EXAM 2 PREP 8th Grade, Week 2, Day 3 July 3, 2013

But first! POP QUIZ! (1) Name two factors that can affect enzyme activity. (2) An irreversible inhibitor binds ____________ to the enzyme. (3) Name one type of enzyme inhibition and what it does to Vmax of the reaction. (4) What is negative feedback? (5) What is an allosteric enzyme?

BONUS: What is the example that we talked about in class and how many subunits does it have?

Announcements • UNITS – you need to include units with every answer if it

needs them • You need to at least TRY on EVERY homework problem • You need to show your work for everything • You need to write NEATLY and LEGIBLY • You need to READ THE QUESTION CAREFULLY • Pay attention to things like positive vs. negative

DISCLAIMER

• Just because it isn’t in this review doesn’t mean it’s not on the test

Thermochemistry

What kind of reaction is this? • How do you know?

Surroundings    

Energy  released  as  heat  

Exothermic Reaction – energy is lost from the system to the surroundings, the

products are at a lower energy than the reactants

What kind of reaction is this? • How do you know?

Surroundings    

Energy gained by the system

Endothermic Reaction – energy is gained from the surroundings to the system, the products are at a higher energy than the

reactants

UNITS • Specific heat capacity – J/g-C (or J/g-K) • Molar heat capacity - J/mol-C (or J/mol-K)

• When using q = ms∆T or q = mc∆T •  q – J or kJ (whichever I ask for!) •  m – in GRAMS (not kilograms) •  T – can be in K or ⁰C,  just reflect that in your final units

• When using ∆G = ∆H - T∆S •  G – in kJ/mol •  H – in kJ/mol •  T – in K •  S – in J/mol (not kJ/mol)

Conversions • Don’t worry about converting between degrees Celsius

and Kelvin – you won’t have to do this on the exam •  BUT your final unit must be correct

•  JOULES à KILLIJOULES •  Move the decimal point 3 places to the left – number should get

smaller

• KILLIJOULES à JOULES •  Move the decimal point 3 places to the right – the number should

get bigger

Practice Conversions •  136978 J = how many kJ?

•  34.56 kJ = how many J?

•  13.3 J = how many kJ?

•  4567 kJ = how many J?

Practice Conversions •  136978 J = how many kJ?

•  136.978 kJ

•  34.56 kJ = how many J? •  34,560 J

•  13.3 J = how many kJ? •  .0133 kJ

•  4567 kJ = how many J? •  4,567,000 J

Specific heat and heat capacity

“Specific Heat Capacity”

• Represented by “s” • Heat needed to raise 1

gram of a substance 1 degree Celsius

•  q = ms∆t •  For this formula, you need m

in GRAMS

“Molar Heat Capacity”

• Represented by “c” • Heat needed to raise 1

mole of a substance 1 degree Celsius

•  q = mc∆t •  For this formula, you need m

in MOLES

Otherwise you need to convert back and forth!

So let’s practice: •  216 J of energy is required to raise the temperature of

aluminum from 15o to 35oC. Calculate the mass of aluminum. (Specific Heat Capacity of aluminum is 0.90 JoC-1g-1).

Answer: •  216 J of energy is required to raise the temperature of

aluminum from 15o to 35oC. Calculate the mass of aluminum. (Specific Heat Capacity of aluminum is 0.90 JoC-1g-1).

q = m x C x DT m= q/C x DT m= 216J/(0.90J/goC x 20oC ) m= 12g

More Practice •  The initial temperature of 150g of ethanol was 22oC. What

will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol? (Specific heat capacity of ethanol is 2.44 JoC-1g-1).

Answer: •  The initial temperature of 150g of ethanol was 22oC. What

will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol? (Specific heat capacity of ethanol is 2.44 JoC-1g-1).

q = m x C x DT DT = q/m x C DT = 3240J/(150g x 2.44J/goC) DT = 8.85oC Tfinal= 22oC +8.85oC= 30.9oC

More Practice •  145 grams of copper begin at a temperature of 36⁰C  and

end at a temperature of 54⁰C.  The c of copper is 24.78 J/mol-K. How much heat was used in this reaction?

Answer: •  145 grams of copper begin at a temperature of 36⁰C  and

end at a temperature of 54⁰C.  The c of copper is 24.78 J/mol-K. How much heat (in kJ) was used in this reaction?

