Narc Flashcards
(8 cards)
Which amino acid is nutritionally essential in children but not in adult
Arginine Glutamine
Histidine
Lysine
2. The nutritional quality of a given protein is given as
A. Essential amino acids per given mass
B. Essential and non-essential amino acids present
C. Protein per given mass
D. Protein source needed for effective growth.
3. Which of the following amino acids don’t have codons for direct incorporation into protein?
A. Hydroxylproline and Hydroxylysine
B. Hydroxytryptophan and peroxiglycine
C. Lysine
D. Tryptophan
4. Which pair of amino acid is not formed from amphibolic intermediate?
A.cystine and serine
B.cystine and thyrosine
C.Hydroxylysine and lysine
D. Phenylalanine and thyrosine
5. Scurvy, the sign of vitamin C deficiency, is as a result of defective
A. Amidation
B. Dehydrogenation
C. Hydroxylation
D. Phosphorylation
6. Positive and negative nitrogen balance are associated with respectively
A) Growth and pregnancy
B) Kwashiorkor and marasmus
C) Kwashiorkor and pregnancy
D) Pregnancy and kwashiorkor
Which amino acid is nutritionally essential in children but not in adults?
Answer: Arginine
Explanation:
Arginine is considered semi-essential. While adults can synthesize enough of it, children, especially during periods of rapid growth, may not produce enough to meet their needs. Therefore, it is nutritionally essential in children.
• Glutamine – conditionally essential in trauma/stress but not age-related. During stress, muscle breaks down glutamine to release nitrogen.
So, even though the body can normally make glutamine, it can’t make enough during stress, making it conditionally essential at that time.
• Histidine & Lysine – both are essential at all ages.
2.A 3A. 4.B
Question 4:
Which pair of amino acids is not formed from amphibolic intermediates?
Your book’s answer: B. Cystine and Tyrosine
Explanation:
• Tyrosine is formed from phenylalanine, not directly from an amphibolic intermediate.
• Cystine is a dimer (2 units) of cysteine, which is made from methionine.
- What is hydroxylation?
Answer:
Hydroxylation is the addition of a hydroxyl group (-OH) to a molecule.
In the body:
• It’s very important in collagen synthesis.
• Vitamin C helps enzymes add hydroxyl groups to proline and lysine.
• These hydroxylated amino acids (hydroxyproline and hydroxylysine) help collagen form stable triple helices. Without hydroxylation, collagen is weak—this causes scurvy.
6.What is positive and negative nitrogen balance?
Positive nitrogen balance = Nitrogen intake > Nitrogen loss
Occurs when the body is building up tissue, like:
• Pregnancy
• Growth in children
• Bodybuilding or recovery from illness
Negative nitrogen balance = Nitrogen loss > Nitrogen intake
Occurs when the body is breaking down more protein than it takes in, like:
• Kwashiorkor (protein malnutrition)
• Burns, trauma
• Starvation
Why is nitrogen used to measure this?
Because protein contains nitrogen, and measuring nitrogen balance helps determine how much protein the body is using or losing.
Amphibolic intermediates refer to molecules in the citric acid (TCA) cycle like oxaloacetate, α-ketoglutarate, etc., which are used to synthesize non-essential amino acids.
So, both cystine and tyrosine are not directly derived from amphibolic intermediates but rather from other amino acids, making option B correct as well.
- The enzyme complex responsible for the degradation of uniquinated proteins is called
A. Protease
B. Exinuclease
C. Endonuclease
D. Proteasome - The essential amino acid needed for attachment of ubiquitin to a target protein
Lysine
Glycine
Proline
Valine
- The amino acids that do not undergo transamination include:
A. Leucine, Methionine, Proline
B. Methionine, Arginine, Leucine
C. Threonine, Arginine, Lysine
D. Threonine, Proline, Lysine - Which of the following amino terminal residues of a given protein confers stability for degradation?
A. Arginine
B. Glycine
C. Leucine
D. Lysine - Which of the following are solely ketogenic?
A. Leucine, Lysine
B. Methionine, Glycine
C. Tryptophan, Phenylalanine
D. Valine, Glycine - The following amino acids are both ketogenic & glucogenic
A. Arginine, Histidine, Glutamate
B. Leucine, Isoleucine, Tryosine
C. Threonine, Isoleucine, Tryosine
D. Valine, Aspartate, Alanine - Which of the following is incorrect about transamination?
A. Involves two amino acids and two keto acids
B. It happens only to a-amino acids
C. It ises pyridoxal phosphate as a coenzyme
D. It is reversible and uses transaminases as enzymes - The conversion of alpha - amino nitrogen to ammonia by the actions of glutamate aminotransferase and glutamate dehydrogenase is by :
A. Transamidation
B. Transamination
C. Transdeamidation
D. Transdeamination - Glutaminase and Asparaginase enzymes are responsible for
A. Transamidation
B. Deamidation only
C. Transdeamidation
D. Amination and transamination
Here are the answers with explanations for questions 7–10:
- The enzyme complex responsible for the degradation of ubiquitinated proteins is called:
Answer: D. Proteasome
Explanation:
The proteasome is a large protein complex that recognizes proteins tagged with ubiquitin and breaks them down into small peptides. It’s a major pathway for regulated protein degradation in cells.
• Protease – general term for enzymes that break down proteins (but not specific to ubiquitinated ones).
• Exinuclease and Endonuclease – act on nucleic acids, not proteins.
- The essential amino acid needed for attachment of ubiquitin to a target protein:
Answer: Lysine
Explanation:
Ubiquitin is attached to proteins via the epsilon-amino group of a lysine residue on the target protein. This lysine becomes the anchor point for the ubiquitin chain.
