Sequence from this region shown below

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NAME________________________________TA______________Section #______

7.013 Spring 2005 Problem Set 1Solutions

FRIDAY February 11, 2004

Problem sets will NOT be accepted late.

Question 1

a) Circle the correct answer.

i) Homo sapiens emerged ____________years ago.

5,000 50,000 500,000 5,000,000 50,000,000

ii) There are __________________billion letters (bases) in the human genome.

one three thirty 3 hundred

iii) An alpha helix is an example of _________________structure.

primary secondary tertiary quarternary

b) What are the four major types of biological molecules discussed in lecture? What monomers make up these molecules? Give one important function of each type of biological molecule in the cell.

Name of molecule Monomer One function of the Polymer

Lipid Fatty Acid,

Triglyceride

Hormones, Membranes, Energy storage

Nucleic Acids Nuceleotide Carry information

Carbohaydrate Sugar Energy, Protein Modification, Protein Amino Acid Enzymes, motility, structural roles

MIT Department of Biology

7.013: Introductory Biology - Spring 2005

Instructors: Professor Hazel Sive, Professor Tyler Jacks, Dr. Claudette Gardel

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c) Name the molecules depicted below.

O CH₂ O P O

N N N N

NH₂

O H H

O^- -

H H

H

O P O

O -

O

H H

O H C₂

H H

adenine

thymine

H

O P O

O -

O

H H

O H C₂

HO

H H

H N

HN O

O

N N N N

NH₂

adenine

CH₃ DNA

O CH₂ O P

O N

N N N

NH2

O H H

O^- -

H H

O P O

O -

O

H H

O H C₂

H H

adenine

uracil

O P O

O -

O

H H

H C₂ O

HO

H H

N HN O

O

N N N N

NH₂ adenine

OH OH

OH

C C C C C C C COO

O

N

H N N N

CH

H H H

O O

(CH ) (CH )

C C

O O NH

+ CH H C CH

H H H H

NH

C O

NH NH +

CH

3 3

2

2 2

2 2 2 2

2

3 3

RNA

Protein

d) Circle one peptide bond if there is any in the figures above.

DNA

RNA

Peptide, protein, polypeptide

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Question 2

a) Draw the energy profile for this reaction. Refer to Chapter 6 in the textbook.

A + B ---> C + D ΔGo' = -3.4 kcal/mol On the diagram be sure to…

1) show relative energy levels of the reactants and the products.

2) label the axes.

3) label reactants and products.

4) indicate the energy of activation.

5) indicate ΔG.

Free Energy (G) kcals/mole

fructose 6-phosphate + ATP

fructose 1,6-bisphosphate + ADP

Energy of activation

ΔG = -3.4 kcals/mole

Reaction Progress

o

b) An enzyme _____________ the activation energy of a reaction.

lowers raises does not affect

c) An enzyme _____________the ΔG of a reaction.

lowers raises does not affect

d) Using a dashed line in the above diagram, draw the energy profile in the presence of an enzyme.

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Question 3

Some receptors are transmembrane proteins found on the cell surface.

