blood cell production and neurological symptoms. The other vitamins are not typically associated with pernicious anemia.

Question 37

A patient presents with dermatitis, diarrhea, dementia, and stomatitis. He is probably deficient in: a. Thiamine b. Niacin c. Vitamin B12 d. Pantothenic acid

Answer 37

b. Niacin Rationale: These symptoms are characteristic of pellagra, which is caused by a deficiency in niacin (vitamin B3). The other vitamins listed do not typically cause this combination of symptoms.

Question 38

All of the following vitamins are soluble in organic solvents used to extract lipids from tissues, EXCEPT: a. Vitamin K b. Vitamin A c. Vitamin C d. Vitamin E

Answer 38

c. Vitamin C Rationale: Vitamin C is water-soluble, whereas vitamins A, K, and E are fat-soluble and can be extracted using organic solvents that dissolve lipids.

Question 39

Specific dynamic action of food: a. Energy used during digestion, absorption, and metabolism of food b. Energy needed to maintain basic physiologic functions under standard conditions c. Energy used during movement and muscle contraction d. Energy used for muscles and growth of body organs

Answer 39

a. Energy used during digestion, absorption, and metabolism of food Rationale: The specific dynamic action (also known as the thermic effect) refers to the energy expended by the body to process food, including digestion, absorption, and metabolism.

Question 40

Basal metabolic rate: a. Energy used during digestion, absorption, and metabolism of food b. Energy needed to maintain basic physiologic functions under standard conditions c. Energy used during movement and muscle contraction d. Energy used for muscles and growth of body organs

Answer 40

b. Energy needed to maintain basic physiologic functions under standard conditions Rationale: Basal metabolic rate (BMR) is the amount of energy required to maintain basic physiological functions, such as breathing, circulation, and cell production, while at rest in a neutrally temperate environment.

Question 41

Pernicious anemia due to lack of intrinsic factor is a classic consequence of a deficiency in: a. Biotin b. Folic acid c. Cobalamin d. Folic acid

Answer 41

c. Cobalamin Rationale: Pernicious anemia is caused by a deficiency in vitamin B12 (cobalamin), which is often due to a lack of intrinsic factor necessary for its absorption in the intestine. The other options, such as biotin and folic acid, do not cause pernicious anemia.

Question 42

If there is a dysfunction in the oxidase enzyme system (like cytochrome oxidase, ascorbic acid oxidase), which mineral is most likely affected? a. Chloride b. Zinc c. Copper d. Selenium

Answer 42

c. Copper Rationale: Copper is an essential trace mineral that acts as a cofactor for several enzymes involved in oxidative processes, including cytochrome oxidase and ascorbic acid oxidase. Dysfunction in these enzymes is likely due to copper deficiency or imbalance.

Question 43

What is an important function of fiber? a. It adds flavor to food b. It can bind toxigenic substances c. It is an important energy source in the absence of lipids d. It aids in the absorption of fat-soluble vitamins

Answer 43

b. It can bind toxigenic substances Rationale: Fiber has several important functions, including binding to toxic substances in the digestive tract, which helps in their excretion. It is not primarily an energy source, does not add flavor to food, and does not aid in the absorption of fat-soluble vitamins.

Question 44

The following are considered macronutrients, except: a. Polysaccharides b. Amino acids c. Triglycerides d. Calcium

Answer 44

d. Calcium Rationale: Macronutrients include carbohydrates (such as polysaccharides), proteins (amino acids), and fats (triglycerides). Calcium is a micronutrient (a mineral) required in smaller amounts compared to macronutrients.

Question 45

A deficiency of vitamin B12 causes: a. Cheilosis b. Beriberi c. Scurvy d. Pernicious anemia

Answer 45

d. Pernicious anemia Rationale: Vitamin B12 deficiency leads to pernicious anemia due to impaired red blood cell production. Cheilosis (a) is caused by riboflavin (vitamin B2) deficiency, beriberi (b) is caused by thiamine (vitamin B1) deficiency, and scurvy (c) is caused by vitamin C deficiency.

