5’ T a c t c a t c g a 3’ ......................… Promotor; transcription right to left.

    D. AAU; UGA. In the beginning of peptide synthesis, the AUG start codon (for met) goes in the P site and the next codon, the one for the second amino acid (AAU for asn in this case) is the first codon that goes in the A site. The special initiator met-tRNA lines up met in the P site and the asn-tRNA lines up asn in the A site to get peptide synthesis started. At the end of peptide synthesis, the CGA codon, the one before the stop codon, is the last codon to be translated, but not the last codon to fit in the A site. Peptide synthesis stops when the CGA codon is in the P site and the stop codon is in the A site because there is no tRNA that can match the stop codon.

2. A. Sense strand is the bottom (= strand that matches mRNA). Top strand is transcribed. Any one gene is always transcribed from the same strand, in the same direction. Note:  It is customary to write the sense strand 5' to 3' and the transcribed strand in the other orientation. (It is also common to just write the sense strand and omit the "other" or transcribed strand.)

    B. Mutant 1 mRNA is GAA; amino acid for wild type and mutant 1 is glu. See the code table for the rest. Note that the triplets listed in the table are the codons -- the triplets in the mRNA, not the ones in the tRNA. (The triplets in the tRNA are  anticodons).

    C-1. In this case, the mutation has simply changed the DNA and mRNA from one codon for glu to a different codon for the same amino acid. The DNA (genotype) is changed, but the protein and its function should be unaltered. So the appearance and properties of the organism (phenotype) should be unaltered.  This is an example of the "degeneracy" of the code -- there is more than one codon for many amino acids.

    C-2. Yes. Both codons specify the same amino acid (that's degeneracy), so it would be reasonable for the same tRNA to read both codons. The same tRNA can read them both because of wobble. The same base in the 'wobble'position' of the tRNA anticodon (the 5' end or first base of the anticodon) can pair with either G or A in the  'wobble position' of the mRNA codon (the 3' end or last base of the codon). Note that wobble is not the same thing as degeneracy -- not all degenerate codons can be read by the same tRNA. Consult your handout or texts to see why the "wobble" base  in the tRNA must be U, and how it pairs with both G and A.

    C-3. No. This codon specifies a different amino acid, so you need a different tRNA.

    C-4. Asp and glu are very similar, so changing one for the other usually doesn't have a big effect on the shape or function of a protein.

    D. 1. Glu and val are quite different, so changing one for the other is more likely to have an effect on shape and/or function than the change of glu to asp. (This is the mutation responsible for sickle cell disease. (See text or lecture #5 notes.)

    E-1. This sort of mutation causes a stop codon where a codon for an amino acid is supposed to be. A stop in the middle of a gene makes no sense; it is usually much more disruptive than a change of one amino acid to another, which merely changes the "sense." The effect of a change in "sense" can be mild or severe, depending on the actual amino acid change. (Compare mutants 2 & 3. )The effect of a change from sense to nonsense is almost always drastic (see below).

    E-2. A misplaced stop leads to production of a shorter protein that almost never works at all. (The shorter protein only works in a few rare cases where the mutation is very close to the 3' end of the translated part of the mRNA, so only a short part of the COOH end of the protein is missing.) A change of one amino acid to another often leads to an abnormal protein that works somewhat, if not as well as the original.

    E-3. None. No new loaded tRNA's can be added, because there is no tRNA with an anticodon to match the stop codons. There is one "loaded" tRNA on the ribosome already -- the one with the growing chain (in the P site), but you can't add any more.

C2005/F2401 '06 -- Recitation Problems #8 -- Hints

1. A. Which end of the chain shown is the 5' end?

    B. Is the carboxyl end the first end made? The last end made?

    C. Which end of the RNA shown corresponds to the 'first 10 bases of the template strand?'

    D. Where does the start codon fit in the ribosome when protein synthesis starts? Where does the stop codon fit when protein synthesis ends?

2. A. Which strand serves as template, the sense strand or its complement?

    B. What is shown in the code table, and what's in the mRNA -- the codons or the anticodons?

    C-1. What is meant by genotype, phenotype, and wild type?

    C-2. Consider the genetic code. Consider both degeneracy -- will the new codon specify the same amino acid? and wobble -- even if the new codon specifies the same amino acid, can the same tRNA be used? If you aren't sure, look up the wobble rules.

