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.

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.

Some of these explanations really doing the most. If the enzyme is able to increase to normal activity with a higher level of the substrate, that means we’re in a competitive inhibitor situation. Competitive always crosses on the graph, so we can eliminate A and D.

Now we analyze Km. We want small Km’s for potent enzymes. Size does not always matter 😉. Except in this case we don’t want that potency, we want to look for a 1/(really big Km), (value that is closer to zero), bc we want a weak big Km to be able to overpower the competitive inhibitor. Comparing the remaining 2 choices, B has a Km value closer to zero so there’s your answer.

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.

So i found this question ridiculous... the explanations provided did not really clear it up for me, but i think i understand it now (please correct me if i'm wrong).

First and foremost, I'm not even sure if this is truly a lineweaver-burke plot, since the X-axis is not 1/Km.

But, more generally, I tried to understand what happened at both extremes of [PLP] and Velocity.

PLP: Ask yourself: where on the graph is [PLP] highest? Well, if the X-axis is equal to 1/PLP, PLP would be at its highest at the value closest to 0 on the x-axis. (think: what is 1 divided by a billion? VERY CLOSE TO ZERO). This would correspond to a point along the y-axis for the enzymes activity. In other words, PLP is at its highest concentration at the graphs Y-intercept.

Remember what we also said: when the mutated enzyme is around lots of PLP, it basically functions like a nml enzyme. This means, that the Vmax of the mutated enzyme is equivalent to the Vmax of the normal enzyme. Therefore, The 1/Vmax of both enzymes are equivalent... at the y-intercept! This is big! If you get this far, then you can eliminate (A) and (D), since these graphs have different Y-intercepts (different V-maxes) for their respective enzymes.

Moving on...

Velocity (V): When V is at its highest (i.e. Vmax), the value of 1/V would be at its lowest. (Again: think of it practically: what is 1 divided by 1 billion? ~0!) Therefore, the value of 1/V is at its lowest when the y-axis is equal to 0. Therefore, The Vmax of this occurs when the y-axis is close to zero... Heres the thing tho.. these enzymes require different levels of PLP to operate at V-max. Remember, the V-max of the MUTATED enzyme will happen at a much HIGHER [PLP]. Or conversely, the V-max of the NORMAL enzyme will happen at a much LOWER [PLP]...

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.

Nobody has answered this question through the lens of a learning objective quite yet so here's my 2 cents: Km is INCREASED bc this patient has an enzyme with an overall LOWER affinity (can be corrected to normal after an increase in substrate) which is characteristic of COMPETITIVE inhibition, which is why the line of the pt. is one of competitive inhibition on a LB plot

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.

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

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.