Question 46

Biotin is involved in which of the following types of reactions? a. Hydroxylations b. Dehydrations c. Carboxylations d. Decarboxylations

Answer 46

c. Carboxylations Rationale: Biotin is a coenzyme involved in carboxylation reactions, which add a carboxyl group to substrates. It acts as a carrier of carbon dioxide in metabolic reactions, such as the carboxylation of acetyl-CoA to malonyl-CoA in fatty acid synthesis.

Question 47

Examples of conditions exhibiting negative nitrogen balance, EXCEPT: a. Debilitating diseases b. Kwashiorkor c. Fasting d. Pregnancy

Answer 47

d. Pregnancy Rationale: Negative nitrogen balance occurs when nitrogen excretion exceeds intake, typically in conditions such as debilitating diseases, kwashiorkor, and fasting. Pregnancy, however, is a condition associated with positive nitrogen balance due to increased protein synthesis for fetal growth.

Question 48

Defined as the minimum daily intake of protein needed to meet the needs for normal maintenance: a. Recommended daily dietary allowance b. Protein requirement c. Protein balance d. Biological value of proteins

Answer 48

a. Recommended daily dietary allowance Rationale: The recommended daily dietary allowance (RDA) defines the minimum daily intake of protein (and other nutrients) needed to meet the physiological requirements for normal maintenance and health.

Question 49

TRUE of triglycerides, EXCEPT: a. Serve as carriers of fat-soluble vitamins b. Provide 7 kcal/gram of energy c. Has a high satiety value d. Supply essential unsaturated fatty acids

Answer 49

b. Provide 7 kcal/gram of energy Rationale: Triglycerides provide 9 kcal/gram of energy, not 7 kcal/gram. The other statements are true: triglycerides serve as carriers of fat-soluble vitamins, have a high satiety value, and supply essential unsaturated fatty acids.

Question 50

Wernicke – Korsakoff syndrome: a. Defective transketolase enzyme b. Seen primarily in alcoholics c. Masked chronic thiamine deficiency d. All of the above

Answer 50

d. All of the above Rationale: Wernicke-Korsakoff syndrome is associated with a defective transketolase enzyme, is seen primarily in alcoholics, and is due to chronic thiamine (vitamin B1) deficiency. Therefore, all the given statements are true.

Question 51

Compounds having the same structural formula but differing in spatial configuration are known as: a. Anomers b. Stereoisomers c. Optical Isomers d. Epimers

Answer 51

b. Stereoisomers Rationale: Stereoisomers are molecules that have the same molecular formula and sequence of bonded atoms (structural formula), but differ in the three-dimensional orientations of their atoms in space. Examples include enantiomers and diastereomers.

Question 52

In glucose, the orientation of the –H and –OH groups around the carbon atom 5 adjacent to the terminal primary alcohol carbon determines: a. D or L series b. α & β anomers c. Epimers d. Dextro or levorotatory

Answer 52

a. D or L series Rationale: The D or L configuration of a sugar is determined by the orientation of the hydroxyl group on the chiral carbon farthest from the carbonyl group, which in glucose is carbon 5. This determines whether the sugar is part of the D-series or L-series.

Question 53

Table sugar or sucrose is made up of which two single sugar units? a. Maltose and glucose b. Glucose and fructose c. Lactose and glucose d. Fructose and galactose

Answer 53

b. Glucose and fructose Rationale: Sucrose is a disaccharide composed of one glucose molecule and one fructose molecule linked by a glycosidic bond.

Question 54

The sugar which forms a major component of nucleic acids is: a. Mannose b. Galactose c. Ribose d. Maltose

Answer 54

c. Ribose Rationale: Ribose is a five-carbon sugar that forms the backbone of RNA (ribonucleic acid). Deoxyribose, a similar sugar, forms the backbone of DNA (deoxyribonucleic acid).

Question 55

Hydrolysis of lactose yields: a. Glucose and fructose b. Galactose and glucose c. Maltose and glucose d. Galactose and fructose

Answer 55

b. Galactose and glucose Rationale: Lactose is a disaccharide composed of one molecule of galactose and one molecule of glucose. Hydrolysis of lactose breaks the glycosidic bond between these two monosaccharides.

