Biochemistry for NEET PG: six pathways and the one blocked step in each

Answer: C. Galactose 1-phosphate

C. Galactose 1-phosphate. Step 2 is the uridyltransferase, which uses galactose 1-phosphate to form UDP-galactose. When it is blocked, galactose 1-phosphate builds up inside cells and injures the liver and kidney tubules. The ill milk-fed baby has classic galactosaemia. Galactose also backs up and is reduced to galactitol in the lens, where it draws in water and causes cataracts.

Why the other options lose: UDP-galactose (A) is the product of the blocked step, not the compound building up behind it. Galactitol (B) explains the cataracts, not the liver and kidney injury. Free galactose (D) also rises, but galactose 1-phosphate is the damaging compound asked for here. Keep the two sites apart: galactose 1-phosphate in liver and kidney, galactitol in the lens.

Recallab question 55811 · Biochemistry › Carbohydrate Metabolism · reference: Harper's Illustrated Biochemistry

Answer: B. Step 3, argininosuccinate synthetase

B. Step 3, argininosuccinate synthetase. Citrulline is made at step 2 inside the mitochondrion and carried into the cytosol. Step 3 joins it to aspartate to make argininosuccinate. A block here lets citrulline build up sharply: citrullinaemia type I. The normal urinary orotic acid in this stem helps distinguish the pattern from an ornithine transcarbamylase block.

Why the other options lose: step 1, carbamoyl phosphate synthetase I (A), lowers citrulline because less carbamoyl phosphate is made. Step 4, argininosuccinate lyase (C), raises argininosuccinic acid. Step 2, ornithine transcarbamylase (D), gives low citrulline and raised orotic acid as unused carbamoyl phosphate enters pyrimidine synthesis. Markedly raised citrulline points to step 3.

Recallab question 55311 · Biochemistry › Amino Acid Metabolism · reference: Harper's Illustrated Biochemistry

Answer: B. Step 5, fumarylacetoacetate hydrolase

B. Step 5, fumarylacetoacetate hydrolase. This is tyrosinaemia type I. A block at the last step backs up fumarylacetoacetate, with succinylacetone formed along the side branch. These compounds damage liver cells and the proximal kidney tubule, explaining liver failure with renal Fanconi syndrome. Succinylacetone in urine or blood is the marker of this defect.

Why the other options lose: step 2, tyrosine aminotransferase (A), causes tyrosinaemia type II, which affects the eyes and skin. Step 4, homogentisate oxidase (C), causes alkaptonuria, with dark urine and later ochronosis. Step 1, phenylalanine hydroxylase (D), causes phenylketonuria, with a risk of intellectual disability if untreated. None explains the succinylacetone marker. Liver and kidney injury with that marker points to step 5.

Recallab question 55411 · Biochemistry › Amino Acid Metabolism · reference: Harper's Illustrated Biochemistry

Answer: C. Medium chain fatty acid oxidation block

C. Medium chain fatty acid oxidation block. The tall bars at eight and ten carbons, with a smaller rise at six, place the block at medium chain lengths. Beta oxidation normally shortens fatty acids and releases acetyl groups for ketone formation. When medium chain oxidation stalls, the liver has less substrate for ketone synthesis. A fasting infant can therefore become severely hypoglycaemic without making enough ketones.

Why the other options lose: a long chain fatty acid oxidation block (A) would raise long chain species, whereas those bars lie within the reference band. Ketotic hypoglycaemia (B) includes ketones, which are absent here. A carnitine uptake defect (D) gives low free carnitine and broadly low acylcarnitines, rather than this rise in medium chain species. The missing ketones point to a fat oxidation problem; the raised chain lengths locate it.

Recallab question 57311 · Biochemistry › Lipid Metabolism · reference: Harper's Illustrated Biochemistry

Answer: C. Step 2

C. Step 2. This is the transporter at the peroxisomal membrane that allows very long chain fatty acids to enter for shortening. Its gene lies on the X chromosome. Failure of transport lets these fatty acids build up, damaging the adrenal cortex and the white matter of the brain. Darkening skin and falling school performance, with an affected maternal uncle, fit X-linked adrenoleukodystrophy. The peroxisomes themselves are present.

Why the other options lose: step 1 (A) activates the fatty acid in the cytosol; it is not the membrane transport step linked to this pattern. Step 3 (B) shortens the chain inside the peroxisome. Defects there can also raise very long chain fatty acids, but do not explain the characteristic X-linked pattern as well. Step 4 (D) carries the shortened chain into the mitochondrion for further oxidation. The adrenal and white matter clues locate the failed door at step 2.

Recallab question 56013 · Biochemistry › Lipid Metabolism · reference: Harper's Illustrated Biochemistry

Answer: D. Coarse basophilic stippling

D. Coarse basophilic stippling. Lead blocks two enzymes of haem synthesis: the one that uses aminolaevulinic acid and the one that inserts iron into protoporphyrin at the end. Aminolaevulinic acid and protoporphyrin therefore accumulate. Lead also inhibits the breakdown of ribosomal RNA in young red cells. The remaining RNA clumps into coarse blue granules, giving the blood film finding shown.

Why the other options lose: hypersegmented neutrophils (A) point to megaloblastic anaemia. Microcytes with pencil cells (B) suggest iron deficiency; lead can cause small red cells, but coarse stippling is the finding that fits this pathway picture. Target cells and acanthocytes (C) can occur in liver disease and do not explain the two blocked haem enzymes. Two pathway blocks with coarse blue granules point to lead and option D.

Recallab question 55113 · Biochemistry › Vitamins & Minerals · reference: Harper's Illustrated Biochemistry