Biol 3054
1.
Worksheet 7
1
You have identified a cDNA that is 2Kbp long. You want to see know long the gene is so you carry out
Southern analysis. You cut genomic DNA with EcoRI and ran it on an agarose gel. The DNA fragments were
then blotted onto a nylon membrane and probed. You used the 2Kbp cDNA as the probe to detect those
genomic fragments which contain the gene. The results of the Southern blot hybridization are shown below.
Read the figure legend to understand the data.
a. Explain what a cDNA is and what it represents
A cDNA is a complementary copy of a mRNA that has been reverse transcribed. cDNAs represent all the genes in a
particular cell, tissue, developmental stage, etc that are being expressed (transcribed).
b. What is the approximate length of the complete gene. __6.5Kbp______________
c. Explain why the apparent size of the gene is larger than the cDNA.
cDNAs correspond only to the mRNA which in turn is only composed of the exons of the gene. Depending on the
length and number of introns, the actual length of the gene will always be somewhat longer than the cDNA,
sometimes 10 to 100 times longer.
d.
This cDNA is not present in persons with a particular disease. The sizes of the bands detected by the cDNA
probe are different in diseased persons. Describe the difference between the size of the gene in healthy
vs diseased persons and how this difference might result in the loss of expression of this gene (and
only this gene).
Because the sequences detected by the probe are smaller (6Kbp vs 6.5Kbp) in the diseased DNA, one can
hypothesize that a portion of the gene has been deleted. If this deletion eliminates the expression of the gene, then
one can further hypothesize that the promoter as well as some of the coding regions of the gene has been lost.
1
Figure 1. Southern blot of
genomic DNA cut with
EcoRI and probed with a
2Kbp cDNA probe. DNA
from a healthy individual
was loaded in lane 1, DNA
from a person with the
disease was loaded in lane 2.
Molecular size standards are
in lane 3. Sizes are given in
Kbp.
2
3
10
8
6
4
3
2
1.55
1.4
1
.75
.5
.4
.3
.2
.1
.05
Biol 3054
2.
Worksheet 7
2
The structure of a gene is given below. The length of each exon is given in the boxes. This gene expresses a
mRNA that is 2871 nt long. The polyA tail of this mRNA is 275 nt long. This mRNA encodes a protein of 673
amino acids.
stop @ 7861
ATG @ 1502
280
/
//
1214
471
/
//
774
522
/
//
987
343
/
//
2177
619
/
//
402
157
/
//
204
521
a.
How many nucleotides are in the ORF of this mRNA? _2019____________
673 aa x 3 nt / aa = 2019 nt
b. How many nucleotides total are in the untranslated regions of this mRNA?__577______
2871nt in total mRNA – 275 nt of pA = 2596 nt of mRNA -2019 nt of ORF = 577 nt of UTR
c. Indicate a plausible position for the start codon and stop codon for this gene on the diagram. Provide
the exact number of the base where the 1 st nucleotide of the start codon is positioned and the 1 st
nucleotide of the stop codon is positioned. Don’t CHEAT or COPY or COLLABORATE or I’ll give
you a BIG FAT ZERO!
Let’s split the UTR ~half so there is 288nt of 5’ UTR and 289 nt of 3’ UTR. Then for the start codon, the 5’ UTR
spans the 1st exon (280 nt) + 8 nt of the 2nd exon + the 1st intron (1214nt ) so the total distance from the 1st base of
the gene is 1502 nt. For the stop codon, the 3’ UTR would span the last exon (204nt) + 85nt of the 6 th intron and the
last intron (521nt). The ORF therefore would extend 72nt into the 6 th exon (157-85). So the stop codon would be at
nt 7861.
3. If molecules like cAMP, Ca2+, & DAG are 2nd messengers, what are the 1st messengers?
the ligands that 1st bind to the receptors – growth factors, hormones, cytokines, neurotransmitters, etc…
4.
You are trying to clone a receptor tyrosine kinase from the genome of an organism whose genome has 3 x 109
bp. The average size of a gene is 50Kbp and you screened a genomic library made in a cosmid vector. How
many cosmid plaques would you need to screen to guarantee that you would find the receptor gene? (show all
your math for any credit)
There are 3x109 bp in the genome, and you can put 50Kbp in one cosmid, therefore you’d need 60,000 cosmids to
carry the entire genome (3x109/5x104). To ensure you find the gene, you’d need to screen at least 60,000 cosmid
plaques ( in reality you’d screen at least 100,000)
5.
Describe in concise detail how the protein kinases A and C would be simultaneously activated within a cell.
Begin your description with the arrival of ligands at the cell.
A ligand would arrive at the cell and be bound by a 7-pass transmembrane receptor. Ligand binding would activate
the receptor and cause it to interact with heterotrimeric G-proteins. Interaction with the G-protein stimulates GDP
exchange in -subunits and dissociation of -subunits from -subunits. The activated GTP-bound -subunits can
move in the membrane to interact with adenyly cyclase and phopholipase C (PLC). Adenylyl cyclase is then
stimulated to produce cAMP and PLC is stimulated to produce diacylglycerol (DAG) and inositol phosphate (IP3).
cAMP is a 2nd messenger that binds to and activates protein kinase A (PKA). IP 3 is a 2nd messenger that binds to
Ca2+ channels in the SER causing them to open and release Ca2+ into the cytoplasm . The released Ca2+ together with
DAG binds to and activates protein kinase C (PKC).
Biol 3054
6.
Worksheet 7
3
In Drosophila melanogaster, a signaling pathway is responsible for the differentiation of the photoreceptor
cells of the eye. Each unit of the compound eye has 8 cells. A mutation was discovered in which the 7 th cell did
not form. This mutation (and the gene) was named sevenless (sev). Flies lacking cell 7 can’t see UV light. After
this discovery a 2nd mutation that had the same phenotype was discovered, although the mutation was in a
different gene. Because of the similarity, the 2nd mutation was named bride of sevenless (boss). It was
postulated that one of these genes encoded a receptor and the other gene encoded a ligand. An elegant
experiment was performed that showed in sev eyes, the mutant cells could not be corrected by transplanting in
wildtype cells. However, in boss eyes, the mutant phenotype could be corrected by transplanting in wildtype
cells. In technical terms, sev behaved in a cell autonomous manner, while boss behaved in a cellnonautonomous manner. With this information and a lot of hard thinking, tell me, which gene encodes the
ligand and which gene encodes the receptor?
Ligand = ___boss____________
Receptor = _____sev___________
Since the boss mutant cells can be rescued by adding normal cells, it implies that the normal cells produce a ligand
that replaces the function of boss, therefore, boss is a ligand. Because the sev mutant cells can’t be helped by added
normal cells, it implies that the protein made by the sev gene is a receptor and is permanently located in the eye cells
rather than diffusing from normal cells to mutant cells.
7.
Diagram the ras signal transduction pathway. Included everything!
See figure 15.9