Question 56

The cells dependent solely on glucose as an energy source are: a. Liver cells b. Brain cells c. Kidney cells d. Muscle cells

Answer 56

b. Brain cells Rationale: Brain cells (neurons) are highly dependent on glucose as their primary energy source because they cannot store glycogen. They rely on a constant supply of glucose from the bloodstream.

Question 57

Which is an example of a triose? a. Erythrose b. Glyceraldehyde c. Xylulose d. Glucose

Answer 57

b. Glyceraldehyde Rationale: Glyceraldehyde is a three-carbon sugar (triose). Erythrose (a) is a tetrose, xylulose (c) is a pentose, and glucose (d) is a hexose.

Question 58

Which of the following is a C-4 epimer of glucose? a. Mannose b. Xylose c. Erythrose d. Galactose

Answer 58

d. Galactose Rationale: Galactose is a C-4 epimer of glucose, meaning they differ in configuration around the fourth carbon atom.

Question 59

Carbohydrates are compounds that contain: a. Hydroxyl groups b. Aldehyde or ketone group c. At least 3 carbons d. All of the above

Answer 59

d. All of the above Rationale: Carbohydrates contain hydroxyl groups (a), an aldehyde or ketone group (b), and at least three carbons (c). These functional groups define the structure and reactivity of carbohydrates.

Question 60

Glycosaminoglycans are large complexes of negatively charged heteropolysaccharides that stabilize and support cellular and fibrous components of tissues while maintaining water and salt balance. The most abundant glycosaminoglycan in the body is: a. Chondroitin sulfate b. Dermatan sulfate c. Keratan sulfate d. Hyaluronic acid

Answer 60

a. Chondroitin sulfate Rationale: Chondroitin sulfate is the most abundant glycosaminoglycan in the body, playing a crucial role in the structure and function of cartilage and other connective tissues.

Question 61

Which process is slow in responding to a falling blood glucose but can provide sustained synthesis of glucose? a. Aerobic Glycolysis b. Anaerobic Glycolysis c. Glycogenolysis d. Gluconeogenesis

Answer 61

d. Gluconeogenesis Rationale: Gluconeogenesis is the process of synthesizing glucose from non-carbohydrate precursors. It is a slow process but provides a sustained source of glucose, especially during prolonged fasting or starvation. Glycogenolysis (c) is faster in response to falling blood glucose but is limited by glycogen stores.

Question 62

Absence of this enzyme causes accumulation of galactose in cells: a. Lactose synthase b. Uridyltransferase c. Aldolase B d. Hexokinase

Answer 62

b. Uridyltransferase Rationale: The enzyme galactose-1-phosphate uridyltransferase (GALT) is necessary for the proper metabolism of galactose. Its absence leads to the accumulation of galactose and its metabolites in cells, causing galactosemia.

Question 63

Sorbitol accumulates in cells where sorbitol dehydrogenase is low or absent. This accumulation causes strong osmotic effects on these cells producing pathologic alterations including the following except one: a. Malignancy b. Nephropathy c. Cataract formation d. Peripheral neuropathy

Answer 63

a. Malignancy Rationale: Sorbitol accumulation in cells, particularly in conditions like diabetes, leads to osmotic stress causing nephropathy, cataract formation, and peripheral neuropathy. Malignancy is not associated with sorbitol accumulation.

Question 64

During digestion of carbohydrates, which organ secretes bicarbonate that neutralizes high acidity of food? a. Stomach b. Pancreas c. Jejunum d. Duodenum

Answer 64

b. Pancreas Rationale: The pancreas secretes bicarbonate into the small intestine to neutralize the acidic chyme that comes from the stomach, facilitating the proper function of digestive enzymes.