• Glycine is present on ubiquitin itself (its C-terminal glycine forms the bond), but lysine on the target is where it attaches.
• Proline and valine are not involved in this process.
- The amino acids that do not undergo transamination include:
Answer: C. Threonine, Arginine, Lysine
Explanation:
Most amino acids undergo transamination, but a few do not, including:
• Threonine
• Lysine
• Proline
• Arginine
These use other pathways for deamination.
• Option C correctly lists three that do not undergo transamination.
• Other options mix some that do and don’t.
- Which of the following amino terminal residues of a given protein confers stability for degradation?
Answer: B. Glycine
Explanation:
According to the N-end rule, the amino acid at the N-terminus of a protein influences its half-life.
• Glycine is a stabilizing residue — proteins starting with glycine are degraded slowly.
• Arginine, lysine, and leucine are destabilizing residues — they target the protein for faster degradation.
Let me know if you’d like a table summarizing the N-end rule residues!
Great questions! Let’s clarify both:
What are ubiquitins?
Ubiquitin is a small regulatory protein (76 amino acids long) found in all eukaryotic cells. Its main job is to tag other proteins for destruction.
How it works:
1. Ubiquitin is attached to a protein that needs to be destroyed (usually one that’s damaged, misfolded, or no longer needed).
2. A chain of ubiquitins is added (polyubiquitination).
3. The proteasome recognizes this tag and breaks down the protein.
Think of ubiquitin as a “recycle me” tag for proteins.
Clarifying Question 10:
Question: Which amino-terminal residue of a given protein confers stability for degradation?
Your note says “lysine”, but the correct answer (based on the N-end rule) should actually be:
Correct Answer: B. Glycine
Explanation:
• Glycine is classified as a stabilizing residue at the N-terminus.
• Lysine is a destabilizing residue — it doesn’t confer stability, it marks the protein for faster degradation.
So if your note says lysine, it might be mistaken or possibly referring to lysine’s role in ubiquitin attachment, which is a different concept.
In Question 8, lysine is correct because it’s where ubiquitin attaches — but not in Question 10.
Would you like a visual of the N-end rule table showing stabilizing vs destabilizing residues?
Here are the correct answers and explanations for questions 11–15:
- Which of the following are solely ketogenic?
Answer: A. Leucine, Lysine
Explanation:
• Leucine and lysine are the only two amino acids that are purely ketogenic, meaning they are broken down to acetyl-CoA or acetoacetate, which can’t be used to make glucose.
• Other amino acids may be both ketogenic and glucogenic.
- The following amino acids are both ketogenic & glucogenic:
Answer: C. Threonine, Isoleucine, Tyrosine
Explanation:
These amino acids give rise to both:
• Glucose (via pyruvate or TCA cycle intermediates)
• Ketone bodies (via acetyl-CoA or acetoacetate)
Correct examples:
• Isoleucine
• Threonine
• Tyrosine
• Also: Phenylalanine, Tryptophan
Other options contain amino acids that are only glucogenic or only ketogenic.
- Which of the following is incorrect about transamination?
Answer: B. It happens only to α-amino acids
Explanation:
• Transamination mostly involves α-amino acids, but not exclusively.
• A few amino acids like lysine, threonine, proline, hydroxyproline do not undergo transamination.
• Hence, saying “only” is incorrect — option B is false.
Other statements (A, C, D) are true:
• Transamination involves 2 amino acids and 2 keto acids.
• Uses pyridoxal phosphate (PLP).
• It is reversible and catalyzed by transaminases (aminotransferases).
- Conversion of α-amino nitrogen to ammonia by glutamate aminotransferase and glutamate dehydrogenase is called:
Answer: D. Transdeamination
Explanation:
This is a 2-step process:
1. Transamination – amino group from other amino acids is transferred to α-ketoglutarate to form glutamate.
2. Deamination – glutamate is deaminated by glutamate dehydrogenase, releasing free ammonia.
This combined process is called transdeamination.
- Glutaminase and Asparaginase enzymes are responsible for:
Answer: B. Deamidation only
Explanation:
• Glutaminase converts glutamine → glutamate + NH₃.
• Asparaginase converts asparagine → aspartate + NH₃.
This process is deamidation — removing the amide (-NH₂) group from the side chain (not from the α-amino group), releasing ammonia.
• Not transamidation or transdeamidation.
Let me know if you’d like a summary chart of glucogenic vs ketogenic amino acids!
20.The following statements occur during Protein synthesis
1.uracil replaces thymine
2.Complementary bonding of condons to anticodons
3.DNA unzips
4.mRna attaches to ribosomes
A.1,2,3,4
B. 1,3,4,2
C. 3,1,4,2
D.3,1,24
- How come eukaryotes are able to replicate dna faster than prokaryotes?
A. Dna pol is faster
B. Use of simultaneous replication forks - Which of the following occurs when substitution of a pair of nucleotide base in DNA is replaced by a different base pair in DNA synthesis?
A. Coding mutation
B. Point mutation
C. Nonsense mutation
C. Transition mutation - The codon for an Amino Acid on the DNA is 5’ AGC 3’. The corresponding codon on the tRNA is …….. in the 5’ to 3’ direction.
A. GAC
B. TCG
C.
D. - The sequence of a coding strand is 5’ GAT-ATC-CAT-TAG-TGA 3’. What is the sequence of the trna?
- The RDA for fibre
A. Adult</= 50years: 38g(men), 25g (women), adults >50 who eat less food 30g(men), 21g(women) - A mutation in the wobble position resulting in the same amino acid is referred to as?