    C-3. How is this case different than the previous one?

    C-4. Consider the properties of the two amino acids found in wild type and mutant 2. What effect would you expect if you substitute one for the other? (If you don't remember the properties of the various amino acids, look them up in your text.)

    D.  Consider the properties of the amino acids involved. Are the two amino acids more or less different than in the previous case? (See C-4 above.)

    E-1. How does each type of mutation change the protein?

    E-2. Why is mutation #4 likely to produce a totally nonfunctional protein?

    E-3. Draw a picture of the situation -- include the ribosome (A & P sites), mRNA (with mutant codon in A site) and growing chain (& any relevant tRNA's).  Now what will happen next?

C2005/F2401 '06  -- Answers to Recitation Questions #9

    A-1. Add lactose.  You need to induce the lac operon by adding lactose and de-repress the trp operon by removing tryptophan.     A-2. The lac operon.  The strength of the promotor determines the rate of synthesis of mRNA. A stronger promotor binds RNA polymerase more strongly so initiation of transcription happens more often, more mRNA is synthesized and the steady state level of mRNA is higher. (In prokaryotes, the mRNA is degraded rapidly, so the rate of synthesis determines the steady state level.) The higher the cellular level of mRNA, the higher the level of protein synthesis and the higher the level of the corresponding protein. Therefore the strength of the promotor (how well it binds RNA polymerase) determines the max. level of the corresponding protein.  We know from the information given  that cells can make more β-galactosidase/cell than trp synthetase. This indicates that the lac promotor is a "stronger" promotor than the trp promotor -- the lac promotor binds RNA polymerase more strongly so initiation of transcription happens more often and the steady state level of mRNA (and protein production) produced from the lac promotor is higher.

    B.  Remove lactose. The gene for enzyme X is under the control of the regulatory system used by the lac operon. Therefore the production of mRNA for enzyme X will be inducible, just like the lac operon.  Production of mRNA will be turned on by the addition of the inducer, lactose, and conversely, production of mRNA will stop when the lactose is removed. When lactose is present, it binds to the repressor protein, forming an inducer-repressor complex. The complex is inactive -- it does not bind to the operator, and the gene for enzyme X is transcribed. When lactose is removed, the repressor protein (no longer part of the complex) becomes active, binds to the operator, and blocks further transcription of the gene for enzyme X.

    C-1. Only lactose. The operator determines which repressor protein will bind, and therefore which co-repressor or inducer must be present to control activity of the repressor protein and turn the operon on or off. In this case, the operator is from the lactose operon, so lactose must be present to inactivate the repressor and allow transcription of the operon, as explained in answer to A. Presence or absence of tryptophan would make no difference.

     C-2. Higher with plasmid 1. The promotor determines how effectively RNA polymerase will bind (when the operator is unblocked) and therefore how much mRNA (and protein) will be made per unit time when the operon is fully induced or derepressed.)  So plasmid 1, with the lac promotor, should allow a higher level of mRNA and enzyme X per cell than plasmid 2, with the trp promotor.

D. This is a case of complementation; if complementation produces active enzyme X, then the bacteria will grow in the presence of sillimycin.  (You need the antibiotic to help you screen large numbers of colonies for ability to make enzyme X. It's very tedious to test individual colonies for presence/absence of the enzyme, but it's easy to see if they grow or not.)    

    D-1. Neither of these. Adding lac wouldn't make any difference here; adding trp would prevent production of enzyme X. You need both plasmids to be transcribed so that one subunit of enzyme X can be made from one plasmid and the other subunit of enzyme X can be made from the other plasmid. Therefore you want both operons/operators to be "on"  -- neither operon should be repressed.  The lac operator is constitutive (always on), so no lactose is needed to induce the operon -- transcription is already turned on and cannot be turned off. The trp operon is a repressible operon that would be turned off if you added trp. So you don't want to add trp -- you have to be sure to leave it out, so that the operon will be transcribed.

    D-2. In different genes. This is a case of complementation -- one plasmid must have a defective gene for one subunit and the other plasmid must have a defective gene for the other subunit. (Each plasmid has a functional gene for the "other" subunit.) This explains why all bacteria that get both plasmids can make functional enzyme X -- they make one good subunit from each plasmid. Since they all make functional enzyme X, they are all resistant to sillimycin, and all produce colonies.  If both plasmids carried mutations affecting the same subunit, then only a few bacteria which got both plasmids would make functional enzyme X. These would be the rare bacteria in which recombination had occurred to produce a recombinant, functional gene. If both plasmids carried identical mutations, there would be virtually no colonies, as there would be no way (except by additional mutations) for bacteria with the two plasmids to make active enzyme X. 