Question 65

The major dietary polysaccharide that is of animal origin. a. Amylopectin b. Cellulose c. Glycogen d. Amylase

Answer 65

c. Glycogen Rationale: Glycogen is the primary polysaccharide stored in animal tissues and is consumed through animal products. Amylopectin (a) and cellulose (b) are plant polysaccharides, and amylase (d) is an enzyme, not a polysaccharide.

Question 66

L-glucose and D-glucose are pairs of structures that are mirror images of each other. These are examples of: a. Isomers b. Enantiomers c. Epimers d. Anomers

Answer 66

b. Enantiomers Rationale: Enantiomers are pairs of molecules that are non-superimposable mirror images of each other, such as L-glucose and D-glucose.

Question 67

Which of the following statements about Fructose Metabolism is correct? a. Entry of fructose into cells is dependent on insulin. b. Fructokinase provides the primary mechanism for fructose phosphorylation. c. Hexokinase phosphorylates glucose in all cells of the body since it has a high affinity for fructose. d. Fructose 1-phosphate is converted to fructose-1, 6 bisphosphate, cleaving aldolase B into DHAP.

Answer 67

b. Fructokinase provides the primary mechanism for fructose phosphorylation. Rationale: Fructokinase phosphorylates fructose to fructose-1-phosphate in the liver, which is the primary mechanism for fructose metabolism. The other statements are incorrect because entry of fructose into cells is not dependent on insulin (a), hexokinase has a low affinity for fructose (c), and fructose-1-phosphate is not converted to fructose-1,6-bisphosphate (d).

Question 68

Both beriberi and Wernicke-Korsakoff syndrome can be reversed by administering: a. Vitamin B12 b. Vitamin B1 c. Vitamin B3 d. Vitamin B5

Answer 68

b. Vitamin B1 Rationale: Vitamin B1 (thiamine) deficiency causes both beriberi and Wernicke-Korsakoff syndrome. Administering thiamine can help reverse these conditions.

Question 69

The following are true in naming fatty acids, except: a. Carbon atoms are numbered from the carboxyl carbon b. The carboxyl carbon is designated as carbon No. 1 c. Carbon number 1 is also known as the α-carbon d. The terminal methyl carbon is known as the ω- or n-carbon.

Answer 69

c. Carbon number 1 is also known as the α-carbon Rationale: In fatty acid nomenclature, the carbon atoms are numbered from the carboxyl carbon (a), the carboxyl carbon is designated as carbon No. 1 (b), and the terminal methyl carbon is known as the ω- or n-carbon (d). The α-carbon is actually the carbon adjacent to the carboxyl carbon, not carbon number 1.

Question 70

Fatty acids with chain lengths 4 to 10 carbons are found in significant quantities in milk. One example of these fatty acids is Capric acid, a 10-carbon fatty acid. What is the classification of Capric acid according to its chain length? a. Short-chain fatty acids (SCFA) b. Medium-chain fatty acids (MCFA) c. Long-chain fatty acids (LCFA) d. Very long-chain fatty acids (VLCFA)

Answer 70

b. Medium-chain fatty acids (MCFA) Fatty Acids Classification - **Short-chain fatty acids (SCFA)** - Fatty acids with aliphatic tails of fewer than six carbons (e.g., butyric acid) - **Medium-chain fatty acids (MCFA)** - Fatty acids with aliphatic tails of 6-12 carbons - Can form medium-chain triglycerides - **Long-chain fatty acids (LCFA)** - Fatty acids with aliphatic tails of 13 to 21 carbons - **Very long-chain fatty acids (VLCFA)** - Fatty acids with aliphatic tails longer than 22 carbons

Question 71

Linoleic acid and α-Linolenic acid are essential fatty acids that have 18 carbons and are classified according to their chain length as: a. Short-chain fatty acids (SCFA) b. Medium-chain fatty acids (MCFA) c. Long-chain fatty acids (LCFA) d. Very long-chain fatty acids (VLCFA)

