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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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submitted by lamhtu(24),

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 ? +  

 +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.

 +1  upvote downvote
submitted by sahusema(12),

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

 +0  upvote downvote
submitted by minhphuongpnt07(0),

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.

 -1  upvote downvote
submitted by stapes2big(3),

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.

 -1  upvote downvote
submitted by usmleuser007(84),

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

 -4  upvote downvote
submitted by hungrybox(174),

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 +  
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. +