CH2

CH₂ CH2

CH2

CH₂ CH2

CH₂ C O O

C C

O O C

CH₂ CH2

CH2

CH₂ CH2

CH2

H H H H₂C

O P O O

O

CH2

CH₂ CH₂ CH2

CH2

CH₂ CH2

CH2

C O O

C C

O O C CH2

CH2

CH₂ CH2

CH2

CH₂ CH₂ CH2

H H H H2C

O P O O

O

CH2

CH₂ CH2

CH2

CH₂ CH2

CH2

C O

O C C O O C CH2

CH2

CH₂ CH2

CH2

CH₂ CH2

H H H H2C

O O P O

O CH2

CH² CH2

CH2

CH₂ CH2

C O

O C C O O C

CH2

CH₂ CH2

CH2

CH² CH2

H H H H₂C

O O P O

O

Membrane

–

– –

– – – –

–

CH₂ CH2

CH2

CH² CH₂ CH2

CH2

CH₂

C O O C C

O O C CH₂ CH2

CH2

CH₂ CH² CH2

CH2

CH₂ CH2

CH₂

H H H H2C

O P O O

O CH2

CH2

CH₂ CH2

CH2

CH² C O O C C

O O C

CH² CH2

CH2

CH₂ CH2

CH2

H H H H2C

O P O O

O

Membrane

– –

CH₂ CH2

CH2

CH₂ CH₂ CH₂ CH2

CH2

C O O C C

O O C

CH2

CH₂ CH₂ CH₂ CH2

CH2

CH₂ H H H H2C

O P O O

O

CH2

CH₂ CH2

CH2

CH₂ CH2

C O O C C

O O C CH2

CH₂ CH2

CH2

CH₂ CH2

CH2

H H H H2C

O P O O

O

CH2

CH₂ CH2

CH2

CH₂

C O

O C C O O C CH₂ CH2

CH2

CH₂ CH2

CH2

H H H H₂C

O O P O

O CH2

CH2

CH₂ CH2

CH2

C O O C C O O C

CH2

CH₂ CH2

CH2

H H H H2C

O O P O

O

–

– –

– – – –

–

CH CH CH CH CH CH CH CH CH CH C O O C C O O C

CH2

CH₂ CH2

CH2

H H H

CH CH CH CH CH CH CH CH CH CH

C O

O C C O O C CH₂ CH2

CH2

CH₂ CH2

CH2

H H H

2 2 2 2 2 2 2 2 2 2

H2C O O P O

O

2 2 2 2 2 2 2 2 2 2

H₂C O O P O

O –

–

– –

intracellular extracellular

growth factor receptors

a) The majority of the molecules that constitute a membrane belong to what

class of macromolecules? _______________________Lipid, Phospholipid_________

b) Explain the important qualities/properties of these molecules that allow them to form membranes. Phospholipids possess hydrophilic “heads” and hydrophobic “tails” that allow them to assemble into a bilayer containing a hydrophobic core when in an aqueous environment.They are Amphipathic. One hydrophilic end that faces the aqueous solution, and one hydrophobhic end that faces another hydrophobic moiety on another molecule.

Receptor proteins

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A smaller schematic of the receptor is shown here.

]

Sequence from this region shown below

= transmembrane region

c) Which stretch of amino acids in the above sequence is within the interior of the

transmembrane region of the receptor? Circle these amino acids and briefly explain your reasoning below.

The circled amino acids have hydrophobic side chains that reside within the hydrophobic interior of the membrane.

When a “LIGAND” binds to the extracellular domain of the receptor, a conformational change occurs in the receptor. Ligand binding causes dimerization of two adjacent receptors in the cell membrane. Upon dimerization, the intracellular domains of the receptors become activated. See schematic below.

active domains intracellular domain

ligand

Receptor 1 Receptor 2

inactive domains plasma

membrane extracellullar ligand- binding domain

ligand

NH3+···Ile-Val-Phe-Leu-Ala-Val-Trp-Met-Phe-Arg-Lys-Thr-His···COO^-

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c) Regions of the two receptors that interact upon dimerization are drawn below. In parts (i - iv) below, name the strongest type of interaction (choose from; hydrogen bond, ionic, covalent, van der Waals) that occurs between the side chains of the amino acids indicated.

O

CH

₂

S

CH

₂

+

CH C

O CH

₃

S

C O

S

Asp68

OH CH

₂

CH

₂

CH

₂

CH

₂

Ser53

Cys36

Ala45

-

Phe50

Cys82

Cys75

CH

₂

CH

₂

H

₂

N

Gln12 CH

₂

CH

₂

CH

₂

CH

₂

H

₃

N

Lys65

H

3

C

H

₃

C Val98

Receptor 2 Receptor 1

H

Interacting Side chains Type of interaction

i) Phe50 : Val98 van der Waals

ii) Asp68 : Lys65 ionic

iii) Cys75 : Cys82 covalent

iv) Ser53 : Gln12 hydrogen

d) Gln12 is the 12^th amino acid in primary sequence. Val 98 is the 98^th amino acid in the primary sequence. Explain how these amino acids are far apart in the primary sequence of the protein yet are close to each other in the region of the protein diagrammed above.

When the protein folds into its final form, amino acid residues that are far apart in the primary structure can be closely aligned to one another.

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e) Molecular interactions between the two receptors are important for dimerization.

Thus, substitution of certain amino acids in the protein can affect receptor dimerization.