    D-3. No. If the bacterium is growing on sillimycin, it must make active enzyme X. However, two different molecules of mRNA are needed to make good enzyme X. Plasmid 3 is being transcribed to make mRNA for one good subunit, and plasmid 4 is being transcribed to make mRNA for the other. If you translate only one molecule of mRNA, you will make only one good subunit, and no active enzyme X.     If the bacteria producing enzyme X were recombinants, then the answer would be "yes." In that case, the bacteria should contain one mRNA, transcribed from the recombinant DNA. That mRNA would contain good information for each subunit. Translation of that recombinant mRNA would produce both subunits and therefore working enzyme X.

C2005/F2401 '06 -- Recitation Problems #9 -- Hints

For these problems it usually helps to draw out all relevant DNA's and indicate all relevant genes for enzymes, repressors, operators, promotors, etc.  Then, for each set of conditions, consider what repressor proteins will be made, what state the proteins will be in, which structural genes will be turned on (or off), etc.

A. What does it take for each operon to be fully "on?"

B. Try drawing the set up. What repressor is controlling production of enzyme X? (Note that there are no repressor genes on the plasmids.)

More hints for B:

     Is production of enzyme X inducible, repressible or constitutive?     If you want the cells to stop making enzyme X, do you want the repressor to bind to the plasmid or not? What will it take to make the repressor "do the right thing?" C-1. What repressor is controlling production of enzyme X this time? (Remember that there are no repressor genes on the plasmids.) Will the promotor and/or operator determine whether you get maximal production of enzyme X?

C-2. What determines the maximum rate of production of enzyme X? The promotor? The operator?

D. What is allowing the bacteria to grow in the presence of sillimycin, recombination or complementation? Be sure to draw the set up. (What do you need the sillimycin for??)

More hints for D:

    D-1. & D-2.  Do both plasmids have to be transcribed? If so, what must be added to the medium to allow both plasmids to be transcribed?

    D-3. Can one molecule of mRNA (from plasmid 3 or plasmid 4) code for good enzyme X?

C2005/F2401 '06 -- Recitation Problems #9

(Modified from Exam #3 of '01) You have bacteria that are completely normal. They have a normal lactose operon (inducible by lactose) and a normal tryptophan operon (repressible by tryptophan). They have no plasmids.  Under optimal conditions, these cells can make 10⁷ molecules of β-galactosidase per cell from the lac operon or 10⁵ molecules of trp synthetase per cell from the trp operon.

    A-1. What are optimal conditions for production of the two enzymes (at the same time)? What do you have to add to the medium? If you want the cell to produce both enzymes, you need to (add lactose) (add trptophan) (add both) (add neither).     A-2. Which operon has a stronger promotor? (lac operon) (trp operon) (neither -- both promotors are the same strength) (can't predict).

In the remaining questions, you transform these bacteria with recombinant plasmids that have the gene for enzyme X (not normally made by these bacteria) next to a promotor and operator. All the plasmids are made correctly so that the enzyme X gene is transcribed under optimal conditions. The promotor and operator on the plasmids come from either the lactose or tryptophan operons.  (See table below for properties of plasmids 1-4).

    B. You transform your bacterial cells with plasmid 1 and the cells start to make enzyme X. To stop the cells from making more enzyme X, you would (add lactose) (add tryptophan) (remove lactose) (remove tryptophan) (add both) (take away both) (doesn’t matter – none of these changes will have any effect).  Explain briefly.

    C. You repeat your experiment, but this time you transform your normal bacterial cells with plasmid 2.     C-1. If you want to get maximal production of enzyme X from cells transformed with this plasmid, what must be present in the medium? (both tryptophan & lactose) (only lactose) (only tryptophan) (neither), AND     C-2. How should maximum production of enzyme X compare in cells with plasmid 1 vs plasmid 2? Maximum enzyme X production should be (higher with plasmid 1) (higher with plasmid 2) (same with either plasmid) (beats me -- can’t tell without more info.)  Explain both answers briefly.