A silent mutation
B nonsense mutation
C sense mutation
D missense mutation - Rifamycin inhibits
A) DNA dependent DNA polymerase
B) DNA dependent RNA polymerase
C) RNA dependent DNA polymerase
D) RNA dependent RNA polymerase
Let’s simplify and order the steps properly:
Step 1: DNA unzips
• To begin transcription, DNA opens up.
Step 2: Uracil replaces thymine
• RNA is built using the DNA template. In RNA, uracil (U) is used instead of thymine (T).
Step 3: mRNA attaches to ribosomes
• After transcription, the mRNA leaves the nucleus and attaches to a ribosome in the cytoplasm.
Step 4: Codons bond with anticodons
• During translation, tRNA brings amino acids and matches its anticodon with the codon on the mRNA.
Final order:
3 (DNA unzips) 1 (Uracil replaces thymine during RNA formation) 4 (mRNA attaches to ribosomes) 2 (Codons bond with anticodons)
Answer: C. 3, 1, 4, 2
- How come eukaryotes replicate DNA faster than prokaryotes?
Answer: B. Use of simultaneous replication forks
Explanation:
Eukaryotes have multiple origins of replication per chromosome, allowing simultaneous replication forks, which speeds up the process — even though their polymerases are not faster than in prokaryotes.
- Substitution of a base pair in DNA synthesis is called:
Answer: B. Point mutation
Explanation:
• A point mutation is a change in a single base pair.
• Other specific types include:
• Silent – no change in amino acid
• Missense – changes the amino acid
• Nonsense – introduces a stop codon
• Transition – purine ↔ purine or pyrimidine ↔ pyrimidine (subtype of point)
- RDA for fibre:
Answer: A. Adult ≤50 years: 38g (men), 25g (women); adults >50: 30g (men), 21g (women)
Explanation:
Fiber needs drop slightly in older adults because they generally eat less food, and energy needs decrease with age.
- Mutation in the wobble position that doesn’t change amino acid is:
Answer: A. Silent mutation
Explanation:
• The wobble position is the 3rd base of a codon.
• A change here often doesn’t alter the amino acid = silent mutation.
- Rifamycin inhibits:
Answer: B) DNA-dependent RNA polymerase
Explanation:
• Rifamycin (or rifampicin) binds to bacterial RNA polymerase, blocking transcription.
• It’s commonly used against Mycobacterium tuberculosis.
- Which step will be affected if Elf2 is inactivated
A. Rna polymerase attaching to rna stand.
B. Ribosomal attachment to endoplasmic recticulum
C. Binding of Met-tRna to p-site
D. Translocation during protein synthesis - In parahemophilia, there is a deletion of 12 base pair in the factor V 5 gene. At which region will this mutation be likely located
3’ untranslated region
5’ untranslated region
First intron
Last exon
- A compound having a pentose sugar has its 1 st carbon attached to a purine and it’s 5 th carbon attached to a phosphate group… The compound can be identified as:
A. Adenosine diphosphate
B. Adenosine monophosphate
C. Thymidine diphosphate
D. Thymidine monophosphate - A new antibiotic inhibits bacterial protein synthesis. When added to a system translating the mRNA sequence 5’-AUGUUUUUUUAG-3’, the product formed is a dipeptide fMet-Phe. What step in protein synthesis is most likely inhibited? [AUG and UUU codes for methionine (met) and phenylalanine (Phe) respectively].
A. Initiation.
B. Binding of charged tRNA to the A-site.
C. Peptidyltransferase activity.
D. Ribosomal translocation.
E. Termination.
- The enzyme that synthesizes glutamate from alpha ketoglutarate requires
A. NAD+ only
B. NADP+ only
C. NAD+ or NADP+
D. NADH or NADPH - Transcription occurs along the___ template of the DNA, producing an mRNA in the___direction.
A. 5’→3’, 5’→3’
B. 5’→3’, 3’→5’
C. 3’→5’, 5’→3’
D. 3’→5’, 3’→5’ - Circle the components in box B which are the breakdown products of the disaccharide A
A. B.
SUCROSE Glucose
Galactose
Fructose
Glucose - All the following are correct about insoluble fiber
A. Cellulose, hemicellulose and lignin
B. They are not soluble in water
C. They are fermented by gut bacteria
D. They are derived from soft parts of vegetable - Mode of action of a-amanitin
A. Inhibits mRNA synthesis
B. Inhibits dna topoisomerase
C. Inhibits helicase
D. Inhibits tRNA
Recommend daily allowance for protein intake in adult males and females without any disease or physiological consideration is
A 0.6gm of protein 1kg of healthy body weight
B 0.7gm of protein 1kg of healthy body weight
C. 0.8gm of protein 1kg of healthy body weight
D. 0.9gm of protein 1kg of healthy body weight
- In signal transduction the formation of ligand-receptor complex results in….
A) Activation and conformational change
B) Activation, amplification and conformational change
C) desensitization, activation and conformational change
D) desensitization and conformational change - Which of the following compounds cannot serve as a second messenger
a. cAMP & cGMP
b. cAMP & peptides
c. cGMP and Ca2+
d. phosphatidylinositol
12) In signal transduction, the ligand that binds appropriate receptor to initiate the process is also referred to as the:
A) first messenger
B) primary messenger
C) second messenger
D) secondary messenger
- Which of the following is true about fatty acid and carbohydrate metabolism?
A. High insulin/glucagon inhibits lipogenesis in the liver
B. High insulin/glucagon activate lipolysis in adipocytes
C. Low insulin/glucagon activate lipolysis in adipocytes
D. Glucose dependent channels are activated when there’s low insulin - The liver is able to produce glucose which is sent to the brain. so why cant muscles produce glucose from glucose 6 phosphate.