Answer 71

c. Long-chain fatty acids (LCFA) Fatty Acids Classification - **Short-chain fatty acids (SCFA)** - Fatty acids with aliphatic tails of fewer than six carbons (e.g., butyric acid) - **Medium-chain fatty acids (MCFA)** - Fatty acids with aliphatic tails of 6-12 carbons - Can form medium-chain triglycerides - **Long-chain fatty acids (LCFA)** - Fatty acids with aliphatic tails of 13 to 21 carbons - **Very long-chain fatty acids (VLCFA)** - Fatty acids with aliphatic tails longer than 22 carbons

Question 72

True of Trans fatty acids, except: a. They are chemically classified as unsaturated fatty acids b. They elevate serum LDL and HDL c. They behave more like a saturated fatty acids in the body d. They increase the risk for coronary heart disease

Answer 72

b. They elevate serum LDL and HDL Rationale: Trans fatty acids elevate serum LDL but lower HDL cholesterol levels, which increases the risk for coronary heart disease. The other statements are correct: they are chemically classified as unsaturated fatty acids (a), behave more like saturated fatty acids in the body (c), and increase the risk for coronary heart disease (d).

Question 73

A reaction involving a gain of electron: a. Oxidation b. Reduction c. Exergonic reaction d. Endergonic reaction

Answer 73

b. Reduction Rationale: Reduction is the process of gaining electrons. Oxidation (a) involves the loss of electrons. Exergonic reactions (c) release energy, and endergonic reactions (d) require energy input. Loss of electron: Oxidation (Reducing agent) Gain of electron: Reduction (Oxidizing agent)

Question 74

Enzymes that change their conformation upon binding of an effector, resulting in an apparent change in binding affinity at a different ligand binding site: a. Allosteric enzymes b. Isozymes c. Modifiers d. Ribozymes

Answer 74

a. Allosteric enzymes Rationale: Allosteric enzymes undergo conformational changes upon binding of an effector (allosteric modulator) at a site other than the active site, leading to changes in their activity and binding affinity for other ligands.

Question 75

Any substance that can diminish the velocity of an enzyme-catalyzed reaction: a. Modifiers b. Inhibitor c. Anti-catalyst d. Effectors

Answer 75

b. Inhibitor Rationale: An inhibitor is a substance that decreases the rate of an enzyme-catalyzed reaction. Modifiers (a) and effectors (d) can either increase or decrease enzyme activity, and anti-catalyst (c) is not a standard term used in enzyme kinetics.

Question 76

Also known as the double-reciprocal plot, this can be used to calculate Km and Vmax, as well as to determine the mechanism of action of enzyme inhibitors. a. Hanes–Woolf plot b. Lineweaver-Burk Plot c. Eadie–Hofstee diagram d. Augustinsson plot

Answer 76

b. Lineweaver-Burk Plot Rationale: The Lineweaver-Burk plot, also known as the double-reciprocal plot, is a graphical representation used to determine Km (Michaelis constant) and Vmax (maximum velocity) of an enzyme-catalyzed reaction. It is also useful for analyzing enzyme inhibition.

Question 77

Transient intermediate in which neither free substrate nor product exists: a. Coupling b. Complex c. Reactant d. Transition State

Answer 77

d. Transition State Rationale: The transition state is a high-energy intermediate state in a chemical reaction where the substrate is converted into the product. It is a temporary state during the conversion process and does not exist as free substrate or product.

Question 78

Enzymes catalyze reactions by: a. Increasing substrate energy b. Altering reaction equilibria c. Decreasing free energy of activation d. Lowering total energy levels of reactants

Answer 78

c. Decreasing free energy of activation Rationale: Enzymes speed up chemical reactions by lowering the activation energy required for the reaction to proceed, making it easier for the reactants to reach the transition state.

Question 79

What kind of curve is exhibited by the Michaelis-Menten reaction of enzymes? a. Parabolic b. Sigmoidal c. Hyperbolic d. Normal distribution

Answer 79

c. Hyperbolic Rationale: The Michaelis-Menten plot of enzyme activity typically exhibits a hyperbolic curve, reflecting the relationship between the rate of reaction and substrate concentration.

