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Solving Problems

Solving Problems


A typical Southern Blot interpretation problem:

You are studying a gene responsible for pigment biosynthesis in plants. RR individuals are red; Rr are pink; and rr are white. The restriction map of the R allele is:



You extract genomic DNA from a red, a pink, and a white plant; digest with EcoRI, HindIII and EcoRI+HindIII; run on a gel, and transfer to nitrocellulose. You have a cDNA for the R gene which you use as a probe. The results of the autoradiogram are as follows:



1) What is the alteration in the R-gene DNA of the mutant r allele?

2) Why are there more bands in the pink strain than in the white or red - what do the fainter bands represent?

- First, it would be useful to correlate the bands on the Southern of the homozygous R strain with the restriction map:

EcoRI: Digestion of this region of DNA should produce 5 fragments in addition to the thousands of others produced from the rest of the genome. These are, from left to right:

e1) A band of unknown size which begins off the left end of the map and ends with the leftmost EcoRI site on the map.

e2) A 1.2kbp band including exon 1 and part of the intron.

e3) An 0.1kbp band from the intron.

e4) A 1.5kbp band including part of the intron and all of exon 2.

e5) A band of unknown size which begins with the rightmost EcoRI site, includes the 0.4kbp EcoRI-HindIII fragment, and extends to an EcoRI site off the right end of the map.

Then why are there only two bands on the southern?

-Because the probe is a cDNA, that is, it only includes the exons. Fragments e1, e3, and e5 do not contain any sequences in common with the probe and therefore will not bind probe and will not appear on the autoradiogram. Fragments e2 and e4 include exons 1 and 2, respectively, and will therefore bind probe and appear on But the probe hybridizes to only part of fragments e2 and e4, why do they appear in their full lengths in the Southern?

- The probe only needs to find a short region of the target (20 - 30 nt) to hybridize to, so the length of the overlap is usually not important. Also, since hybridization occurs after the DNA fragments have been separated by molecular weight, the length of hybridization overlap is irrelevant to the size that the fragments migrate. (This is a subtle, but very important point!)

HindIII: Digestion of this region of DNA should produce 5 fragments in addition to the thousands of others produced from the rest of the genome. These are, from left to right:

h1) A band of unknown size which begins off the left end of the map, includes the 0.2kbp EcoRI-HindIII fragment, and ends with the leftmost HindIII site on the map.

h2) A 1.75kbp band including exon 1, the intron, and part of exon 2.

h3) A 0.3kbp band from exon 2.

h4) A 0.95 kbp band including part of exon 2 and the 0.4kbp EcoRI-HindIII fragment.

h5) A band of unknown size which begins with the rightmost HindIII site and extends off the right end of the map.

Since only fragments h2, h3, and h4 contain sequences found in the probe, only they will appear on the autoradiogram.

EcoRI+HindIII: Digestion with both enzymes will result in cleavage of the DNA at all EcoRI sites and at all HindIII sites. This should produce 9 fragments in addition to the thousands of others produced from the rest of the genome. These others produced from the rest of the genome. These are, from left to right:

eh1) A fragment of unknown size which begins off the left end of the map.

eh2) A 0.2kbp EcoRI-HindIII fragment.

eh3) A 1.0kbp HindIII-EcoRI fragment.

eh4) A 0.1kbp EcoRI fragment.

eh5) A 0.65kbp EcoRI-HindIII fragment.

eh6) A 0.3kbp HindIII fragment.

eh7) A 0.55 kbp HindIII-EcoRI fragment.

eh8) A 0.4kbp EcoRI-HindIII fragment.

eh9) A fragment of unknown size extending off the right end of the map.

Fragments eh3, eh5, eh6, and eh7 will bind the probe and show up on the autoradiogram.

1) To find the difference between r and R, find the bands that are altered between the two DNA's.

The differences are:

EcoRI: The 1.5kbp band (e4) has disappeared and a 1.25kbp band has appeared. The simplest explanation for this is that 0.25kbp has been deleted from this region in the R allele to produce the r allele.

HindIII: The 0.3kbp band (h3) has disappeared and a 0.05kbp band has appeared. The simplest explanation for this is that 0.25kbp has been deleted from this region of the R allele to give the r allele.

EcoRI+HindIII: The 0.3kbp band (eh6) has disappeared and a 0.05kbp band has appeared. The simplest explanation for this is that 0.25kbp has been deleted from this region of the R allele to give the r allele.

Thus, all three digests agree with the model that a 250bp deletion has occurred in the 0.3kb HindIII-HindIII region of exon 2 in the R alleles given rise to the r allele. A deletion of this type would be likely to cause an inactive protein to be produced by the r allele, leading to reduced red pigment production.

Note: You cannot be certain that alterations have not occurred in the intron regions as well - without an intron sequence probe, you cannot tell for sure. Also, it is possible that other models may be able to explain this data. However, this is the simplest reasonable explanation - you could be more certain if you did a Southern using: other restriction enzymes; a probe to exon 2 only; or even by sequencing the DNA of the R and r alleles to find the exact differences.

2) What about the pink (Rr) strain?

There are two types of bands in this part of the blot. This is because this strain is a heterozygote - it contains one copy of the R allele and one copy of the r allele.

i) Heavy bands at positions which are not altered between the R and r alleles. These represent fragments that are present in two copies per cell. They are the same intensity as the bands from the RR and rr strains because in the RR and rr, the alleles are present in two identical copies per cell.

ii) Light bands at positions representing both R and r alleles. These represent fragments that are present in only one copy per cell. They are lighter and thinner than the other bands because there is half as much DNA in each band.