    D. Suppose that enzyme X has quaternary structure – it consists of two separate subunits. Both subunits are needed to form active enzyme X. All plasmids involved here carry the structural genes for both subunits. Both genes are transcribed as part of a single polycistronic mRNA. Assume that enzyme X is needed for growth in the presence of sillimycin (an antibiotic). You transform cells with two plasmids at the same time -- you use plasmids 3 & 4. Cells transformed with either plasmid alone do not produce active enzyme X. (See table.)     D-1. You mix your normal bacteria with a solution containing both plasmids and place the bacteria on medium containing sillimycin. All of the bacteria which get both plasmids produce colonies. To get this result, it is necessary to include (lactose) (tryptophan) (both of these ) (neither of these) in the growth medium AND     D-2. The structural mutations in plasmids 3 & 4 are (identical) (in different base pairs but in the same gene) (in different genes) (different, but can't tell if they're in the same gene or not).     D-3. Suppose you could isolate one molecule of polycistronic mRNA (for enzyme X) from a bacterium that is growing on sillimycin. (This is one of the bacteria transformed with both plasmids.) You translate the mRNA in a test tube using added tRNA, ribosomes, etc. Do you expect the protein you make to be enzymatically active? (yes) (no) (beats me). Explain your choices briefly.

Plasmid	Promotor	Operator	Enzyme X gene(s)
Plasmid 1	lac	lac	normal
Plasmid 2	trp	lac	normal
Plasmid 3	lac	constitutive lac operator	Substitution mutation in structural gene
Plasmid 4	trp	trp	Substitution mutation in structural gene

All promotors and operators are normal except in plasmid 3. "lac" means from the lactose operon; "trp" means from the tryptophan operon.

C2005/F2401 '06 -- Answers to Recitation Problems #10 updated

1. See below. The * indicates the points of cutting. You assume the original sequence is a palindrome, so there must be a G on the 5' end in the original sequence.

5' G*T C G A  C	-->	G	+	TCGAC
3' C A G C T* G		CAGCT		G

2.     A. (i) Digest both DNAs with the same enzyme, either Cut 1 or Cut 2. To make a recombinant DNA, you must cut the two DNA's so that the pieces will have overlapping complementary ends. In other words, you need overhanging ('sticky") ends that will pair up. In this case, the DNA cut by one enzyme will not pair up to DNA cut by the other. The sticky ends of the DNA cut by the two enzymes look the same at first glance, but they aren't equivalent. If you cut two DNA's with the two enzymes, the ends will look like this:

	Cut 2				Cut 1
5' ...C	3'	5' TTAA	G ... 3'	5' ...G	AATT 3'	5'	C ...3'
3' ...G	AATT 5'	3'	C ... 5'	3' ...C		3' TTAA	G ...5'

It looks like you can just flip one fragment over, say the right-most  fragment from Cut 2, and ligate it to the left-most fragment from Cut 1. But that won't work because of problems with the 3' and 5' ends. You have to connect a 3' to a 5' -- you can't connect two 3' ends or two 5' ends. Try it and see.     (ii). You need ligase.  First you mix the pieces and let them hybridize. This is the equivalent of pairing up say, the left fragment from Cut 1 (or Cut 2) with the right fragment from a different DNA cut with the same enzyme. The hydrogen bonds between the "sticky" overlapping ends will hold the two fragments together temporarily, but you need ligase to join the fragments permanently (covalently). 

    B. (i). a. No, because the plasmid will not have a functional origin of DNA replication. In this case, the insert goes in the middle of the ori, not in the middle of a Drug resistance gene or other selectable marker.              b. Drug sensitive. Since the plasmid can't replicate, only one of the descendants of a single transformed cell will have a plasmid. All the rest will have no plasmid and no drug resistance gene.

        (ii) (a). Plasmids must  be integrated into the chromosome. (b) Integrated plasmids must contain inserts of human DNA. (c) Both. Explanation: You selected colonies that contain the human DNA and are drug resistant.  Each original cell that got a plasmid (with a human DNA insert) must have divided many times to produce a colony -- without losing the plasmid (& insert). So we know that the plasmid DNA and the  human DNA have been replicated and passed on at each cell division. However insertion of the  human DNA into the plasmid destroys the origin of replication of the plasmid. So the simplest solution is that you have selected cells in which the plasmid (plus insert) has become integrated into the chromosome by crossing over. Therefore the plasmid doesn't need an origin -- it (along with the human DNA insert) is replicated using the chromosomal origin. In the presence or absence of the drug, the integrated plasmid will continue to be replicated and passed on, just like the rest of the bacterial chromosome. If the plasmid were not integrated, it would probably be lost in the absence of the drug, since there would be no selection against cells that had lost their plasmids (& lost their drug resistance). But since the plasmid is integrated, it and its insert, will both be faithfully replicated and passed on, and so both should be detectable by hybridization.

