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A level Biology: Post your doubts here!
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The difference between the 2 graphs is the availability of Oxygen. In the 1st one it is present but not in 2nd.
The rate of uptake of 3C sugar is the same, with or without Oxygen which means that active transport is not involved and the uptake is by diffusion...the reverse goes for 6C sugar, the rate of which has significantly decreased in the absence of O2, so it is transported by active transport.
HI guys, I was solving this question. The "less turgid" aspect is clear to me. However, I don't understand why it isn't "less negative" as in option D.
The mark scheme says B. The water needs to move out of the cells to ensure that final length is less than initial. and water moves from less negative value-> more negative value right? I don't know where I am messing up.
The mark scheme says B. The water needs to move out of the cells to ensure that final length is less than initial. and water moves from less negative value-> more negative value right? I don't know where I am messing up.
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can anyone plz tell me the answer and why it is the answerSo, first of all, you can eliminate B and C because substrate concentration will never increase in an enzyme reaction, it would decrease since it is being used up to produce the products (whose conc is increasing). Now the correct answer is D but it is easy to confuse A to be the answer. You would initially think of it as A saying that enzymes are denatured so substrate concentration will remain as how much it was just before. But if you think practically, we aren't varying temperature during the course of the reaction, we're carrying out the whole reaction AT different temperatures. There the correct answer is D where: as temperature increases, enzyme activity increase, so substrate conc. decreases. And then the optimum temperature is exceeded so, at any temperature greater than that, the reaction would not take place, leaving substrate concentration to be how it was initially, in the very beginning.
I hope this cleared it
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Thank youSo, first of all, you can eliminate B and C because substrate concentration will never increase in an enzyme reaction, it would decrease since it is being used up to produce the products (whose conc is increasing). Now the correct answer is D but it is easy to confuse A to be the answer. You would initially think of it as A saying that enzymes are denatured so substrate concentration will remain as how much it was just before. But if you think practically, we aren't varying temperature during the course of the reaction, we're carrying out the whole reaction AT different temperatures. There the correct answer is D where: as temperature increases, enzyme activity increase, so substrate conc. decreases. And then the optimum temperature is exceeded so, at any temperature greater than that, the reaction would not take place, leaving substrate concentration to be how it was initially, in the very beginning.
I hope this cleared it
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The active transport is enabled by cell surface membrane rather than the plasmodesmata
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Ugh omg I guess I confused a bit with the wordings
Tysmm
It's wrong to say the ke is the highest because even above the given temperature the ke will increase, point B was rather supposed to be that both the enzyme and substrate posses the activation energy, therefore, they are at their optimum working.
And it's obvious why the rest except A is wrong.
A is right because what's happening is that your substrate molecule will form hydrophobic interactions and hydrogen bonds with the active site of the enzyme to form enzyme-substrate complex ,
Hope this helped.
When you define endocytosis, we say it is the inwards folding of the cell surface membrane with the help of the cytoskeleton, which will form a vesicle containing the molecule in and fusion by cytokinesis using ATP.
It is basically through the phospholipid bilayer,
and in such cases when you overthink and get confused give yourself examples.
Like facilitated is glucose, osmosis is water, active transport are like mineral ions, K+ for instance. All these molecules are very small and therefore it makes sense.