Must first convert grams of copper to moles. (145g)/(63.55g/mol) = 2.28 moles copper Then q = mc∆T = (2.28)(24.78)(54-36) = (2.28)(24.78)(18) = 1016.971 J = 1.016 kJ

More Practice • An iron bar of mass 869 grams cools from 94⁰C to 74.6⁰.

Calculate the heat released (in kJ) by the metal. Specific heat for iron is 0.444 J/g·⁰C.

Answer: •  An iron bar of mass 869 grams cools from 94⁰C to 74.6⁰.

Calculate the heat released (in kJ) by the metal. Specific heat for iron is 0.444 J/g·⁰C.

q = mC∆T (869g)(0.444 J/g·⁰C)(-19.4⁰C) = -7.49 x 10^3 J = -7.49 kJ

More Practice • Consider two metals, A and B, each having a mass of 100

g and an initial temperature of 20⁰C. The specific heat of A is larger than that of B. Under the same heating conditions, which metal would take longer to reach a temperature of 21⁰C?

Answer: • Consider two metals, A and B, each having a mass of 100

g and an initial temperature of 20⁰C. The specific heat of A is larger than that of B. Under the same heating conditions, which metal would take longer to reach a temperature of 21⁰C?

Metal A Why?

More Practice • A 6.22 kg piece of copper metal is heated from 20.5⁰C to

324.3⁰C.Calculate the heat absorbed (in kJ) by the metal. Specific heat for copper is 0.385 J/g·⁰C.

Answer: • A 6.22 kg piece of copper metal is heated from 20.5⁰C to

324.3⁰C.Calculate the heat absorbed (in kJ) by the metal. Specific heat for copper is 0.385 J/g·⁰C.

q = mC∆T q = (6,220g)( 0.385 J/g·⁰C)(304.3⁰C) = 728.7 kJ

More Practice • At what T is the following reaction spontaneous? Br2(l) à Br2(g) Where ∆H = 30.91 kJ/mol, ∆S = 93.2 J/mol.K.

Answer: • At what T is the following reaction spontaneous? Br2(l) à Br2(g) Where ∆H = 30.91 kJ/mol, ∆S = 93.2 J/mol.K. ∆G = ∆H - T∆S = 0, set equal to 0, for when a reaction will become negative, and thus spontaneous. Rearrange the equation to solve for T. T = ∆H/∆S = (30.91 kJ/mol)/(93.2 J/mol.K) = .332 K

Enthalpy, Entropy, & Free Energy • Enthalpy = H

•  Measure of heat flow, determines endothermic vs. exothermic •  ∆H > 0, endothermic reaction •  ∆H < 0, exothermic reaction

• Entropy = S •  Measure of disorder

•  Free Energy = G •  Energy available to do work, determines spontaneous vs. non-

spontaneous •  ∆G > 0, not spontaneous reaction •  ∆G < 0, spontaneous reaction

Enzyme Kinetics and Inhibition

Estimate Km

Answer:

Km = 1.6-1.7 mM

Enzyme Kinetics • High Km means _________ affinity

•  Why?

•  Low Km means _________ affinity •  Why?

Lineweaver-Burke Plot •  Taking the inverse of the Michaelis-Menten hyperbola

graph gives us the equation of a straight line • Vmax can be found where on the graph? • Km can be found where on the graph?

Lineweaver-Burke Plot •  Taking the inverse of the Michaelis-Menten hyperbola

graph gives us the equation of a straight line • Vmax can be found where on the graph? • Km can be found where on the graph?

Enzyme Inhibition

What’s changing with each type and how?

What would the Lineweaver-Burke plot of all 3 look like?

Answer:

Negative Feedback •  The end product of a metabolic pathway affects its own

production •  A product later in the pathway will stop the earlier reactions that

lead to its own production

A à B à C à D à E

Why would you want to do this?

What kind of enzymes does this show?

Why is the purple line a different shape than the blue line?

Effectors Regulate Allosteric Enzymes • Negative effectors inhibit the enzyme activity – the curve

will shift to the right • Positive effectors increase enzyme activity – the curve will

shift to the left

Equations you will need to be able to use: •  q = ms∆T or q = mc∆T •  ∆Hreaction = H(products) – H(reactants) •  ∆G = ∆H - T∆S • Vo = Vmax[S]/(Km + [S])