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.

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