A. Muscles lack Glucose 6 phosphate
B. The liver is the only organ needed for that
C. Muscles cannot produce glucose
D.
- What is the consequenc of an activated receptor that fails to desensitize?
A. Hampered amplification
B. Increased specificity
C. Prolonged expected effect
D. Heightened signal adaptation
19)The involvement of secondary messengers to several signaling pathways leads to features of …..
a) amplification and specificity
b) amplification and integration
c)
d) crosstalk and desensitization
Here are the correct answers with clear and simple explanations:
- Which step will be affected if eIF2 is inactivated?
Answer: C. Binding of Met-tRNA to P-site
Explanation:
eIF2 (eukaryotic initiation factor 2) helps bring the Met-tRNA to the P-site of the small ribosomal subunit during the initiation of translation.
Without eIF2, Met-tRNA can’t bind → initiation is blocked.
- Mutation in parahemophilia likely affects?
Answer: D. Last exon
Explanation:
A 12 base pair deletion affecting protein function is most likely in the last exon, which is part of the coding region.
• 3′ and 5′ UTRs are untranslated → don’t code protein
• Introns are spliced out
• Exons are what form the protein → mutation here affects protein structure
- Compound with pentose sugar, purine at C1, phosphate at C5?
Answer: B. Adenosine monophosphate
Explanation:
• Pentose sugar: Ribose
• C1 attached to purine (adenine) → Adenosine
• C5 attached to one phosphate → monophosphate
→ So it is Adenosine Monophosphate (AMP)
- Product stops at dipeptide fMet-Phe, what’s blocked?
Answer: D. Ribosomal translocation
Explanation:
Translation starts with fMet (AUG), then Phe (UUU).
If it stops after just two amino acids, it means the ribosome can’t move to the next codon. That’s a translocation problem.
So:
• Initiation happened
• Peptide bond formed
• But ribosome failed to shift → translocation step blocked
- Enzyme that synthesizes glutamate from α-ketoglutarate needs?
Answer: C. NAD⁺ or NADP⁺
It’s d not c
Explanation:
Glutamate dehydrogenase (the enzyme) can use either NAD⁺ or NADP⁺ as a coenzyme to reduce α-ketoglutarate to glutamate.
- Transcription template and direction of mRNA synthesis?
Answer: C. 3′→5′ template, 5′→3′ mRNA
Explanation:
• DNA template is read 3′ to 5′
• mRNA is synthesized 5′ to 3′
Think of it like reverse copying — enzyme moves down 3′→5′, building the new strand in 5′→3′.
- Disaccharide: SUCROSE – what are its breakdown products?
Answer: Glucose & Fructose
Explanation:
• Sucrose = Glucose + Fructose
• Galactose is found in lactose, not sucrose.
So in Box B, circle: Glucose and Fructose
- All true about insoluble fiber EXCEPT?
Answer: C. They are fermented by gut bacteria
Explanation:
• Insoluble fibers (e.g., cellulose, lignin) are not easily fermented
• Soluble fibers (like pectin) are the ones fermented
So option C is incorrect.
- α-Amanitin action?
Answer: A. Inhibits mRNA synthesis
Explanation:
• α-Amanitin (from poisonous mushrooms) inhibits RNA polymerase II
→ Blocks mRNA synthesis in eukaryotes
→ Lethal because mRNA = no proteins
Would you like a summary table of these for memorization or revision?
Great questions! Let’s break down both:
- Enzyme that synthesizes glutamate from α-ketoglutarate requires…?
Your note says D. NADH or NADPH, but let’s clarify:
• The enzyme is glutamate dehydrogenase.
• It catalyzes:
α-ketoglutarate + NH₄⁺ + NAD(P)H → glutamate + NAD(P)⁺ + H₂O
So actually, your note is correct — the enzyme uses NADH or NADPH (the reduced forms) as cofactors to donate electrons for the reductive amination of α-ketoglutarate.
Therefore:
Correct answer: D. NADH or NADPH
- All the following are correct about insoluble fiber — are the rest true?
Let’s go through the options:
• A. Cellulose, hemicellulose, and lignin
True – These are classic examples of insoluble fiber.
• B. They are not soluble in water
True – That’s what defines insoluble fiber.
• C. They are fermented by gut bacteria
False – Insoluble fiber is minimally or not fermented; this is true for soluble fiber, like pectin.
• D. They are derived from soft parts of vegetables
False or misleading – Insoluble fiber is more abundant in the tough, fibrous parts of plants, like bran, stalks, and skins, not the soft parts.
So actually, both C and D are incorrect. The best false option among them is C because it contradicts the definition most clearly.
Would you like a quick table summary or memory tip for insoluble vs soluble fiber?
Here are the correct answers with easy explanations:
- Recommended daily allowance for protein in healthy adults:
Correct answer: C. 0.8 g of protein per 1 kg of healthy body weight
• This is the standard RDA set for healthy adults.
• It provides enough protein to meet basic metabolic needs.
- What happens when ligand-receptor complex forms in signal transduction?
Correct answer: B. Activation, amplification, and conformational change
• Binding causes a shape change (conformational).
• It activates the receptor.
• It often triggers a cascade, leading to signal amplification.
- Which of the following cannot serve as a second messenger?
Correct answer: B. cAMP & peptides
• cAMP, cGMP, Ca²⁺, phosphatidylinositol = all are valid second messengers.
• Peptides (like insulin) are first messengers — they bind to receptors.
- The ligand that starts the signal transduction process is called the…?
Correct answer: A. First messenger
• First messenger = ligand (e.g., hormone or neurotransmitter).