Question 80

What is Km? a. It refers to the rate of the reaction when the substrate concentration is one-half the maximum level b. It refers to substrate concentration at which a reaction is proceeding at half the maximum velocity c. It indicates the capacity of the substrate to bind to the enzyme's allosteric site d. None of the above

Answer 80

b. It refers to substrate concentration at which a reaction is proceeding at half the maximum velocity Rationale: Km (Michaelis constant) is the substrate concentration at which the reaction rate is at half of its maximum velocity (Vmax). It provides an indication of the affinity of the enzyme for its substrate; a lower Km indicates higher affinity.

Question 81

In our diet, 90% of the fats are in this form: a. Cholesterol b. Glycolipids c. Triglycerides d. Phospholipids

Answer 81

c. Triglycerides Rationale: The majority of fats in our diet, about 90%, are in the form of triglycerides. Triglycerides are the main form of stored energy in animals.

Question 82

Which statement more appropriately defines lipids? a. Lipids are a homogeneous group of compounds b. Lipids are all hydrophobic compounds c. Lipids are essential components of all living organisms d. Lipids are water-soluble organic compounds

Answer 82

c. Lipids are essential components of all living organisms Rationale: Lipids are a diverse group of compounds that are essential components of all living organisms, playing key roles in energy storage, cellular structure, and signaling.

Question 83

The essential fatty acids are examples of this class of fatty acids: a. Saturated fats b. Monounsaturated fats c. Polyunsaturated fats d. None of the Above

Answer 83

c. Polyunsaturated fats Rationale: Essential fatty acids, such as linoleic acid and alpha-linolenic acid, are polyunsaturated fats. They contain multiple double bonds in their hydrocarbon chains and are vital for various physiological functions.

Question 84

The following statements are true, except: a. When double bonds are present, they are nearly always in the trans configuration b. As the number of double bonds increases, the melting point decreases c. When double bonds are present, they are always spaced at three-carbon intervals d. The introduction of a cis double bond causes the fatty acid to bend at that position

Answer 84

a. When double bonds are present, they are nearly always in the trans configuration Rationale: Naturally occurring double bonds in fatty acids are usually in the cis configuration, which introduces a bend in the fatty acid chain. Trans configuration is less common and usually results from industrial processing.

Question 85

The following statements about fatty acids are true, except: a. As the number of carbon chain length increases, the melting point also increases b. The omega system of naming fatty acids involves numbering the carbon atoms beginning at the carboxyl end c. When double bonds are present, they are always spaced at three-carbon intervals d. When double bonds are present, they are nearly always in the cis rather than a trans configuration

Answer 85

b. The omega system of naming fatty acids involves numbering the carbon atoms beginning at the carboxyl end Rationale: The omega system of naming fatty acids involves numbering the carbon atoms starting from the methyl (omega) end, not the carboxyl end. This system identifies the position of the first double bond relative to the methyl end.

Question 86

With only a few exceptions, natural fatty acids: a. Have a carboxyl group (-COOH) at one end b. Contain an even number of carbon atoms c. Are arranged in a branched line d. Have a methyl group (CH3) at the other end

Answer 86

b. Contain an even number of carbon atoms Rationale: Natural fatty acids typically have an even number of carbon atoms because they are synthesized by the sequential addition of two-carbon units. They also have a carboxyl group (-COOH) at one end (a) and a methyl group (CH3) at the other end (d), but they are not arranged in a branched line (c).

Question 87

True of geometric isomerism among fatty acids: a. Occurs in saturated fatty acids b. Depends on the orientation of the groups around the axes of the double bonds c. Trans configuration entails Acyl chains to be on the same side of the double bonds d. Cis configuration entails Acy

Answer 87

b. Depends on the orientation of the groups around the axes of the double bonds Rationale: Geometric isomerism in fatty acids is due to the different spatial arrangements of groups around the double bonds. In the cis configuration (d), acyl chains are on the same side, while in the trans configuration (c), acyl chains are on opposite sides. Geometric isomerism does not occur in saturated fatty acids (a), as they lack double bonds.

