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NBME 21 Answers

nbme21/Block 3/Question#37

A 5-year-old boy who has homocystinuria improves ...

The line on the graph labeled ‘B’

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 +6  upvote downvote
submitted by lamhtu(31),

Needing "higher concentrations" of the B6 for enzyme activity is another way of saying Km is higher since more is required for 1/2 vmax activity. Increased Km values result in 1/Km being smaller (closer to 0 on the x-axis), which is demonstrated in answer choice B.

A "normal" enzyme activity in the presence of higher B6 concentrations means that Vmax is not changing, but Km is.

d_holles  In other words, Normal enzyme activity = Vmax Adding more B6 to a mutated enzyme will improve it back to normal (Vmax stays constant, so Km will decreased due to ↑ affinity of the enzyme). +  
wolvarien  why the answer is no C ? +  




 +3  upvote downvote
submitted by sahusema(28),

GUYS! You really have to look at the axis labels. The question says the pt.'s cystathionine synthase activity (y-axis) is NORMAL with an INCREASED amount of pyridoxal phosphate (x-axis). Y-axis is the same as NORMAL, x-axis shifts to the right showing INCREASED pyridoxal phosphate





The Lineweaver Burk Plot is mainly interpreted by its X and Y intercepts. The question states that giving B6 increases the activity to normal levels, which means the activity should be the same as the normal. Hence, their Y intercept should be the same, so we are between choices B and C. A and D do not have the same Y intercept as normal. Higher concentrations of B6 caused activity to become normal, and so Vmax will not be changing and we can cancel A and D. A has a lower Vmax and D has a much higher Vmax. The difference between B and C is in their Km. Moving to the left on the X axis makes the Km lower and tells you that affinity is higher so you would not need more B6. But in our case affinity of the enzyme for B6 is really low, which is why we need a ton more B6.

In summary, we want the same Vmax and a higher Km. We want the "normal" activity (same vmax as normal) and we need higher amounts of B6 for success so affinity of the enzyme for B6 is probably very low.

Choice B has a (-1/Km) value closer to 0 which means Km is lower and affinity of the enzyme for its substrate is super low. This makes sense as giving higher amounts of the "competitive" substrate B6 is helping.

maleehaak  amazing explanation +  




 +1  upvote downvote
submitted by navevan3(1),

For this question, I was thinking that they asked you compare to a “normal” person, so the Vmax wouldn’t change since they state that his activity returns to a normal level. Only the x-axis would since it would be at increased concentrations in comparison.





Remember 1) competitive inhibitors ~ lines Cross at y-intersect 2) Non-competitive inhibitors ~ no line cross





be aware that the x-axis for this question is not "Km" (vitB6 is not the subtrate that's 1/[VitB6]. to simpify, because the Vmax is gonna stay the same, you just need to increase [vitB6] to get the same Vmax (aka shift to right on the x-axis). Don't be fooled around by the question writer.





The question is basically setting up a situation similar to that of a competitive inhibitor. They say that higher concentrations of pyridoxal phosphate increases the activity of the enzyme to normal levels. Similarly, when you have a competitive inhibitor and increase the concentration of substrate, the enzyme can achieve the same Vmax as if without the inhibitor.





 -5  upvote downvote
submitted by hungrybox(250),

Long answer ahead, but bear with me.

HINT: v looks kind of like y, whereas k looks more like x.

y-intercept = 1/Vmax

  • Vmax is the upper limit on how fast a reaction is catalyzed by enzymes.

x-intercept = 1/Km

  • Km is a ranking of how good an enzyme is at binding its substrate. An enzyme with a ranking of 1 is better at binding its substrate than an enzyme with a ranking of 5. (Lower Km = better enzyme)

Note that Vmax, as a measure of performance, can be altered through many things. Meanwhile, Km is a set characteristic of the enzyme, and cannot be altered.

In this example, the enzyme performance (Vmax) is increased by increasing the vitamin cofactor so that it reaches a "normal" activity. However, the enzyme is still inherently shitty due to a congenital defect, so the Km stays the same.

mnemonia  Awesome. +  
ht3  wait line B shows the vmax doesn't change and that the km is getting larger (enzyme is still shitty so larger km) so -1/km would be a smaller number and would approach 0 +1  
lamhtu  You say Km cannot be altered and its staying the same, but the answer of the graph demonstrates a higher Km value. Needing "higher concentrations" of the B6 for enzyme activity is another way of saying Km is higher since more is required for 1/2 vmax activity +1  
sbryant6  Yeah this explanation is wrong. +