Anatomy
1 questionsWhich type of glial cell is derived from mesodermal origin?
NEET-PG 2015 - Anatomy NEET-PG Practice Questions and MCQs
Question 321: Which type of glial cell is derived from mesodermal origin?
- A. Macroglial cells
- B. Microglial cells (Correct Answer)
- C. Oligodendrocytes
- D. Ependymal cells
Explanation: ***Microglial cells*** - **Microglial cells** are unique among glial cells as they originate from **mesoderm**, specifically from **monocyte/macrophage precursors** in the bone marrow [1]. - They function as the **immune cells of the central nervous system (CNS)**, scavenging for plaques, damaged neurons, and infectious agents [1]. *Macroglial cells* - This is a broad category that includes **astrocytes, oligodendrocytes, and ependymal cells**, all of which are derived from **neuroectoderm**, not mesoderm [1]. - They perform various supportive roles but are distinct in origin from microglial cells [1]. *Oligodendrocytes* - **Oligodendrocytes** are derived from **neuroectoderm** and are responsible for forming the **myelin sheath** around axons in the CNS [2]. - Myelination is crucial for rapid and efficient nerve impulse conduction. *Ependymal cells* - **Ependymal cells** are derived from **neuroectoderm** and line the **ventricles of the brain** and the **central canal of the spinal cord**. - They play a role in the production and circulation of **cerebrospinal fluid (CSF)**.
Biochemistry
8 questionsWhat are Okazaki fragments?
What is the first purine nucleotide synthesized in de novo purine biosynthesis?
How many molecules of Acetyl CoA are produced from β-oxidation of palmitic acid?
What is the most important tool used in genetic engineering?
If the content of adenine (A) is 15%, what is the percentage of guanine (G) in the DNA?
The gaps between Okazaki fragments on the lagging strand during DNA replication are rejoined and sealed by:
What primarily forms the core of chylomicrons?
Which of the following is not classified as a chaperone protein?
NEET-PG 2015 - Biochemistry NEET-PG Practice Questions and MCQs
Question 321: What are Okazaki fragments?
- A. Long pieces of DNA on the lagging strand.
- B. Short pieces of DNA on the lagging strand. (Correct Answer)
- C. Short pieces of DNA on the leading strand.
- D. Long pieces of DNA on the leading strand.
Explanation: ***Short pieces of DNA on the lagging strand.*** - **Okazaki fragments** are the short, newly synthesized DNA fragments that are formed on the **lagging strand** during DNA replication. - The lagging strand is synthesized discontinuously because DNA polymerase can only add nucleotides in the **5' to 3' direction**, requiring it to move away from the replication fork as the DNA unwinds. *Long pieces of DNA on the lagging strand.* - The lagging strand is synthesized discontinuously in **short fragments**, not long continuous pieces. - The enzyme **DNA ligase** eventually joins these short fragments together to form a continuous strand. *Short pieces of DNA on the leading strand.* - The **leading strand** is synthesized continuously in one long stretch, moving towards the replication fork. - It does not require the synthesis of short fragments like the lagging strand. *Long pieces of DNA on the leading strand.* - While the leading strand is synthesized in a continuous, long piece, this statement does not accurately describe Okazaki fragments, which are specific to the lagging strand. - The leading strand's continuous synthesis is due to its **3' to 5' template orientation**, allowing DNA polymerase to proceed uninterrupted.
Question 322: What is the first purine nucleotide synthesized in de novo purine biosynthesis?
- A. AMP
- B. GMP
- C. IMP (Correct Answer)
- D. UMP
Explanation: ***IMP (Inosine Monophosphate)*** - **IMP** is the first complete purine nucleotide synthesized during the **de novo purine biosynthesis pathway**. - It serves as a branch point, from which **AMP** and **GMP** are subsequently synthesized through separate pathways. *AMP (Adenosine Monophosphate)* - **AMP** is a derivative of **IMP**, synthesized by the addition of an amino group from **aspartate** to IMP. - This step occurs after the formation of the complete purine ring structure in IMP. *GMP (Guanosine Monophosphate)* - **GMP** is also derived from **IMP**, through a pathway involving the oxidation of IMP to **XMP** (xanthosine monophosphate) and subsequent amination. - Its synthesis occurs downstream from IMP. *UMP (Uridine Monophosphate)* - **UMP** is a **pyrimidine nucleotide**, not a purine, and is synthesized via a completely different de novo pathway. - Pyrimidine biosynthesis involves forming the ring structure first, then attaching it to ribose-phosphate, unlike purine synthesis which builds the ring on a pre-existing ribose-phosphate.