Question 88

True about lipids: a. Melting point decreases with increasing carbon number b. Their fluidity decreases with chain length and increases according to the degree of unsaturation. c. Melting point of a saturated fatty acid is lower than in an unsaturated fatty acid with the same number of carbons d. Double bonds increase the melting point relative to a saturated acid

Answer 88

b. Their fluidity decreases with chain length and increases according to the degree of unsaturation. Rationale: The fluidity of lipids decreases as the chain length increases and increases with the degree of unsaturation. Longer chains and saturated fatty acids pack more tightly and have higher melting points, thus decreasing fluidity. The other options are incorrect: melting point increases with carbon number (a), saturated fatty acids have higher melting points than unsaturated fatty acids with the same number of carbons (c), and double bonds decrease the melting point relative to a saturated acid (d).

Question 89

These fatty acids lower both the plasma LDL and HDL: a. ω-3 Fatty Acids b. ω-6 Fatty Acids c. Saturated Fatty Acids d. Monounsaturated Fatty Acids

Answer 89

b. ω-6 Fatty Acids Rationale: Omega-6 fatty acids can lower both LDL and HDL cholesterol levels. Omega-3 fatty acids (a) are known to lower triglycerides and have different effects on LDL and HDL, saturated fatty acids (c) generally raise LDL, and monounsaturated fatty acids (d) typically lower LDL and maintain or slightly increase HDL.

Question 90

Structural lipids and triacylglycerols contain primarily fatty acids of at least 16 carbons such as that found in palmitic acid. These are classified according to their chain length as: a. Short-chain fatty acids (SCFA) b. Medium-chain fatty acids (MCFA) c. Long-chain fatty acids (LCFA) d. Very long-chain fatty acids (VLCFA)

Answer 90

c. Long-chain fatty acids (LCFA) Fatty Acids Classification - **Short-chain fatty acids (SCFA)** - Fatty acids with aliphatic tails of fewer than six carbons (e.g., butyric acid) - **Medium-chain fatty acids (MCFA)** - Fatty acids with aliphatic tails of 6-12 carbons - Can form medium-chain triglycerides - **Long-chain fatty acids (LCFA)** - Fatty acids with aliphatic tails of 13 to 21 carbons - **Very long-chain fatty acids (VLCFA)** - Fatty acids with aliphatic tails longer than 22 carbons

Question 91

What is the bond that makes up the primary level of protein structure? A. peptide B. hydrogen C. ionic D. van der Waals

Answer 91

A. peptide Rationale: The primary structure of a protein is composed of a sequence of amino acids linked together by peptide bonds. These bonds form between the carboxyl group of one amino acid and the amino group of another.

Question 92

Which of the following bonds stabilizing the quaternary level of protein structure has a very stable feature? A. hydrogen B. covalent C. ionic D. hydrophobic

Answer 92

B. covalent Rationale: Covalent bonds, such as disulfide bonds, provide significant stability to the quaternary structure of proteins. While hydrogen bonds, ionic bonds, and hydrophobic interactions also contribute to protein stability, covalent bonds are particularly strong and stable.

Question 93

Which of the following is the common nitrogen acceptor for all reactions undergoing transamination? A. alpha-ketoglutarate B. pyruvate C. oxaloacetate D. alpha-ketobutyrate

Answer 93

A. alpha-ketoglutarate Rationale: Alpha-ketoglutarate is a common nitrogen acceptor in transamination reactions, forming glutamate upon accepting the amino group. This reaction is central to amino acid metabolism.

Question 94

A basic amino acid that is commonly found making up the active site of the enzyme: A. Lysine B. Arginine C. Histidine D. Tyrosine

Answer 94

C. Histidine Rationale: Histidine is a basic amino acid frequently found in the active sites of enzymes. Its side chain can donate and accept protons, making it versatile in catalyzing biochemical reactions.