A typical Northern Blot problem:

This is the restriction map of an autosomal gene in Drosophila.



You extract RNA from male and female flies and run duplicate samples of both on several northern blots. You have the entire 3.85kbp EcoRI-HindIII fragment containing the gene on a plasmid, from which you generate probes by restriction enzyme digestion. Assuming that the DNA is the same in both sexes (How could you tell this?), explain the difference between the mRNA for this gene in males and females using the data below.

	Probe		Restriction Fragment

	A			1.0kbp EcoRI-HindIII

	B			0.6kbp HindIII-EcoRI

	C			0.1kbp EcoRI-EcoRI

	D			0.5kbp EcoRI-HindIII 



- You could tell if males and females have identical copies of this gene by doing a Southern blot of DNA from males and from females, probed with cloned fragments from this region, and make sure that there are no differences in the restriction maps of the regions in both sexes. Note that this cannot guarantee that the sequences are identical throughout - that would require sequencing the DNA from both - but it can show that the sequences are very similar.

- There are at least two ways to analyze this data:

i) First, what do the blots show? Since the target molecule is mRNA, only exon regions of the probes will form hybrids. Each probe will then detect the presence of a different exon or exons in the final mRNA.

		Probe	Exon(s) Detected

		A		1

		B		1 and 2

		C		2

		D		3

From this, we can determine which exons are present in the mature mRNA's.

Blot A shows that exon 1 is present in both male and female mRNA for this gene. Blot B shows that exon 1 and/or exon 2 are present in both mRNA species.

Blot C shows that exon 2 is present in the female mRNA but not in the male mRNA - note that this does not conflict with Blot B's result.

Blot D shows that exon 3 is present in both male and female mRNA for this gene.

Summarizing: female mRNA for this gene contains exon 1, 2, and 3 - this is expected pattern of splicing; male mRNA for this gene contains only exon 1 and 3 - exon 2 is skipped. These models are confirmed by the sizes of the mRNA's in both sexes. We would predict that in females, mature mRNA = 0.5 + 0.2 + 0.6 = 1.3knt; and in males, mature mRNA = 0.5 + 0.6 = 1.1knt.

Note that the size of the mRNA observed in each sex does not change even with different probes.

ii) You could also look at this another way:

You would predict that the mature mRNA for this gene would be 0.5 + 0.2 + 0.6 = 1.3knt. This size is observed in females and represents apparently normal splicing of the pre- mRNA. Therefore, in females, the gene is expressed as we would expect.

But what about the males?

In males, the mature mRNA is 0.2knt shorter. There are three possible models that can explain this:

i) Transcription in males starts 0.2kbp to the right of where it normally starts in females (2/5 of the way through exon 1).

ii) Transcription in males ends 0.2kbp to the left of where it normally ends in females (2/3 of the way through exon 3).

iii) Exon 2 is skipped in males - the spliceosome starts with the start of intron 1 and uses the end of intron 2.

In the case of i) and ii), complete or partial sequences from all three exons would be present in the mature mRNA. This is not observed - blot C shows that males lack exon 2. This confirms iii) and we must conclude that exon 2 is missing in the mature mRNA from this gene in males.

In either method of analysis, you are forced to a surprising conclusion: that splicing can skip an intron under certain circumstances. You are not required to know that this process occurs, nor would you be expected to come up with it on your own. However, when given this data, you are forced to conclude - even with out knowing a mechanism - that all of exon 2, and only exon 2 is somehow missing from the mature mRNA in males.


A typical Western blot problem:

Protein W isolated from wild-type E. coli can be detected by anti-W antibodies. On Western blots, protein W has a molecular weight of 40kDa.

Draw the Western blot that would result from running samples from the following E. coli strains and explain your drawing:

Lane		Protein 

Isolated From:

1		wild-type

2		a strain with a mutation in the gene W promoter that
		inactivates the promoter 

3		a strain with a nonsense mutation 3/4 of the way from
		the 5' end of gene W 

4		a strain  with a missense mutation 3/4 of the way from
		the 5' end of gene W 

5		the strain used in lane 3 with an added F'-plasmid
		containing a wild-type copy of gene W

6		a strain with a single nucleotide deletion at the gene W
		stop codon which causes translation to continue for
		another 90nt along the mRNA. (the average amino acid is 0.1kDa)

- Lane 1: Wild-type protein is 40kDa, so there will be one band at 40kDa.

- Lane 2: No gene W mRNA will be produced, so no protein W will be produced, so the lane will be blank.

- Lane 3: Translation will stop at the new stop codon generated by the mutation, resulting in the production of a 3/4-size protein. This will show a band at 30kDa.

- Lane 4: This may change the molecular weight slightly (especially if the change was from glycine to phenylalanine) but not usually enough to detect on a blot. One band at 40kDa.

- Lane 5: There will be two types of protein W made: wild-type, from the mRNA made by the F' of 40KDa; and mutant, from mRNA made from the chromosome of 30kDa. This will result in two bands on the gel: one at 30kDa and one at 40kDa.

- Lane 6: If translation continues for 90nt, that is 30 codons or 30 amino acids added to protein W. Thirty amino acids will add roughly 3kDa to the molecular weight of this mutant protein W. One band at 43kDa.

This is shown below:




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