• It binds the receptor to start the signal.
- Which is true about fatty acid & carbohydrate metabolism?
Correct answer: C. Low insulin/glucagon activates lipolysis in adipocytes
• Low insulin, high glucagon → fat breakdown (lipolysis)
• High insulin inhibits lipolysis and promotes lipogenesis.
- Why can’t muscles release glucose from glucose-6-phosphate?
Correct answer: A. Muscles lack glucose-6-phosphatase
• This enzyme is needed to convert glucose-6-phosphate into free glucose.
• Only the liver and kidneys have this enzyme, so muscles can’t do it.
- What happens if a receptor fails to desensitize?
Correct answer: C. Prolonged expected effect
• Desensitization normally reduces response after activation.
• Without it, the signal continues, leading to prolonged effects.
- Role of second messengers in signal pathways leads to…?
Correct answer: B. Amplification and integration
• Amplification: one ligand → many second messengers.
• Integration: different pathways can use the same second messenger.
Let me know if you want a summary table for memorization!
Great follow-up!
For Question 8:
Your note says A: “Activation and conformational change”
This is partially correct, but let’s break it down clearly:
• When a ligand binds its receptor, it:
• Activates the receptor
• Causes a conformational change (shape change)
• Often triggers amplification (e.g., via second messengers like cAMP)
So technically:
• A is correct (basic changes: activation + shape change)
• B is more detailed and includes amplification, which happens in most pathways
Your note isn’t wrong, just less detailed.
For Question 16: What do muscles have?
Muscles have:
• Glucose-6-phosphate
• Glycogen phosphorylase (to break down glycogen)
• Glycolytic enzymes (to use glucose for energy)
But muscles do NOT have:
• Glucose-6-phosphatase, the enzyme that converts glucose-6-phosphate to free glucose, which can leave the cell.
So result:
Muscles can use glucose for their own energy needs but cannot release it into the bloodstream like the liver can.
Would you like a diagram to help with the liver vs. muscle glucose pathways?
What is the maximum charge on a zwitterionic amino acid in an acidic medium
A. +2
B. +1
C. -1
D. -2
A scenario was given about dissolving hemoglobin and myoglobin in an anionic detergent, SDS, which disrupted non covalent bonds followed by dithrothreitol (not sure), a reducing agent which reduced disulfide bridges And then they were ran on electrophoresis.
- With the electrophoresis, based on what will they separate on the gel?
A. Charge and mass
B. Mass
C. Charge
D.
E. - Which structure in haemoglobin and myoglobin will be lost respectively
A. Primary, secondary
B. Secondary, tertiary
C. Tertiary, quartenary
D. Quartenary, tertiary - The use of dithrothreitol in the above experiment *ligated(not sure)
A. Cystine
B. Cysteine
C. Dipeptide
D. Tetrapeptide - Chronic stress leads to the following except
A. Elevated levels of cortisol and insulin
B. Suppression of certain catabolic hormones
C. Overeating and abdominal obesity
D. Systemic inflammation - One of the following is not true
A. Emotions are electrochemical signals that affect the chemistry and electricity of every cell in the body
B. Fear or anger creates certain peptide chains that are reintroduced into the living cells of the body through repeated DNA replication
C. Emotional states only affect the world inside the body, they don’t affect the world outside
D. Unexpected and suppressed emotional collect in the molecules of the brain with potential detrimental effects. - Which of the following amino acids is used in both purine and pyrimidine synthesis.
A) aspartate
B) Carbon dioxide
C) glutamate
D) Tetrahydrofolate - RNA is converted to complementary DNA( cDNA) by
A. Taq polymerase
B. RNA polymerase
C. Reverse transcriptase
D. DNA polymerase - When you want to determine the presence of albumin which method can you use to test it
A. Southern
B. Western
C. Eastern
D. Northern - Phe-try-ile-ser-glu have the codons(please look it up). It’s Changed to phe-try-phe-cys. What happened?
A. Deletion of G
B. Deletion of TG
C. Insertion of A
D. Insertion of AG
Let’s go through each question clearly and simply, with explanations:
Q: What is the maximum charge on a zwitterionic amino acid in an acidic medium?
Answer: B. +1
Explanation:
• In acidic medium, there’s excess H+ (protons).
• So the COO⁻ becomes COOH (neutral) and NH₂ becomes NH₃⁺.
• Resulting in +1 net charge on a normal amino acid.
• Zwitterion means +1 and -1 cancel out (net 0), but in acidic pH, it becomes fully protonated → +1.
Q: With SDS and electrophoresis, on what will they separate?
Answer: B. Mass
Explanation:
• SDS gives all proteins a uniform negative charge by binding to them.
• This removes the effect of native charge.
• So during electrophoresis, they’re separated by size (mass) only.
Q: Which structure in hemoglobin and myoglobin will be lost respectively?
Answer: D. Quaternary, Tertiary
Explanation:
• SDS breaks non-covalent bonds → destroys 3D shape.
• Dithiothreitol breaks disulfide bonds (important for 3D shape).
• Hemoglobin has quaternary structure (4 subunits).
• Myoglobin is one chain → only has up to tertiary structure.
Q: Dithiothreitol ligates what in the experiment?
Answer: A. Cystine
Explanation:
• Cystine = 2 cysteine residues joined by a disulfide bridge (-S-S-).
• Dithiothreitol (DTT) breaks these disulfide bonds.
• So it targets cystine, not cysteine.
Q: Chronic stress leads to the following except?
Answer: B. Suppression of certain catabolic hormones
Explanation:
• Chronic stress increases:
• Cortisol (a catabolic hormone)
• Insulin
• Inflammation
• Appetite and fat storage
• So catabolic hormones are not suppressed — they are elevated.