Question 95

The fatty acid abbreviation 18:1 (9) represents the following, except: A. The fatty acid has 18 Carbons B. The fatty acid has 1 double bond C. Position of the double bond is at Carbon 9 from the methyl end D. This represents Oleic Acid

Answer 95

C. Position of the double bond is at Carbon 9 from the methyl end Rationale: The abbreviation 18:1 (9) indicates an 18-carbon fatty acid with one double bond at the 9th carbon from the carboxyl end, not the methyl end. This represents oleic acid, which fits the description except for the incorrect statement regarding the double bond's position from the methyl end.

Question 96

Which of the following statements about the competitive inhibition of an enzyme-catalyzed reaction is correct? A. The Vmax for a reaction remains unchanged in the presence of a competitive inhibitor. B. The Vmax is halved in the presence of a competitive inhibitor. C. A competitive inhibitor and substrate can bind simultaneously to the enzyme. D. The Km for a reaction remains unchanged in the presence of a competitive inhibitor.

Answer 96

A. The Vmax for a reaction remains unchanged in the presence of a competitive inhibitor. Rationale: Competitive inhibitors bind to the active site of the enzyme, competing with the substrate. This type of inhibition increases the Km (apparent substrate affinity decreases) but does not affect the Vmax, as the inhibition can be overcome by increasing substrate concentration.

Question 97

Which of the following statements about the mechanism of allosteric control of enzyme activity is correct? A. Allosteric enzymes reaction is represented by a sigmoid curve B. Allosteric enzymes show reduced sensitivity to changes in substrate concentration compared to classical type enzymes with hyperbolic kinetics. C. Allosteric enzymes are typically single-subunit enzymes. D. Allosteric enzymes show less sensitivity to changes in substrate concentration compared to classical type enzymes with hyperbolic kinetics.

Answer 97

A. Allosteric enzymes reaction is represented by a sigmoid curve Rationale: Allosteric enzymes often show a sigmoidal (S-shaped) curve in their reaction rates when plotted against substrate concentration. This reflects cooperative binding, where the binding of substrate to one active site affects the binding properties of other active sites.

Question 98

Which of the following statements about non-competitive inhibition of an enzyme-catalyzed reaction is correct? A. A competitive inhibitor can bind to the enzyme-substrate complex. B. Competitive inhibition can be overcome by the addition of large amounts of substrate to a reaction. C. The addition of large amounts of substrate to a reaction cannot overcome the effect of a non-competitive inhibitor. D. The Vmax of a reaction decreases in the presence of a competitive inhibitor.

Answer 98

C. The addition of large amounts of substrate to a reaction cannot overcome the effect of a non-competitive inhibitor. Rationale: Non-competitive inhibitors bind to an allosteric site on the enzyme, not the active site. This binding changes the enzyme's conformation, reducing its activity regardless of substrate concentration. Therefore, increasing substrate concentration cannot overcome non-competitive inhibition, and the Vmax decreases while the Km remains unchanged.

Question 99

Which of the following statements about Michaelis-Menten kinetics is correct? A. The Michaelis constant (Km) of an enzyme increases when the enzyme concentration is increased. B. A high Michaelis constant (Km) indicates a high affinity of an enzyme for its substrate C. The Michaelis constant (Km) of an enzyme decreases when the enzyme concentration is increased. D. A low Michaelis constant (Km) indicates a high affinity of an enzyme for its substrate.

Answer 99

D. A low Michaelis constant (Km) indicates a high affinity of an enzyme for its substrate. Rationale: Km is inversely related to the affinity of an enzyme for its substrate. A low Km indicates high affinity because the enzyme can reach half its maximum velocity at a lower substrate concentration.

Question 100

An uncompetitive inhibitor of an enzyme-catalyzed reaction A. binds to the Michaelis complex and decreases Vmax. B. is without effect at saturating substrate concentration C. can actually increase reaction velocity in rare cases D. all of these

Answer 100

A. binds to the Michaelis complex and decreases Vmax. - Rationale: Uncompetitive inhibitors bind only to the enzyme-substrate complex, not to the free enzyme. This binding decreases both the Vmax and the apparent Km (substrate concentration required to reach half Vmax). This type of inhibition is most effective when the substrate concentration is high and the enzyme-substrate complex is prevalent.