Chapter 5 homework
Terms in this set (16)
Ovalbumin is the major protein of egg white. The chicken ovalbumin gene contains eight exons separated by seven introns. Should ovalbumin cDNA
or ovalbumin genomic DNA be used to form the protein in E. coli? Why?
cDNA should be used to form the protein because cDNA is synthesized from mRNA. mRNA is absent of introns therefore it will only synthesize the codons necessary to produce the Ovalbumin in the egg. cDNA is not the whole genome, it is only the segment you want.
I would suggest that the DNA phosphorous back-bone binds to the Nitrogenous base at the center of the molecule. This is because DNA has a negative charge, and this positive N would neutralize the charges. This molecule is also planar so it imbeds in-between the paired bases of the dna.
The restriction enzyme AluI cleaves at the sequence 5'-AGCT-3', and NotI cleaves at 5'-GCGGCCGC-3'. What would be the average distance between cleavage sites for each enzyme on digestion of double-stranded DNA? Assume that the DNA contains equal proportions of A, G, C, and T.
So from this we can see that the first one cleaves only 4 base pairs that are in this sequence. The possibility that this could reoccur once again is (1/4)^4 or 1/256 base pairs. So every 250 kb, and if it is the next one it would be (1/4)^8 or 1/66000 or every 66 kb.
Suppose that a human genomic library is
prepared by exhaustive digestion of human DNA with the EcoRI restriction enzyme. Fragments averaging about 4 kb
in length would be generated. Is this procedure suitable for cloning small genes? Large genes? Why or why not?
A genomic library consists of many fragments of the entire genome. A genomic library is when the genes are cleaved at specific sites in the genome and dispersed in a solution. A genomic library can also be in bacterial or viral vectors. For this question 4 kB of DNA fragments are not too large to be replicated. This is because the lambda viral DNA that is useless for viral replication is 48 kB large, and many viruses can carry the different fragments that can be later identified by preparing specific mRNA probs. Large genes are too big for this to occur so it would be unsuitable, human genome is 3,200,000 kB.
What are probes?
Probes are specific areas of the cDNA that can be synthesized from mRNA molecules that are complimetary to the fragment of interest. Normally these cDNA are labeled so they can be tracked inside the vectors. They can be marked by 32 P at the 5' end.
Sickle-cell anemia arises from a mutation in the gene for the b chain of human hemoglobin. The change from GAG to GTG in the mutant eliminates a cleavage site for the restriction enzyme MstII, which recognizes the target sequence CCTGAGG. These findings form the basis of a diagnostic test for the sickle-cell gene. Propose a rapid procedure for distinguishing between the normal and the mutant gene. Would a positive result prove
that the mutant contains GTG in place of GAG?
A possible method of testing the posibility of a mutant gene would be to first grab a sample of genomic DNA then place it into a solution of MstII. Take the fragments and place them into a test tube with Ecoli vectors and Calcium so that the DNA fragments can enter the bacteria. Allow the bacteria to replicate. Once the bacteria replicates make a complimentary probe and insert it into the solution and examine the solution under UV light and if it glows, bravo! You have a possibility of sickle-cell anemia. Southern blotting would be useful because it could identify if there is a longer line or fragmented line; not definitive because there could be a mutation somewhere else in the restriction area.
The restriction enzymes KpnI and Acc65I recognize and cleave the same 6-bp sequence. However, the
sticky end formed from KpnI cleavage cannot be ligated
directly to the sticky end formed from Acc65I cleavage. Explain why. There is an image on p
This does not occur because the cleavage areas do not generate the same over-hang. The KpnI creates a 3' overhang while the Acc65I creates a 5' hang.
DNA sequences that are highly enriched in G-C base pairs typically have high melting temperatures.
Moreover, once separated, single strands containing
these regions can form rigid secondary structures. How
might the presence of G-C-rich regions in a DNA template affect PCR amplification?
