module 2 Flashcards
Why is it that you can overcome enzyme inhibition of a competitive inhibitor by adding an excess of substrate, although this will not have any effect on a non-competitive inhibitor
In competitive inhibition, the structurally similar molecule to the substrate competes with the substrate for binding at the active site of the enzyme. If you add more substrate to the solution, there will be more substrate than the competing molecule and therefore substrate will be more likely to bind at the active site.
The KM for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8 x 10-2 M, and the KM for the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.4 x 10-4 M. For which substrate does the enzyme have the higher affinity
- The lower Km the higher the affinity, because KM is the substrate concentration at which the reaction is running ath ½ Vmax. Therefore, lower KM means that we need less substrate to get to ½ of Vmax.
- Therefore, the enzyme has a higher affinity to N-acetyltyrosine ethyl ester
Feedback inhibition typically alters the activity of the first enzyme of a metabolic pathway, rather than one of the later enzymes of the pathway. Why is this adaptive?
By inhibiting the initial step of the reaction, we make sure that we do not waste energy producing intermediates that are not going to be used.
In the reaction A + B <—> C + D:
How might the reaction occur in the cell if the DG is very positive?
G is via coupling. This reaction might occur in the cell if it is coupled with the reaction that is very exergonic (has very negative DG) and leads to the overall negative DG
What would ATP synthase do if H+ ions were raised to a high concentration in the mitochondrial matrix and simultaneously decreased to a low concentration in the intermembrane space
- Concentration of H+ ions is higher in the intermembrane space and would normally be pumped into the matrix since the concentration is lower there through the ATP synthase that would
- in the given situation, ATP synthase would pump H+ ions from matrix into the intermembrane space to reach the equilibrium. Simultaneously, ATP would be hydrolyzed and ADP would be produced
Determine which of the substances PQH2, P680+, and A-0 is:
a. The strongest oxidizing agent
b. The strongest reducing agent
c. A mobile electron carrier
- P680+
- is A-0
- PQH2 is the mobile electron carrier that carries electrons between PSII and cyt b6
An investigator radioactively labels CO2 with 14C. He exposes a plant from a temperate climate to the radiolabeled CO2.
In what molecule does the 14C first appear to accumulate stably
- looking at C3 plants
- radiolabelled C from CO2 will be combined with initial receptor (RuBP) and the newly formed molecule will be rapidly converted into PGA (3 C molecule; also an intermediate in glycolysis).
- C+RuBP->PGA
An investigator radioactively labels CO2 with 14C. He exposes a plant from a temperate climate to the radiolabeled CO2.
- Where does the 14C appear in a plant that is well adapted for life in a hot, dry habitat?
- c4 plants
- intermediate produced is a 4-carbon molecule (oxaloacetate or malate
- This intermediate allows C4 plants to open and close their stomata in response to the climate
What do plants do to prevent water loss in a hot, dry climate
- C4 plants
- producing four-carbon intermediate that allows them to open and close their stomata
- closed stromata-> no gas exchange-> no h2o loss/no CO2 diffusion
- can fix CO2 even when its concentration is very low (via PEP carboxylase in mesophyll cells
- CAM plants
* stomata is closed during the day (to avoid water evaporation) and is open at night
photorespiration in C3 plants in hot climate
- RuBP has higher affinity to O2 in hot and dry climate allowing O2 fixation
- O2 concentration would be increasing in the hot
- rubisco uses O2 as a substrate leading to photorespiration in C3 plants
why are C4 plants immune to
photorespiration
C4 plants are protected from photorespiration due to the following adaptations:
- Use different enzyme for CO2 fixation (PEP carboxylase) found in mesophyll cells
- PEP carboxylase has no affinity for O2 and can work on lower CO2 concentration level
- PEP carboxylase enables C3 plants to fix CO2 at the lower levels of CO2 present when the stomata are closed during the heat day
- Fixed CO2 is then transferred to bundle sheath cells ->splits from carrier producing high level of CO2 ->Rubisco produces carbs in Calvin cycle
- Having spatial arrangements enables PEP pathway to fix CO2 at much lower CO2 concentration and then send this fixed CO2 to another area that contains Rubisco (bundle sheath cells) that function at higher CO2 levels
- In bundle-sheath cells Rubisco is not exposed to O2
types of rxn dep on Gibbs
- ΔG<0: exergonic rxn; spontaneous; proceeds towards the lower state of E
- ΔG>0:endergonic rxn; nonspontaneous ; thermodynamically unfavourable
how t would impact rxn rates
- At lower t rxn rates rise w/ increasing t due to higher E of the reactants
- At higher t, rxn rates decrease due to denaturation of the enzymes
enzyme inhibition
- competitive: V max const, increase in Km
- noncompetitive: lower V max, same Km
- uncompetitive: decrease V max and Km
- mixed: decrease in V max, increase/decrease in Km
Enzymes can be very specific in the reactants they can bind and the reaction they catalyze. How are they capable of doing this? Why is this specificity important?
- Enzymes are capable of binding one or a small number of closely related biological molecules b/c the strucute of the active site is very specific
- Specificity is most important in terms of making sure that enzymes only work on one molecule of a very small group of molecules. It is crucial to maintain order within the cell at the particular time