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Hard biology question

2. Melvin Calvin studied the light-independent reaction (Calvin cycle) in plant cells.
He used radiolabelled 14CO2 to measure the production of organic molecules in chloroplasts.

He placed an aquatic plant in water.
The plant was given light for 20 minutes.
The light was then turned off (dark conditions) for a further 30 seconds.

He measured the radioactivity of the solutions produced and used these values to calculate the number of
molecules of triose phosphate (TP) and ribulose bisphosphate (RuBP) present.

The results are shown in the table below.

Activity of 14C (x10^27 Bq)
After 20 minutes light / 30 seconds dark conditions
Molecule

TP 5.5 10.1
RuBP 4.9 0.6


Assuming 8.5 × 10^18 Bq are generated by each 14C atom in the molecule, how many new TP molecules are
produced after 30 seconds in the dark


A 6.47 × 10^8
B 1.80 × 10^8
C 1.83 × 10^27
D 3.37 × 10^27

(by the way ^ means to the power of in this context)
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Reply 2
Helpp i need this answer too
Reply 3
Original post by inc0gnegr0
2. Melvin Calvin studied the light-independent reaction (Calvin cycle) in plant cells.
He used radiolabelled 14CO2 to measure the production of organic molecules in chloroplasts.

He placed an aquatic plant in water.
The plant was given light for 20 minutes.
The light was then turned off (dark conditions) for a further 30 seconds.

He measured the radioactivity of the solutions produced and used these values to calculate the number of
molecules of triose phosphate (TP) and ribulose bisphosphate (RuBP) present.

The results are shown in the table below.

Activity of 14C (x10^27 Bq)
After 20 minutes light / 30 seconds dark conditions
Molecule

TP 5.5 10.1
RuBP 4.9 0.6


Assuming 8.5 × 10^18 Bq are generated by each 14C atom in the molecule, how many new TP molecules are
produced after 30 seconds in the dark


A 6.47 × 10^8
B 1.80 × 10^8
C 1.83 × 10^27
D 3.37 × 10^27

(by the way ^ means to the power of in this context)

I figured it out! Bq = becquerel, so is the unit. Therefore 10.1-5.5 = 4.6x10^27 Bq, which is the radiation produced from only the NEW atoms in that 30 seconds (find the difference in radiation basically). Now we need to figure out how many atoms we have, so we take the 4.6x10^27 Bq and we divide it by the 8.5 x10^18 Bq to find out how many atoms that had produced that Bq, as we know that each atom produces that much radiation.. so there is 541000000 atoms (new) rounded, then we divide that by 3 as TP is a 3C molecule (so has 3x C atoms) and that should get 1.8 x10^8 :smile:
Reply 4
Original post by isla21!
I figured it out! Bq = becquerel, so is the unit. Therefore 10.1-5.5 = 4.6x10^27 Bq, which is the radiation produced from only the NEW atoms in that 30 seconds (find the difference in radiation basically). Now we need to figure out how many atoms we have, so we take the 4.6x10^27 Bq and we divide it by the 8.5 x10^18 Bq to find out how many atoms that had produced that Bq, as we know that each atom produces that much radiation.. so there is 541000000 atoms (new) rounded, then we divide that by 3 as TP is a 3C molecule (so has 3x C atoms) and that should get 1.8 x10^8 :smile:

thank you sooo much i was struggling so hard on it

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