Question 323: How many molecules of Acetyl CoA are produced from β-oxidation of palmitic acid?
- A. 3 acetyl CoA
- B. 16 Acetyl CoA
- C. 6 acetyl CoA
- D. 8 acetyl CoA (Correct Answer)
Explanation: ***8 acetyl CoA*** - Palmitic acid is a **16-carbon saturated fatty acid (C16:0)**. During β-oxidation, each cycle cleaves two carbons as **acetyl CoA**. - The formula for acetyl CoA produced is **n/2**, where n = number of carbons. For palmitic acid: 16/2 = **8 acetyl CoA molecules**. - Alternatively: Palmitic acid undergoes **7 cycles of β-oxidation** [(n/2) - 1 = 7], each producing 1 acetyl CoA (7 total), plus the final 2-carbon fragment forming the 8th acetyl CoA. *3 acetyl CoA* - This number is too low for a 16-carbon fatty acid. **Short-chain fatty acids** would produce fewer acetyl CoA molecules. - This value corresponds to β-oxidation of a **6-carbon fatty acid** (hexanoic acid), not palmitic acid. *6 acetyl CoA* - This number is also too low for a 16-carbon fatty acid. - This quantity would be produced from a **12-carbon fatty acid** (lauric acid), not palmitic acid. *16 Acetyl CoA* - This number is too high and would incorrectly imply that each carbon forms an acetyl CoA independently. - Sixteen acetyl CoA molecules would be produced from a **32-carbon fatty acid**, which is extremely rare in biological systems.
Question 324: What is the most important tool used in genetic engineering?
- A. Topoisomerase
- B. DNA Ligase
- C. Restriction endonuclease (Correct Answer)
- D. Helicase
Explanation: ***Restriction endonuclease*** - **Restriction endonucleases** are crucial for genetic engineering as they specifically cut DNA at particular recognition sites, allowing the insertion or deletion of genes. - This precise cutting ability is fundamental for creating **recombinant DNA** molecules. *Helicase* - **Helicase** is primarily involved in unwinding the DNA double helix during processes like DNA replication and transcription. - While essential for cellular functions, it does not directly manipulate DNA for gene insertion or modification in the way restriction enzymes do. *Topoisomerase* - **Topoisomerase** enzymes are responsible for managing DNA supercoiling, preventing tangling during DNA replication and transcription by cutting and rejoining DNA strands. - It plays a role in DNA structure but is not directly used for targeted gene editing or insertion. *DNA Ligase* - **DNA ligase** is essential for joining DNA fragments, which is a critical step in genetic engineering after restriction endonucleases have cut the DNA. - However, while it acts as a "molecular glue" to seal nicks and re-form phosphodiester bonds, it cannot initiate the precise cutting required to isolate genes.
Question 325: If the content of adenine (A) is 15%, what is the percentage of guanine (G) in the DNA?
- A. 15%
- B. 85%
- C. 70%
- D. 35% (Correct Answer)
Explanation: ***35%*** - According to **Chargaff's rules**, in a DNA molecule, the amount of **adenine (A) is equal to the amount of thymine (T)**, and the amount of **guanine (G) is equal to the amount of cytosine (C)**. - If A = 15%, then T must also be 15%. This means A + T = 30%. Since the total percentage of all bases is 100%, G + C must be 100% - 30% = 70%. As G = C, then G = 70% / 2 = 35%. *15%* - This would only be correct if guanine paired with adenine, which it does not; guanine pairs with **cytosine**. - This answer incorrectly assumes that all four bases are present in equal proportions, or that G equals A, which violates **Chargaff's rules**. *85%* - This percentage would imply an incorrect base pairing or an imbalanced ratio of purines and pyrimidines, violating the fundamental structure of DNA. - An 85% guanine content would mean that G + C far exceeds 100% or that T is extremely low, which is biologically impossible. *70%* - This represents the combined percentage of **guanine and cytosine**, not guanine alone. - While it correctly acknowledges the remaining proportion of bases, it fails to divide this sum between the two equal components, **G and C**.