Q: One of the following is not true?
Answer: C. Emotional states only affect the world inside the body, they don’t affect the world outside
Explanation:
• Emotions can affect external behavior, social interactions, and outcomes.
• All other options are based on known mind-body connections.
Q: Which amino acid is used in both purine and pyrimidine synthesis?
Answer: A. Aspartate
Explanation:
• Aspartate contributes nitrogen in both pathways.
• Glutamine is also involved, but Aspartate is used in both as a backbone or donor.
Q: RNA is converted to complementary DNA (cDNA) by?
Answer: C. Reverse transcriptase
Explanation:
• Reverse transcriptase = RNA → DNA.
• Used in retroviruses and lab methods (e.g., making cDNA libraries).
Q: To test presence of albumin, which method?
Answer: B. Western blot
Explanation:
• Western blot = detects proteins (like albumin).
• Southern = DNA
• Northern = RNA
• Eastern = post-translational modifications.
Q: Phe-Tyr-Ile-Ser-Glu → changes to Phe-Tyr-Phe-Cys. What happened?
Answer: D. Insertion of AG
Explanation:
Let’s decode:
• Original codons:
• Phe = UUU / UUC
• Tyr = UAU / UAC
• Ile = AUU / AUC / AUA
• Ser = UCU / UCC / UCA / UCG
• Glu = GAA / GAG
• Mutated:
• Phe = UUU / UUC
• Tyr = UAU / UAC
• Phe = UUU / UUC
• Cys = UGU / UGC
If AG is inserted somewhere early, it shifts the reading frame and results in different amino acids downstream (frameshift mutation).
So the best choice is D. Insertion of AG.
Let me know if you’d like a diagram or codon breakdown for the last one!
- What happens to a zwitterion in a basic medium?
In a basic (alkaline) medium:
• There’s less H⁺, so deprotonation occurs.
• The amino group (NH₃⁺) loses a proton → becomes NH₂ (neutral).
• The carboxyl group (COO⁻) stays negative.
• So the net charge becomes −1.
Answer in basic medium: −1
So in acidic medium → +1,
Neutral pH → 0 (zwitterion),
Basic medium → −1
Let me explain both questions step by step, using easy examples so you truly understand:
- What happens to a zwitterion in basic medium?
Zwitterion means the molecule has both positive and negative charges but is overall neutral.
Let’s take glycine, the simplest amino acid:
In neutral (pH ~7):
• Amino group (–NH₃⁺) → positive
• Carboxyl group (–COO⁻) → negative
• Total charge = +1 + (−1) = 0
Now in a basic medium (high pH):
• There are fewer H⁺ ions in the solution.
• So the NH₃⁺ loses a hydrogen → becomes NH₂ (no charge).
• The COO⁻ stays negative.
So now:
• NH₂ → 0 charge
• COO⁻ → −1 charge
• Total charge = −1
Answer: In basic medium, the zwitterion becomes −1.
- Why is “Insertion of AG” the correct answer?
Let’s first look at the original and mutated amino acids:
Original:
• Phe = UUU or UUC
• Tyr = UAU or UAC
• Ile = AUU, AUC, AUA
• Ser = UCU, UCC, UCA, UCG, AGU, AGC
• Glu = GAA, GAG
Mutated:
• Phe = UUU or UUC
• Tyr = UAU or UAC
• Phe = UUU or UUC
• Cys = UGU or UGC
We are told that the original mRNA codes for:
Phe–Tyr–Ile–Ser–Glu → now changed to Phe–Tyr–Phe–Cys
Let’s imagine the mRNA (codons) for the original:
Just an example:
• Phe = UUU
• Tyr = UAU
• Ile = AUU
• Ser = UCU
• Glu = GAG
So:
UUU UAU AUU UCU GAG
Now they say it changed to:
• Phe = UUU
• Tyr = UAU
• Phe = UUU
• Cys = UGU
That could be:
UUU UAU UUU UGU
See how the codons shifted? To go from AUU UCU GAG to UUU UGU, a frameshift happened.
What kind of mutation causes a frameshift?
• Insertion or deletion of 1 or 2 bases, not 3.
• In this case, inserting AG causes the ribosome to read the mRNA differently, starting at a new place.
This is how AG inserted in the right spot makes:
• New codons → UUU UGU
• Which code for Phe-Cys
That’s why the answer is Insertion of AG – it causes a specific frameshift that explains the new amino acids.
Would you like me to draw a simple codon table or show how the codons shift step by step?
What should be administered to someone who has hereditary orotic aciduria
A. Oral thymine
B. Oral uridine
C. Oral xanthine
D. Oral adenine
- Which of the following is the most appropriate definition for a promoter in the context of prokaryotic gene expression
A. Cluster of genes regulated by a single promoter
B. DNA binding protein that regulates gene expression
C. Noncoding region on a DNA sequence that is bound by a repressor protein
D. Non coding region on DNA sequence that is bound by a DNA polymerase
Which of the following
The diathesis stress model is also known as
A. Biopsychosocial approach
B. Maintenance approach
C.
D.
We can define emotions as
A. Electrochemical signals that affect the chemistry and electricity of every cell
B. Electrical signals that affect the chemistry and electricity of every cell
C. Electrochemical signals that affect the chemistry and electricity of the brain
D. Chemical signals that affect the chemistry and electricity of every cell.
Someone has anxiety, when we do a blood work up we would find
A. High immune cells
B. Low immune cells
C. High body water
D. Low body water
Chronic stress is directly associated with all of the following except
A. High cortisol
B. High insulin
C. High adiposity
D. Low cytokines ns
Someone is exposed to a disease that causes accumulation of proteins that further causes deregulation of other proteins. The pathogen probably
A. Causes desensitization of the receptor
B. Cause the signaling molecule to remain bound to the receptor
C.