    C. (i). The 4 bands are of equal intensity, but that doesn't mean the DNA (or length of exon) in each of them is the same size. It means the amount of probe sticking to the material in each band is the same. (The DNA fragments in different bands must be different lengths to end up in different places on the gel/blot.)  The equal intensity means the DNA in each band "captures" or hybridizes to the same amount of labeled probe. In this experiment, each fragment of genomic DNA hybridizes to one cDNA molecule. The length of the hybridized (hydrogen bonded double stranded) region varies, depending on the length of the exon(s) in the fragment, but there is always one probe molecule "stuck to" each genomic fragment. If the genomic DNA were cut up to give several different pieces, and some of the pieces were the same length (ended up in the same band), then the bands would not be of equal intensity. Also if the probe were cut up into many pieces, then different exons might hybridize to (trap) different amounts of probe.

        (ii).  3 introns and 4 exons. The DNA must be cut in 5 places, to produce 4 different pieces of DNA. In won't be cut in any of the exons, because the cDNA ( = exons only) has no sites for the restriction enzyme.  The two "outside" cuts are not in introns -- they are in spacers outside the gene. The other three cuts must be in introns. (Only spacers between exons, inside genes, qualify as introns.) Each of the 4 pieces contains at least one exon, and so hybridizes to the cDNA probe. All the pieces hybridize to the same probe, but each piece hybridizes to a different stretch of bases in the probe.

    D. (i). One radioactive band. (ii). Band will be longer than 200 bp.     Why one band? The DNA fragment that contains the exon corresponding to the probe will hybridize to the radioactive probe. None of the other fragments will hybridize. You will get the same number of DNA fragments on the gel or blot as before, but only one will trap probe and form a radioactive band.     Why longer than 200 bp? The piece of genomic DNA that hybridizes to the probe will be longer than the probe. The probe is a single exon (of 200nt), but the genomic DNA fragment includes some intronic sequences on each end, because all the restriction sites are in introns.     The length of the DNA fragments is given in bp (base pairs), because the DNA fragments are double stranded when they are separated on the gel. The length of the probe is given in nt (nucleotides), because the probe is single stranded. The double stranded DNA fragments are separated first, and then denatured and hybridized to the single stranded probe.

C2005/F2401 '06  -- Recitation Problems #10 -- Hints

Note: For all restriction enzyme problems it is usually easier to write one strand 5' to 3',  left to right, as usual, and then to write the complementary strand with the bases upside down. This makes it clear that the sequence of the complementary strand goes from right to left, 5' to 3', antiparallel to the other strand. Writing the complementary strand upside down makes it much easier to keep track of the 5' and 3' ends (because it's obvious) and it makes it easier to see what "sticky end" will or will not pair up with another "sticky end."

1. The uncut site should be a palindrome. What was the palindrome? If it isn't obvious, consider the half of the DNA that is not shown. (When the original piece was cut, there were two halves. What did the other half look like?) Now put the two pieces back together.

2. A. (i). What sticky ends will you get with each enzyme? Draw out a sample DNA with sticky ends for each case. Now try and pair them up. (If you are having trouble, this is a good place to try writing one of the strands "upside down" as explained in the note above.)         (ii). How do you construct a hybrid plasmid? How do you "glue" the parts together?

    B. (i).  a. Where is the inserted piece of human DNA located? Will the drug resistance gene still work? The origin of DNA replication?                b. Will most of the bacteria in the colony contain plasmids?

        (ii).  a. Do the selected cells still contain plasmids? If so, how is this possible, given your answers to part (i)?                b. What properties did you select for? What must be present for the cell to have those properties?                c.  Where are the segments you are interested in? (On the chromosome? On an independent plasmid?) Do you need selection to keep the segments from being lost?

