1
Question 1
You are studying the synthesis of the amino acid tryptophan in bacteria. The enzymes
TrpA, TrpB, TrpC, TrpD, TrpE and AroH are all required for tryptophan synthesis. In the presence of tryptophan, wild-type bacteria do not synthesize any of these enzymes; however, in the absence of tryptophan, all of these enzymes are synthesized at high levels.
a) Theoretically speaking, if the synthesis of the above enzymes is negatively regulated i) what change to the repressor protein would cause the enzymes to be synthesized even in the presence of tryptophan?
Inactivation of the functional repressor so that it cannot bind the operator. ii) what change in the operator sequence would cause the enzymes to be synthesized even in the presence of tryptophan?
A mutation in the sequence of the operator such that it no longer binds the repressor. iii) what change in the repressor protein would cause the inhibition of enzyme synthesis, even in the absence of tryptophan?
A change that increases the affinity of the repressor for the operator, i.e., a constitutively active repressor.
b) If the synthesis of the above enzymes is positively regulated, i) what change in the activator protein would cause the enzymes to be synthesized even in the presence of tryptophan?
A change that increases the affinity of the activator for the operator, i.e., a constitutively active activator.
OR…A change in the activator such that tryptophan can no longer bind. ii) what change in the activator protein would prevent synthesis of the enzymes, even in the absence of tryptophan?
A mutation in the sequence of the activator gene, such that a nonfunctional activator protein is produced. iii) what change in the operator sequence would prevent synthesis of the enzymes, even in the absence of tryptophan?
A mutation in the operator sequence rendering it unable to bind activator.
c) By mutational analysis you identify two regions of DNA that are important in the regulation of tryptophan synthesis. The first of these regions, called trpR , is a gene that encodes a DNA-binding protein. The second region is a DNA sequence to which the trpR gene product binds, called trpO . Analysis of three bacterial strains with different genotypes at the trpR and trpO loci yields the following results:

i) Is the control of tryptophan synthesis likely an example of positive or negative regulation?
The synthesis of tryptophan is under negative regulation. ii) Is the protein made from the trpR gene an activator or a repressor?
The trpR protein acts as a repressor.
d) Experiments to monitor the presence or absence of enzyme made from the trpC , trpD , trpE and aroH genes in some isolated mutants yields the following results:
2 i) List all mutants that have mutations which affect production of any functional TrpC? mutant 1, mutant4, and mutant 5 ii) List all mutants that have mutations which affect production of any functional TrpD? mutant 2, mutant 4, and mutant 5 iii) List all mutants that have mutations which affect production of any functional AroH? mutant 3 and mutant 5 e) Is the above data consistent with the theory that any of these genes are in the same operon (and therefore under the control of a single promoter)? Why? Which genes would be contained in this operon?
The absence of functional enzymes from three loci as in the case of mutant 4 suggests that several genes could be contained within a single operon. All genes of the operon would be under control of the same promoter, which in mutant 4 would is mutated and nonfunctional, resulting in no synthesis of any of these enzymes. This operon would contain the trpC, trpD and trpE genes. f) How could you explain mutant 5 in terms of regulatory mutations?
Though in different regions of the DNA, both the trpC/trpD/trpE operon and the aroH gene could be under regulation of the same repressor protein trpR. Mutant 5 would have a mutation in the trpR gene that knocked out functional trpR repressor. Therefore all genes would be expressed both in the presence and absence of tryptophan.