D.
Pathogens that have a nuclear receptor is probably
A. Lipophilic
B. Hydrophobic
C.
D.
Someone has severe combined immunodeficiency disease. This means the enzyme adenosine deaminase is deficient. This enzyme is necessary to catalyze the reaction that converts
A. Adenosine to inosine
B. Adenosine to inosine monophosphate
C. Dihydrofolate to tetrahydrofolate
D.
To specifically know whether DNA is being synthesized we measure serum
A. Adenine
B. Cytosine
C. Guanine
D. Thymine
The RDA for fibre
A. Adult</= 50years: 38g(men), 25g (women), adults >50 who eat less food 30g(men), 21g(women)
B.
C.
D.
Which feature pf signal transduction is similar to molecular medicine
A. DNA polymerase
B. RNA polymerase
C. Helicase
D. PCR
Glycaemic index is determined by all the following
A. Starch content
B. Non fibre content
C. Cooking method employed
D. Processing before consumption
A growth hormone will bind to which of the following receptors
A. Plasma membrane
B.
C.
D. Cytoplasmic
Here are the correct answers with brief explanations for each:
- What should be administered to someone with hereditary orotic aciduria?
Answer: B. Oral uridine
Why? Hereditary orotic aciduria is due to a defect in UMP synthase. Giving uridine bypasses the defect and provides UMP directly. - Most appropriate definition for a promoter in prokaryotic gene expression:
Answer: D. Noncoding region on DNA sequence that is bound by a DNA polymerase
Why? A promoter is a noncoding DNA region where RNA polymerase binds to begin transcription. (DNA polymerase is for replication, but question likely meant transcription initiation.)
The diathesis-stress model is also known as:
Answer: A. Biopsychosocial approach
Why? The model explains how biological vulnerability (diathesis) interacts with environmental stressors to cause disorders, aligning with the biopsychosocial framework.
We can define emotions as:
Answer: A. Electrochemical signals that affect the chemistry and electricity of every cell
Why? This describes the modern view of how emotions impact both brain and body-wide cellular functions.
Someone has anxiety. Blood work would show:
Answer: B. Low immune cells
Why? Chronic anxiety/stress often leads to immunosuppression, reducing immune cell count.
Chronic stress is directly associated with all of the following except:
Answer: D. Low cytokines
Why? Chronic stress increases pro-inflammatory cytokines, not decreases.
Accumulation of proteins causing deregulation suggests the pathogen:
Answer: B. Causes the signaling molecule to remain bound to the receptor
Why? This leads to constant activation, over-signaling, and misregulation.
Pathogens with nuclear receptors are likely:
Answer: A. Lipophilic
Why? Only lipophilic (fat-soluble) substances can pass through membranes to reach nuclear receptors.
SCID from ADA deficiency affects conversion of:
Answer: A. Adenosine to inosine
Why? ADA (adenosine deaminase) converts adenosine → inosine. Without it, toxic buildup occurs in lymphocytes.
To check if DNA is being synthesized, measure serum:
Answer: D. Thymine
Why? Thymine (or its derivatives like thymidine) is only in DNA, not RNA, so it’s a marker for DNA synthesis.
RDA for fibre:
Answer: A. Adult ≤50 years: 38g (men), 25g (women); >50 years: 30g (men), 21g (women)
Why? Correct values from dietary guidelines.
Signal transduction feature similar to molecular medicine:
Answer: D. PCR
Why? PCR (polymerase chain reaction) amplifies DNA and is used in diagnostics, often involving signaling pathway genes.
Glycaemic index is determined by all EXCEPT:
Answer: B. Non-fibre content
Why? Glycaemic index is influenced by starch type, cooking, processing, and fiber content — not “non-fiber content”.
Growth hormone binds to which receptor?
Answer: A. Plasma membrane
Why? GH is a peptide hormone, so it cannot cross membranes and binds to surface receptors.
Would you like me to turn these into a summarized study sheet or flashcards?
What is deregulation?
Deregulation in biology typically refers to:
• The loss of control over a normal biological process.
• It often means that a signaling pathway, gene expression, or protein function is no longer being properly regulated, leading to abnormal or excessive activity.
In the context of your question (protein accumulation causing deregulation of other proteins), it means:
• The buildup interferes with normal protein control — e.g., overactivating or inhibiting key pathways, leading to disease.
Why isn’t the growth hormone answer D. Cytoplasmic?
Because:
• Growth hormone (GH) is a peptide hormone, and peptides are not lipophilic.
• They cannot cross the cell membrane to reach cytoplasmic or nuclear receptors.
Instead:
• GH binds to a receptor on the plasma membrane (Answer A).
• This receptor then activates intracellular signaling pathways, like JAK-STAT.
So, D. Cytoplasmic would be true for steroid hormones, not for GH.
Would you like a table of hormone types and their receptor locations?
Which of the following bacterial trunsport methods is energy indepe likely uninvolved in the transport of antibioties into the cell?
ATP-dependent active transport.
Facilitated diffusion.
Group translocation.
Simple diffusion.
Membrane carriers resemble enzymes except for the fact that cal
are not selective for particular ligands.
change conformation when they bind to a ligand.
do not chemically change their ligands.
do not have binding sites.
Which of the following lipoproteins removes surplus cholesterol from the peripben and transports it to the liver for disposal?
Low density lipoprotein.
High density lipoprotein.
Very Low density lipoprotein.
Very High density lipoprotein.