    C. (i).  How can one probe hybridize to 4 different fragments?  Is the same amount of probe present in each band? (Be sure you understand how this works -- ask your TA or Dr. M if you can't see it.)          (ii). If  there are no restriction sites in the cDNA, how could you cut up the gene into 4 pieces? Where are the restriction sites in the gene? (introns? exons? Both?) You can't do this one without drawing it out. Keep fiddling with your picture (adding or removing introns and exons) until it matches the results.

    D. Note that the DNA fragments separated on the gel are the same as in C, but the probe is different.         (i). What do the fragments look like? Do they contain exonic sequences, intronic sequences or both?         (ii). Consider where the DNA is cut relative to the ends of the exons. (Consult your picture.)

C2005/F2401 '06 Answers to Recitation Questions #11-- Last Update

1. A-1. This gene has two introns. There are spacers on the ends of each exon, but these are considered spacer regions between genes, not introns. Introns are  always "intervening sequences" that interrupt genes, not the spacer regions between genes. A gene must begin and end with an exon. (Note: Untranslated regions on the ends of mRNA (UTR's) are encoded by exons. They are not encoded by introns. The exons correspond to the entire mRNA, not just to the coding part.)

    A-2. The DNA shown is the sense strand. When one strand alone is given, it is always the sense strand unless it says otherwise. Why? Because the sense strand corresponds to the RNA, and it's the sequence of the RNA transcript that matters, not the sequence of the DNA that codes for it. If the DNA codes for a mRNA, it's easy to use the code table to "translate" the sense strand. It's much harder to start with the "other" strand and figure out what protein the DNA codes for.

    B-1. (somewhere inside) of (exon 1). One expects a 5' untranslated region (5' UTR) of unknown length that precedes the ATG initiation codon. The initiation codon must lie within exon 1 in order to accommodate a coding length of 405 nt (derived from the 135 amino acid length of protein X). Exons 2 + 3 could code for 400 nts at most, so exon 1 is needed as well.

    B-2. (at the right end) of (exon 3). The polyadenylation site (point where poly A addition begins) defines the end of the final exon, and is the point where the transcript is cleaved. The polyadenylation signal sequence (which triggers action of polyA polymerase) is located a little before the right end of the exon; it comes about 25 nucleotides before (upstream of) the actual spot where poly A is added.

    B-3. (Somewhere inside) of (exon 3). You don't know where in the exon the coding sequence ends and the 3' UTR begins.

    C-1.  You'll get 3 labeled bands of lengths 300 , 500 , and >600 bp.      Explanation: DNA will be cut into 5 pieces. The DNA will be cut at the E and H sites shown,  and at additional sites in the DNA (in areas to right or left of that shown). The length of the pieces from the right and left ends of the section shown will depend on where the next site for E or H is -- how far off it is to the left or right.     One piece does not hybridize to the probe -- that's the piece to the left of (upstream of) the left-most EcoRI site. (We don't know how long this piece is, but it doesn't matter as it doesn't hybridize to the probe.) The remaining 4 pieces will hybridize to the cDNA, as they each contain part of at least one exon. Two pieces are the same size, so they end up on the same place in the gel. Therefore you get only 3 bands: 300 , 500 , and >600 bp as follows.     The 300 bp band is from the DNA that goes from EcoRI site upstream of exon 1 to the HindIII site in the middle of exon 1.     The 500 bp band contains of two types of DNA molecules, one that goes from the HindII site in exon 1 to the EcoRI site in exon 2, and one that goes from the EcoRI site in exon 2 to the HindIII site in exon 3.     The >600 band originates from DNA that starts at the HindII site in exon 3 and goes to some site (E or H) that must exist at some point downstream of the SalI site on the right side of the map (in a region not shown, off the right edge of the picture). C-2. The 500 bp band, as it contains two types of DNA molecules complementary to the probe. Doubling the number of target molecules compared to the other bands should double the amount of probe hybridized. (In addition, each 500 nt piece may trap more than one molecule of probe, because each fragment has two separated sequences complementary to the probe.) Note: DNA fragments separated on gels are double stranded; when denatured, the DNA is single stranded (until it traps probe). That's why the terms bp and nt are both used here.

D-1. Sal I because all the other enzymes cut within the gene. You want the plasmid to have one Sal I site so you can open it up at one point and insert a fragment of genomic DNA without cutting the plasmid anywhere else.