Question 326: The gaps between Okazaki fragments on the lagging strand during DNA replication are rejoined and sealed by:
- A. DNA Ligase (Correct Answer)
- B. DNA Helicase
- C. DNA Phosphorylase
- D. DNA Topoisomerase
Explanation: ***DNA Ligase*** - **DNA ligase** forms a **phosphodiester bond** between the **3'-OH group** of one Okazaki fragment and the **5'-phosphate group** of the adjacent fragment, effectively sealing the nicks. - After **DNA polymerase I** removes the **RNA primers** and fills in the gaps, DNA ligase completes the synthesis of the **lagging strand** during DNA replication. - This enzyme is essential for maintaining the **integrity of the DNA backbone**. *DNA Helicase* - **DNA helicase** functions to **unwind the DNA double helix**, separating the two strands to create a replication fork. - It does not participate in joining DNA fragments. *DNA Phosphorylase* - **DNA phosphorylase** is not a standard enzyme involved in the direct sealing of DNA fragments during replication. - This is not the enzyme responsible for ligating Okazaki fragments. *DNA Topoisomerase* - **DNA topoisomerase** relieves the **supercoiling tension** that builds up in the DNA double helix ahead of the replication fork due to unwinding. - It does not have a role in forming phosphodiester bonds between newly synthesized DNA fragments.
Question 327: What primarily forms the core of chylomicrons?
- A. Triglycerides and Cholesterol together
- B. Triglycerides (Correct Answer)
- C. Free fatty acids
- D. Triglyceride, Cholesterol and Phospholipids
Explanation: ***Triglycerides*** - Chylomicrons are primarily responsible for transporting **dietary triglycerides** from the intestines to other tissues. - Their large core, composed mainly of **triglycerides**, allows efficient transport of these hydrophobic molecules. *Triglycerides and Cholesterol together* - While **cholesterol** is present in chylomicrons, it is less abundant than **triglycerides** and primarily exists as **cholesterol esters** in the core. - The core is not an equal mixture; **triglycerides** overwhelmingly dominate the volume. *Free fatty acids* - **Free fatty acids** are transported in the blood primarily bound to **albumin**, not within the core of chylomicrons. - Chylomicrons typically carry **esterified fatty acids** as part of triglycerides. *Triglyceride, Cholesterol and Phospholipids* - **Phospholipids** form the outer monolayer of the chylomicron, along with apoproteins, making them **amphipathic**. - They do not constitute a core component but rather the **surface interface** with the aqueous environment.
Question 328: Which of the following is not classified as a chaperone protein?
- A. Calnexin
- B. Protein disulfide isomerase
- C. Calreticulin
- D. Calbindin (Correct Answer)
Explanation: ***Calbindin*** - **Calbindin** is a **calcium-binding protein** that helps regulate intracellular calcium levels, particularly in the brain and intestines. - It does not assist in **protein folding** or assembly like chaperone proteins. *Calnexin* - **Calnexin** is a **chaperone protein** located in the endoplasmic reticulum (ER). - It assists in the proper folding and quality control of newly synthesized **glycoproteins**. *Protein disulfide isomerase* - **Protein disulfide isomerase (PDI)** is an ER enzyme that **catalyzes the formation and rearrangement of disulfide bonds** in newly synthesized proteins, which is crucial for proper folding. - Due to its role in enabling correct protein folding, it is considered a **chaperone-like protein**. *Calreticulin* - **Calreticulin** is another **calcium-binding chaperone protein** found in the endoplasmic reticulum. - It works synergistically with calnexin to ensure the **proper folding of glycoproteins**.
Pathology
1 questionsWhat is the definition of lipofuscin?
NEET-PG 2015 - Pathology NEET-PG Practice Questions and MCQs
Question 321: What is the definition of lipofuscin?
- A. Wear and tear pigment (Correct Answer)
- B. Fat deposits
- C. Blood pigment
- D. Form of calcification
Explanation: ***Wear and tear pigment*** - Lipofuscin is known as **wear and tear pigment** that accumulates in cells over time, especially in aging cells [1]. - It is a byproduct of **cellular lipid peroxidation** and protein degradation, indicative of oxidative stress [1]. *Form of calcification* - Not to be confused with calcification, lipofuscin is a **pigment** and not related to calcium deposition [1]. - Calcification usually occurs in response to tissue injury or necrosis, which differs fundamentally from lipofuscin accumulation. *Fat deposits* - Lipofuscin is made up of **an insoluble complex** and is not classified simply as fat or fat deposits [1]. - It is the result of the **degradation of cellular components**, rather than the accumulation of unutilized fats [1]. *Blood pigment* - Lipofuscin is not derived from **hemoglobin** or any blood components, distinguishing it from true blood pigments like **bilirubin**. - It is associated with **cellular aging** rather than with any specific blood function or metabolism [1]. **References:** [1] Kumar V, Abbas AK, et al.. Robbins and Cotran Pathologic Basis of Disease. 9th ed. Cellular Responses to Stress and Toxic Insults: Adaptation, Injury, and Death, pp. 75-77.