• 25. Which of the following disease conditions below is not due to storage of lipids?
Niemann-Pick disease.
Krabbe’s disease.
TaySach’s disease.
Zellweger syndrome.
26. Which of the following is NOT true about eicosanoids?
it was first discovered in prostate gland secretions.
it is a 20 carbon compound generated from lipoxins. © synthesized in all tissues.
D. acts as local hormones.
27. The first step in ß - oxidation is a/an reduction reaction.
oxidation reaction.
hydration reaction.
hydroxylation reaction.
- Which of the following forms part of the prosimal row of the carpal bones
Capitate
Hamate
Triquetrum
D. A and B - The following are observed in individuals with carpal tuanel syndrome
B and C
Reduction in sensation over central part of the palm
There is difficulty in adducting the thumb
There is difficulty in opposing the thumb
Question 1:
Which of the following bacterial transport methods is energy independent and likely uninvolved in the transport of antibiotics into the cell?
Correct Answer: Simple diffusion.
Explanation:
Simple diffusion is passive, energy-independent, and does not involve carrier proteins or energy input. Most antibiotics require active transport or facilitated diffusion to enter cells, so simple diffusion is typically not involved unless the antibiotic is very small and non-polar.
• ATP-dependent active transport – Requires energy; often used for transporting antibiotics.
• Facilitated diffusion – Passive but uses specific carriers; some antibiotics use this route.
• Group translocation – Involves chemical modification and energy input; used by bacteria.
• Simple diffusion – No energy, no carrier; unlikely to be used by most antibiotics.
Question 2:
Membrane carriers resemble enzymes except for the fact that they:
Correct Answer: Do not chemically change their ligands.
Explanation:
Membrane carriers bind specific ligands and undergo conformational change to transport substances across the membrane, but unlike enzymes, they don’t alter the chemical structure of the molecule being transported.
• Are not selective for particular ligands – False: They are highly selective.
• Change conformation when they bind to a ligand – True: This is how they function.
• Do not chemically change their ligands – Correct: This is what differentiates them from enzymes.
• Do not have binding sites – False: They do have binding sites for specific substrates.
Would you like a few more practice questions like this?
Here are the correct answers followed by brief explanations for each question:
- Which of the following lipoproteins removes surplus cholesterol from the periphery and transports it to the liver for disposal?
Answer: High density lipoprotein (HDL)
Explanation: HDL is known as “good cholesterol” because it removes excess cholesterol from tissues and transports it to the liver for excretion (reverse cholesterol transport).
- Which of the following disease conditions is not due to storage of lipids?
Answer: Zellweger syndrome
Explanation: Zellweger syndrome is a peroxisomal biogenesis disorder, not a lipid storage disease.
The others—Niemann-Pick, Krabbe’s, and Tay-Sachs—are lysosomal storage diseases involving lipid accumulation.
- Which of the following is NOT true about eicosanoids?
Answer: It is a 20 carbon compound generated from lipoxins.
Explanation:
• True: Eicosanoids were first found in prostate gland secretions.
• True: They act as local hormones.
• True: They are synthesized in almost all tissues.
• False: They are not generated from lipoxins; rather, lipoxins are a type of eicosanoid, derived from arachidonic acid (a 20-carbon compound).
- The first step in β-oxidation is a/an:
Answer: Oxidation reaction
Explanation:
The first step of β-oxidation involves oxidation of a fatty acyl-CoA by acyl-CoA dehydrogenase, forming a double bond (trans-enoyl-CoA).
- The following are observed in individuals with carpal tunnel syndrome:
Answer: B and C
(That is: Reduction in sensation over central part of the palm and Difficulty in opposing the thumb)
Explanation:
• Carpal tunnel syndrome affects the median nerve.
• It leads to sensory loss in the lateral palm and difficulty opposing the thumb.
• Adduction of the thumb is by the ulnar nerve, so it’s usually not affected.
- Which of the following forms part of the proximal row of the carpal bones?
Answer: Triquetrum
Explanation:
Proximal row of carpal bones (lateral to medial): Scaphoid, Lunate, Triquetrum, Pisiform
Capitate and Hamate are in the distal row.
Would you like me to turn these into flashcards or a quiz format for practice?
Which of the following molecules is not a gluconcogenic or glycolytic intermediat
Glucose
phosphoenolpyruvate
© Phosphoglycerate
I. Oxaloacetate
Classify the fatty acid given:
Structure: CH₃–CH₂–CH=CH–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–COOH
Question 1: Which of the following molecules is NOT a gluconeogenic or glycolytic intermediate?
Correct Answer: Glucose
Explanation:
• Glycolytic and gluconeogenic intermediates refer to the molecules in the steps between glucose and pyruvate (and reversed in gluconeogenesis).
• Glucose is the starting substrate of glycolysis, but not considered an intermediate — it’s the input.
• The others (phosphoenolpyruvate, phosphoglycerate, oxaloacetate) are true intermediates in either pathway.
Question 2: Classify the fatty acid given:
Structure: CH₃–CH₂–CH=CH–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–COOH
Steps to classify:
1. Count total carbon atoms: 14 carbons (including the –COOH and –CH₃ ends)
2. Count double bonds: 1 double bond (CH=CH)
3. Double bond position (Δ system): From the carboxyl (COOH) end, the double bond is between C11 and C12(note that this time the carboxyl group isn’t at the left it’s at the right and you count from the front and the front is always where the carboxyl group is and the back is where the methyl group is. So this one if we were even naming via omega, we’ll count from the methyl group so this is omega 3)
So, this is a 14-carbon monounsaturated fatty acid.
C. 14:1 (Δ¹¹)