Predict whether the receptors will or will not be able to dimerize given the substitutions (i - iv) below. EXPLAIN your reasoning.

i) Asp68 → Arg

The receptors will not be able to dimerize because this substitution replaces a negatively charged amino acid with a positively charged amino acid. The ionic bond between Asp68 and Lys65 is disrupted, and a repulsion occurs.

ii) Ser53 → Thr

The receptors will be able to dimerize because this substitution replaces a polar amino acid that can participate in hydrogen bonds with another such amino acid.

iii) Phe50 → Asn

The receptors will be able to dimerize even though this substitution replaces a hydrophobic acid with a polar amino acid because the van der Waals forces remain.

(Within the region diagrammed, the close proximity of the charged species makes it unlikely that hydrophobic interactions are a key force in the interaction between the two receptors.)

iv) Val98 →^Ile

The receptors will be able to dimerize because this substitution replaces a

hydrophobic acid with another hydrophobic amino acid and the van der Waals forces remain.

f) Substitution of one amino acid, Cys75 → Gly, leads to dimerization of the receptors with or without ligand. Provide a brief explanation for this observation.

This substitution positions two cysteine residues opposite each other. These two residues can form a disulfide bond and thus covalently link the two receptors together.

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Question 4

You have identified a new, rapidly reproducing species of dog that you have named the “mitbull”. The mitbull is diploid and two autosomal chromosomes are shown. One chromosome carries gene A with two alleles designated A and a. The other chromosome carries gene B with two alleles designated B and b.

a) Diagram mitosis in a heterozygous (AaBb) mitbull cell shown below. Draw your diagrams in the outlines of the ovals adjacent to the stages. Include the alleles and the mitotic spindle in your diagrams.

Prophase

Metaphase Mitbull cell

A

A a a

A a

B b

B B b b

A

a

B A

a

B

b b

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b) Indicate the genotype of cells that would result after telophase. _____AaBb_______

Anaphase

Telophase phase

A

a

A a

B B

b b

A

a

B

b aA

B b

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c) A cell in Metaphase I is shown below where a crossover event takes place. Diagram the indicated stages in meiosis. Designate the alleles and the spindle.

Metaphase II

Telophase I

A A a a

B B b b

Telophase II

B B b

A a A a

b b

B B b

a a

b b

A A

B B

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d) You determine another gene, gene D, maps 20 cM or map units away from gene A.

i) Given the cell with chromosome configuration below, what would be the genotypes of all the gametes if a recombination takes place between A and D?

A d

A D a d A d a D

a D

ii) Circle the genotypes of the recombinant gametes above.

iii) At what frequency do you expect each of the recombinant genotypes to occur? ___10%___

C C

O O H

NH3

CH2CH2CH2CH2 NH3+

ALANINE

(ala) ARGININE

(arg) ASPARAGINE

(asn)

+

-

ASPARTIC ACID (asp)

CYSTEINE (cys)

-

+

C N CH2

CH2

H H

H - C

O O

+

GLUTAMIC ACID (glu)

GLUTAMINE (gln)

GLYCINE (gly)

HISTIDINE (his)

ISOLEUCINE

(ile) LEUCINE

(leu)

LYSINE (lys)

METHIONINE

(met) PHENYLALANINE

(phe) PROLINE

(pro)

SERINE (ser)

THREONINE

(thr) TRYPTOPHAN

(trp) TYROSINE

(tyr)

VALINE (val)

STRUCTURES OF AMINO ACIDS at pH 7.0

C N

H H H

C C

O O H

NH3

CH3

-

+

C C

O O H

NH3

CH2CH2CH2 N H

C NH2

NH2

-

+

C C O O H

NH3

CH2 C O NH2

-

+

C C O O H

NH3

CH2 C O O -

+

C C O O H

NH3

CH2 SH -

+

C C

O O H

NH3

CH2CH2

O O C -

+

C C

O O H

NH3

CH2CH2

O C

NH2

-

+

C C

O O H

NH3

H -

+

C C

O O H

NH3

CH2

C N N C H

H H

H -

+

C C

O O H

NH3

C H

CH3

CH2CH3

-

+

C C

O O H

NH3

CH2 C H

CH3

-

+

-

+

C C

O O H

NH3

CH₂CH₂ S CH₃ -

+

C C

O O H

NH3

CH2

H H

H

H H -

+

C C

O O H

NH3

CH2 OH -

+

C C O O H

NH3

C H

OH CH3

-

+

C C

O O H

NH3

CH2

-

+

C C

O O H

NH3

CH2 OH

H H

H H -

+

C C

O O H

NH3

C H

CH3

-

+