Carbohydrate metabolism: trapping of fluorodeoxyglucose: practice question

A Biochemistry question in the style of the NEET PG paper. Answer it first, then read why the right option is right and why each of the others is wrong.

Try it

Biochemistry · Carbohydrate metabolism: trapping of fluorodeoxyglucose

Sample item

Fluorodeoxyglucose labelled with fluorine-18 is injected for a PET scan and accumulates in metabolically active tumour cells. Why does it remain trapped inside the cell?

Original practice item, not a recalled exam question. Nothing here is recorded.

The answer and the reasoning

A. It is phosphorylated by hexokinase and the product is not further metabolised

The analogue enters through glucose transporters and is phosphorylated by hexokinase to the 6-phosphate. This product is not a substrate for phosphoglucose isomerase and is a poor substrate for glucose-6-phosphatase in tumours, so the charged molecule accumulates in proportion to glucose uptake.

Option by option

  • A. It is phosphorylated by hexokinase and the product is not further metabolised (correct)

    Correct. Hexokinase phosphorylates it, and the charged 6-phosphate cannot proceed through glycolysis or leave the cell.

  • B. It is oxidised in the pentose phosphate pathway to a charged acid

    It does not enter the oxidative arm of the pentose phosphate pathway as a substrate for glucose-6-phosphate dehydrogenase.

  • C. It is converted to glycogen by glycogen synthase

    The trapped species is the phosphorylated analogue itself; it is not converted to glycogen to any appreciable extent.

  • D. It binds irreversibly to the GLUT1 transporter

    It crosses the membrane via GLUT1 and does not bind that transporter irreversibly.

Source note: Original NextPrep sample item — pending expert review. This item is original practice material awaiting independent expert review, so confirm important facts in a standard textbook.

More Biochemistry questions

Questions about this Biochemistry question

What is the answer to this Biochemistry question?

The best answer is A, It is phosphorylated by hexokinase and the product is not further metabolised. The analogue enters through glucose transporters and is phosphorylated by hexokinase to the 6-phosphate.

Why is "It is oxidised in the pentose phosphate pathway to a charged acid" not the answer?

It does not enter the oxidative arm of the pentose phosphate pathway as a substrate for glucose-6-phosphate dehydrogenase.

Why is "It is converted to glycogen by glycogen synthase" not the answer?

The trapped species is the phosphorylated analogue itself; it is not converted to glycogen to any appreciable extent.

Why is "It binds irreversibly to the GLUT1 transporter" not the answer?

It crosses the membrane via GLUT1 and does not bind that transporter irreversibly.

What concept does this question test?

It tests Carbohydrate metabolism: trapping of fluorodeoxyglucose in Biochemistry. This product is not a substrate for phosphoglucose isomerase and is a poor substrate for glucose-6-phosphatase in tumours, so the charged molecule accumulates in proportion to glucose uptake.

How hard is this question?

We tag it as an application question: it asks you to use a fact on a patient or a finding. The tag is our own editorial judgement, not an official difficulty rating, so read the reason for every option, not only the answer.

How should I study Carbohydrate metabolism: trapping of fluorodeoxyglucose?

Build one table of enzyme deficiency, disease, inheritance and key finding, then revise it by spaced recall. Then use the flashcards and the other Biochemistry questions on this site to test what stuck.

Is this a real NEET PG question?

No. It is an original practice item (editorial status: sample, version 1). NextPrep does not publish recalled or leaked questions, and independent faculty review of this item has not been completed yet, so confirm important facts in a standard textbook.