Steps in PCR:
1) Heating the DNA to separate the strands
2) Cool solution to add primers to the 3' ends of both strands.
3) The solution is then heated again and this will allow the polymerase to synthesize the daughter strand, that will extend past the target sequence.
-For this question the problem would occur when it comes time to add the primers because the rigid secondary structure would hinder the possibility of attachment or even recognition. If it was primed, there woulds still be problems with the polymerase going down the strand.
The stringency of PCR amplification can be controlled by altering the temperature at which the primers and the target DNA undergo hybridization.
How would altering the temperature of hybridization
affect the amplification? Suppose that you have a particular yeast gene A and that you wish to see if it has a counterpart in humans. How would controlling the stringency of the
hybridization help you?
Stringency is the required closeness of the match between primer and target; controled by temperature and salt. If the hybridization temperature was changed this would alter the amount of target sequences amplified because then all of the strands would not be ss or primed. Some of the strands may renature. Reducing the temperature may even yield the wrong amplified targets because there would be mix-matched hybridization.
-If the human genome is mixed with primers for the yeast and there is no yield then this means that the surrounding seqences may not be the same. Repeat the experiment at lower temperatures of hybridization to discover the optimal temperature.
PCR is typically used to amplify DNA that lies between two known sequences. Suppose
that you want to explore DNA on both sides of a single known sequence. Devise a variation of the usual PCR protocol that would enable you to amplify entirely new genomic terrain.
1.) Heat the DNA in order to yield the two different strands of DNA.
2.) After this create a primer that is similar to the sequence of the known area. Cool the solution and put in the primers that are complementary for both the template and coding strands.
3) Heat up the solution in order to then bind the polymerase. Allow the synthesis to begin.
4) Use PCR to identify the synthesized areas.
Which of the following amino acid sequences would yield the most optimal oligonucleotide probe?
Because certain protein sequences can be generated from different codons, sequences that have tryptophan and methionine are preferred because they are made from 1 specific sequence.
-the best sequences would be the first and second.
Why might the genomic analysis of dogs be particularly useful for investigating the
genes responsible for body size and other physical
This would be particuarly useful because of the wide range of dog species that can be seen in the Dog species. These genes afect the dog's hair color, length of the hair, height, behaviors, and many other specific characteristics. When dogs are breed together they become homozygous for the characteristics making them very useful when studying phenotypic/genotypic inheritance.
You have identified a gene that is located on human chromosome 20 and wish to identify its location within the mouse genome. On which chromosome would you be most likely to find the mouse counterpart of this gene?
Based on the figure on page 151, chromosome 20 best resembles chromosome 2 in a mouse.
You wish to amplify a segment of DNA from a plasmid template by PCR with the use of the following primers: 5' GGATCGATGCTCGCGA-3'
and 5'AGGATCGGGTCGCGAG-39. Despite repeated attempts, you fail to observe a PCR product of the expected
length after electrophoresis on an agarose gel. Instead, you observe a bright smear on the gel with an approximate length of 25 to 30 base pairs. Explain these results
The primers are essentially prime oneanother and this would lead to the inability to prime the DNA. The line would be due to the 25-30 bp sequence corresponding to the overlapped primers.
Question number 19.
B-This person is a carrier, and from the DNA you can see the person has the defective gene due to the extra line. But if you look at the Northern and Southern blots they appear normal and this indicates that the individual does not display the traits.
C-This person has a defective gene, therefore does not make protein Y, leading to the disease.
D-This person has a normal gene however there is a problem with transcribing the mRNA and this leads to absence of the Y protein
E-This person has a normal gene, and mRNA but no protein leading to the idea that there is a problem translating
F-This indicates that the person has a normal DNA, mRNA, and protein synthesizer but is suffering from external problems.
Explain why there would be two peaks that seem to code for T or C.
This person does have a mutation and actually is heterozygous for this allele. One is wild the other is mutated.