ACS BIOCHEMISTRY TESTED QUESTIONS
AND COMPLETE SOLUTIONS
◉ Mixed Inhibition Graph
Answer: Allosteric inhibitor that binds either E or ES.
Pivot point is between X-intercept and Y-intercept.
◉ Non-Competitive Inhibition Graph
Answer: Form of mixed inhibition where the pivot point is on the
x-axis. Only happens when K1 is equal to K1'.
◉ Ionophore
Answer: Hydrophobic molecule that binds to ions and carries
them through cell membranes. Disrupts concentration gradients.
◉ ΔGtransport Equation
Answer: ΔGtransport = RTln([S]out / [S]in) + ZFΔΨ
◉ Pyranose vs. Furanose
Answer: Pyranose is a 6-membered ring.
Furanose is a 5-membered ring.
◉ Mutarotation
,Answer: Conversion from α to ß forms of the sugar at the
anomeric carbon.
◉ Anomeric Carbon
Answer: Carbon that is cyclized. Always the same as the aldo or
keto carbon in the linear form.
◉ α vs. ß sugars
Answer: α form has -OR/OH group opposite from the -CH2OH
group.
ß form has -OR/OH group on the same side as the -CH2OH group.
◉ Starch
Answer: Found in plants. D-glucose polysaccharide. "Amylose
chain". Unbranched. Has reducing and non-reducing end.
◉ Amylose Chain
Answer: Has α-1,4-linkages that produce a coiled helix similar to
an α-helix. Has a reducing and non-reducing end.
◉ Amylopectin
Answer: Has α-1,4-linkages. Has periodic α-1,6-linkages that
cause branching. Branched every 24-30 residues. Has reducing
and non-reducing end.
,◉ Reducing Sugar
Answer: Free aldehydes can reduce FeIII or CuIII. Aldehyde end is
the "reducing" end.
◉ Glycogen
Answer: Found in animals. Branched every 8-12 residues and
compact. Used as storage of saccharides in animals.
◉ Cellulose
Answer: Comes from plants. Poly D-glucose. Formed from ß-1,4-
linkage. Form sheets due to equatorial -OH groups that H-bond
with other chains.
◉ Chitin
Answer: Homopolymer of N-acetyl-ß-D-glucosamine. Have ß-1,4-
linkages. Found in lobsters, squid beaks, beetle shells, etc.
◉ Glycoproteins
Answer: Carbohydrates attached to a protein. Common outside of
the cell. Attached at Ser, Thr, or Asn residues.
◉ Membrane Translayer Flip-Flop
Answer: Typically slow, but can be sped up with Flippase,
Floppase, or Scramblase.
, ◉ Membrance Fluidity
Answer: Membrane must be fluid. Cis fats increase fluidity, trans
fats decrease fluidity.
◉ Type I Integral Membrane Protein
Answer: Membrane protein with C-terminus inside and N-
terminus outside
◉ Type II Integral Membrane Protein
Answer: Membrane protein with N-terminus inside and C-
terminus outside
◉ Type III Integral Membrane Protein
Answer: Membrane protein that contains connected protein
helices
◉ Type IV Integral Membrane Protein
Answer: Membrane protein that contains unconnected protein
helices
◉ Bacteriorhodopsin
Answer: Type III integral membrane protein with 7 connected
helices.
◉ ß-Barrel Membrane Protein
AND COMPLETE SOLUTIONS
◉ Mixed Inhibition Graph
Answer: Allosteric inhibitor that binds either E or ES.
Pivot point is between X-intercept and Y-intercept.
◉ Non-Competitive Inhibition Graph
Answer: Form of mixed inhibition where the pivot point is on the
x-axis. Only happens when K1 is equal to K1'.
◉ Ionophore
Answer: Hydrophobic molecule that binds to ions and carries
them through cell membranes. Disrupts concentration gradients.
◉ ΔGtransport Equation
Answer: ΔGtransport = RTln([S]out / [S]in) + ZFΔΨ
◉ Pyranose vs. Furanose
Answer: Pyranose is a 6-membered ring.
Furanose is a 5-membered ring.
◉ Mutarotation
,Answer: Conversion from α to ß forms of the sugar at the
anomeric carbon.
◉ Anomeric Carbon
Answer: Carbon that is cyclized. Always the same as the aldo or
keto carbon in the linear form.
◉ α vs. ß sugars
Answer: α form has -OR/OH group opposite from the -CH2OH
group.
ß form has -OR/OH group on the same side as the -CH2OH group.
◉ Starch
Answer: Found in plants. D-glucose polysaccharide. "Amylose
chain". Unbranched. Has reducing and non-reducing end.
◉ Amylose Chain
Answer: Has α-1,4-linkages that produce a coiled helix similar to
an α-helix. Has a reducing and non-reducing end.
◉ Amylopectin
Answer: Has α-1,4-linkages. Has periodic α-1,6-linkages that
cause branching. Branched every 24-30 residues. Has reducing
and non-reducing end.
,◉ Reducing Sugar
Answer: Free aldehydes can reduce FeIII or CuIII. Aldehyde end is
the "reducing" end.
◉ Glycogen
Answer: Found in animals. Branched every 8-12 residues and
compact. Used as storage of saccharides in animals.
◉ Cellulose
Answer: Comes from plants. Poly D-glucose. Formed from ß-1,4-
linkage. Form sheets due to equatorial -OH groups that H-bond
with other chains.
◉ Chitin
Answer: Homopolymer of N-acetyl-ß-D-glucosamine. Have ß-1,4-
linkages. Found in lobsters, squid beaks, beetle shells, etc.
◉ Glycoproteins
Answer: Carbohydrates attached to a protein. Common outside of
the cell. Attached at Ser, Thr, or Asn residues.
◉ Membrane Translayer Flip-Flop
Answer: Typically slow, but can be sped up with Flippase,
Floppase, or Scramblase.
, ◉ Membrance Fluidity
Answer: Membrane must be fluid. Cis fats increase fluidity, trans
fats decrease fluidity.
◉ Type I Integral Membrane Protein
Answer: Membrane protein with C-terminus inside and N-
terminus outside
◉ Type II Integral Membrane Protein
Answer: Membrane protein with N-terminus inside and C-
terminus outside
◉ Type III Integral Membrane Protein
Answer: Membrane protein that contains connected protein
helices
◉ Type IV Integral Membrane Protein
Answer: Membrane protein that contains unconnected protein
helices
◉ Bacteriorhodopsin
Answer: Type III integral membrane protein with 7 connected
helices.
◉ ß-Barrel Membrane Protein