D-2. Sal I or Bam HI. All the others cut within the cDNA = region with exons. You want to cut the recombinant plasmid with the same enzyme -- one that will cut the sequences at the two ends of the cDNA and release the cDNA from the plasmid. You need to cut the sequences that form the joints between the cDNA and plasmid DNA. The joints were made by sticking matching "sticky ends" together. To cut them, you will need to use the enzyme that generates the same sticky ends.

2. Karyotype should have 6 chromosomes. All chromosomes should look double.  The type A are upside down V shaped; the type B are like an X but asymmetric -- the connection is 1/3 down; type C are X shaped. Note that karyotypes are made from squashes of dividing cells (usually at metaphase of mitosis) so there are always 2 chromatids per chromosome. It is customary to put the end nearest the centromere on top. (So shorter arm, if there is one, is on top,  and longer arm is on bottom.)     Note: If the two sister chromatids stick tightly to each other, as they do in some cases, then all chromosomes will look single (even though there are 2 chromatids per chromosome). Chromosomes will differ by where the centromeres are. (On the end, 1/3 of the way down, or in the middle.) Centromeres will be visible in banded chromosomes as a darkly staining area and/or as a constriction (even without banding).

    A. N = 3; C = 12 X 10^-11 g. 

    N = the number of kinds of chromosomes (which is 3 in this case).     C = the DNA content of one set of unreplicated (haploid) chromosomes = 3  X [4 X 10^-11g/chromosome]     Note C and N are properties of the organism or species; they do not change with the state of the individual cell. DNA content per cell or number of chromosomes per cell may change at different stages of the cell or life cycle. For example, # of chromosomes per cell may change from 2N to 4N or DNA content per cell may change from 2C to 4C, etc. But N and C are constants.

    B-1. There should be two V shaped chromosomes of the type with the centromere on the end (type A).  Chromosomes are doubled in metaphase; each chromosome has 2 chromatids. If the centromere is on the end, the two chromatids will be attached to form a V. There will be two "V's" as the cell is diploid, so there should be two of each type of chromosome at metaphase of mitosis.

    B-2. There should be four V shaped chromosomes of the type with the centromere in the middle (type C).  Sister chromatids separate at anaphase. The chromatids with the centromere in the middle are pulled away from each other toward the poles, forming a V shape. (The chromosomes that were V shaped in metaphase are rod shaped in anaphase.) Before anaphase starts, there are two X shaped chromosomes per cell. Each chromatid of the "X" shaped chromosomes forms a V at anaphase.  Note that the two halves of an X shaped chromosome are not considered "V shaped chromosomes" -- they are V shaped chromatids as long as they are connected to each other. Once the two chromatids separate, after centromeres split, each chromatid is then considered an individual chromosomes. As long as the two chromatids are still connected, they are considered part of the same chromosome.

    C. 6 Chromosomes, 12 DNA molecules, and 48 X ^-11g of DNA.  There are 6 chromosomes because all somatic cells are diploid (2N). There are 2 chromatids per chromosome in G-2, and always one double stranded DNA molecule per chromatid.

C2005/F2401 '06 -- Recitation Problems #11 -- Hints

1. A-1. What is the definition of an intron? Are all nontranscribed regions of the DNA introns?     A-2. If you had a choice, which strand would you show??

    B-1. How long is the peptide, and how long are the exons? Do all 3 exons have to contain sequences that code for the peptide? Could translation start in the second exon?     B-2. What is the relationship between the position of the polyA addition site and the end of transcription?     B-3. See B-1. Could translation end in the second exon?

    C-1. How many pieces will you get from the region shown?  How many of those pieces will hybridize to cDNA? When you consider the lengths of the pieces, remember that the DNA shown is part of a longer DNA molecule.     C-2. How could one band be darker than another? How could it trap more probe?

    D-1. Do you want to cut inside the DNA you are trying to clone?     D-2. What enzyme will cut up the genomic DNA but NOT the cDNA? Can you use it?

2. For the karyotype: What does each type of chromosome look like at metaphase? How many chromatids, and how are they connected? How do you arrange the chromosomes in a karyotype?     A. Check out the definitions of C and N.     B-1. How many chromatids/chromosome at metaphase? How are they attached to the spindle?     B-2. How many chromatids/chromosome at anaphase? How are they attached to the spindle?     C. Has the DNA replicated in G-2? How many DNA molecules per chromosome and/or chromatid?