3
While working with mice as a UROP, you discover a very important catabolic enzyme.
You make mutations in the gene encoding this enzyme such that the mutant enzyme is always active. Mice carrying this mutant gene show rapid and permanent weight loss with no other side effects. YOU HAVE DISCOVERED THE MAGIC BULLET THAT
MILLIONS OF PEOPLE WORLDWIDE HAVE BEEN SEEKING FOR ENERATIONS!
But… you need the human version of this gene.
To begin, you choose to make a human genomic DNA library. a) In the space below, briefly outline how you would make a genomic DNA library in bacteria. Include the terms: genomic DNA, restriction enzyme, plasmid, ligation, transformation, petri plates, probe, and hybridize.
1. Cut human genomic DNA with a restriction enzyme.
2. Cut the plasmid vector that has a selectable marker (like ampicillin resistance) with the same restrictionenzyme.
3. Ligate the genomic DNA into the vector.
4. Transform the ligation mix into E. coli.
5. Plate the transformed E. coli onto solid medium in petri plate that contains ampicillin.
6. Make a probe and hybridize it to DNA from the library
You are successful in finding the human homolog. You have pinpointed the colony that contains a plasmid, called plib1 , with the human homolog. You isolate the plasmid, clone the gene and designate it the svelte gene. Ideally, you want to express lots of human enzyme so that you can do a complete biochemical study of the protein. Therefore, you need to move the svelte gene from the plib1 plasmid into an expression vector. You have a great expression vector, p7.01MIT
, for use in E. coli. A diagram of the expression vector p701MIT and the plib1 are shown below with their unique restriction enzyme sites. b) You want to insert a DNA fragment containing both the svelte gene and the kanamycin resistance gene ( kan
R
) into the p701MIT expression vector. i) What enzyme(s) would you use to cut plib1 to obtain a single fragment containing both the svelte gene and the kanamycin resistance gene ( kan
R
)? What size fragments would you get?
Choice 1: SalI and BamHI: 1129 + 650
Choice 2: StyI and BamHI: 1028 + 751 ii) What enzyme(s) would you use to cut p701MIT ? What size fragments would you get?
Choice 1: SalI and BamHI: 2996 + 205
Choice 2: StyI and BamHI: 2495 + 706

4 c) After ligation, you transform your new vector p701MIT containing both the svelte gene and the kan
R gene into an E. coli strain. i) Prior to transformation, this strain of E. coli should be... (circle all that apply.) ampicillin resistant. ampicillin sensitive.
kanamycin resistant. kanamycin sensitive.
ii) How would you detect which cells have the vector containing the svelte gene?
Plate the transformed cells on medium containing both ampicillin and kanamycin. iii) What is the advantage of including the kan
R gene on the fragment that you cloned into p701MIT ?
This allows you to select against any cell that was transformed with the vector alone (i.e., with a vector that did not carry any insert), or with the vector carrying the wrong insert.
a) On the gel below, mark places where you expect to find bands. Label each band with the number of bases that the fragment would contain.
For the BamHI and StyI cloning:

For the SalI and BamHI cloning: b) primer:
5’ GAGTATCAGT 3’ c) What is the sequence of the newly synthesized DNA?
5’ GAGTATCAGTAGCGGGCCTTTACGCCCAGCAT
primer
5

You are a genetic counselor at a major Boston hospital. A couple (shown in the pedigree below as 1 and 2) comes to you and wants to know what the probability is that the child they are expecting will have big ears.
Recently, the gene that results in big ears was cloned and sequenced. The region flanking the ear size gene is shown below.
a) You want to amplify this gene from each parent using PCR. Give the sequence of two primers for this PCR reaction. Label 5’ and 3’ ends.
primer A: 5’ GTCCTGATTT 3’ primer B: 5’ ATTGGACCTA 3’ b) Below is a schematic of the same region. On the diagram below draw where each primer would bind.
6 c) In a schematic like that in b) draw the DNA strands that would be present after 2 rounds of PCR. Include the primers where appropriate on each strand.

A restriction map of this region is shown below:
The PCR-amplified DNA from each family member was digested with Hind III, Nde I, and Bgl II. The digested DNA was separated by gel electrophoresis and the results are shown below.
Individual
7 d) Is the big-eared phenotype dominant or recessive over the normal phenotype? How did you determine this?
The big-eared phenotype is recessive over the normal phenotype. From the gel, you can see that individual two has both the wild-type gene and the mutant gene. Individual two does not show the big-eared phenotype. e) Given this family pedigree and gel, what are the chances that individuals 1 and 2 will have a child with big ears? Please explain.
Individuals 1 and 2 have a 50% chance of having a child with big ears.
Ee X ee: 1ee (affected) : 1Ee (carrier, normal ears)