Ch. 41 textbook problems

Differentiate between a cDNA library and a genomic library

1. The cDNA library is known as the DNA representation of mRNA whereas genome library is a collection of DNA fragments 2. The cDNA library is expressed in a particular tissue under a specific set of physiological conditions. Whereas genome library is known as the entire genome of an organism. 3. The cDNA library is different even in the same organism as they differ from different cell types. Whereas genome libraries that are prepared for any diploid tissue in an organism are identical.

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.

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 cleave 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 rapid procedure to distinguish between the normal and mutant gene would be southern blotting. When we subject the mutant and the normal gene to digestion by the restriction enzyme MstII; the normal gene will give one set of fragments while the mutant gene will give another set of fragments.

The power of PCR can create problems. Suppose someone claims to have isolated dinosaur DNA by using PCR. What questions might you ask to determine if it is indeed dinosaur DNA?

After sequencing the dna, we could compare it with the other genomes that we have sequenced. If it shows similarities with human, bacterial, or fungal dna, then it means that the isolated dna is not that of a dinosaur. If it shows similarity to the dna of birds or crocodiles, it could help confirm the dna is that of a dinosaur. That's because these species are evolutionarily similar.

polymerase chain reaction

Amplifies specific DNA sequences

restriction enzyme

Cleave double-helical DNA at specific sequences

reverse transcriptase

Copies RNA into DNA

cDNA library

DNA sequences representing mRNA

Ovalbumin is the major protein of egg white. The chicken ovalbumin gene contains eight exons separated by seven introns. Which should be used to form the protein in E. Coli: ovalbumin cDNA or ovalbumin genetic DNA? Why?

E. coli being prokaryotic does not have the genes for splicing the mRNA that has exons and introns. Therefore we need to use ovalbumin cDNA instead of ovalbumin genetic DNA to form the protein in E. coli.

Look at 16

FIGURE OUT

Suppose that you have isolated an enzyme that digests paper pulp and have obtained its cDNA. The goal is to produce a mutant that is effective at high temperature. You have engineered a pair of unique restriction sites in the cDNA that flank a 30-bp-coding region. Propose a rapid technique for generating many different mutations in this region.

First, design oligonucleotide primers that will create small mutations in the gene of interest. Then, use these oligonucleotides in oligonucleotide-directed mutagenesis to create a library of mutated versions of the gene of interest. Using the restrictions sites flanking the 30 base pair-coding regions, each mutant gene can be inserted into plasmid containing the same restriction site. Protein can then be expressed from these mutant genes and tested for its resilience to high temperatures.

genomic library

Fragments of DNA, housed in phages, that represent an entire genome

DNA ligase

Joins two DNA molecules

DNA microarray

Measures the expression of many genes

Look at 18

Person A is determined to be normal with 2 normal copies of the gene. And they act as a control in this experiment. Thus, he produces a normal mRNA (through northern blot) and a normal protein (through western blot.) Person B has one normal allele, while the other allele is mutated and defective. The mutated allele is smaller in size, suggesting that there has been a deletion. As, he has one normal copy of the gene, he produces the enzyme; hence, does not show any defects. Thus, he produces a normal mRNA (through northern blot) and a normal protein (western blot). And he does not show any symptoms. Person C has no normal allele. He has only mutated genes. He has only the smaller allele. As he has only the defective gene, he produces no enzyme; hence, shows defects. Thus, he does not produce a normal mRNA (through northern blot) or a normal protein (through western blot). And he shows the symptoms of a defective gene. Person D has a normal gene, but the transcription or the translation does not occur. This is seen in the absence of any bands in the northern blot and western blot. Hence, there may be a mutation in the promoter region of the gene, which prevents transcription and further translation. Thus, he does not produce a normal mRNA (through northern blot) or a normal protein (western blot). And he shows the symptoms of a defective gene. Person E has a normal gene and even produces a normal mRNA (northern blot). But the mRNA has not been translated (absence in western blot). Hence, there may be a mutation that will introduce a stop codon in the mRNA. This has prevented the translation of the mRNA and the subsequent absence of the protein. Thus, he does produce an mRNA (through northern blot), but he does not produce a normal protein (western blot). And he shows the symptoms of a defective gene. Person F has a normal gene and produces the mRNA and the protein. But still he shows the signs of a defective gene. Hence, there may be a mutation, which affects the activity of the protein and makes it inactive. This may be a mutation in the active site of the enzyme, which is vital, for the functioning of the protein. Thus, he shows the symptoms of a defective gene because of this mutation.

vector

Piece of DNA taken up and replicated by bacteria

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 temperate 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 human beings. How would controlling the stringency of the hybridization help you?

Synthesise primers corresponding to the ends of the yeast gene, used these primers and human DNA as the target,. If nothing amplified at 54C, then human gene is different from yeast gene. Repeat experiment at lower temperature of hybridization

why is Taq polymerase especially useful for PCR

The Taq polymerase is known as the DNA polymerase that are from thermophilic bacterium. And the thermophilic bacteria are the bacteria that live in hot springs. The PCR is known as polymerase chain reaction that occurs in high temperature. So, Taq polymerase is useful in PCR because it can easily stand without denaturing in high temperatures that are required for PCR.

what is a cDNA library? How is it constructed?

The cDNA or complementary DNA library is known as the collection of DNA sequences that represents all of the mRNA, which is expressed by a cell. The cDNA library is constructed, when by the use of reverse transcriptase and DNA polymerase enzymes the mRNA is isolated and converted into DNA. The structure of DNA is double stranded and it is also bound chemically by linkers that are inserted into some kind of vector.

A gel pattern displaying PCR products shows four strong bands. The four pieces of DNA have lengths that are approximately in the ratio of 1:2:3:4. The largest band if cut out of the gel and PCR is repeated with the same primers. Again, a ladder of four bands is evident in the gel. What does this result reveal about the structure of the encoded protein?

The visibility of 4 bands in the gel indicates that the encoded protein contains four repeats of the specific sequence in the largest band.

Suppose that a human genomic library is prepared by exhaustive digestion of human DNA with the EcoRI restriction enzyme. Fragments averaging about 4kb in length will be generated. Is this procedure suitable for cloning large genes? Why or why not?

These 4kb fragments are not suitable for cloning large genes This is because most of the human genes are much larger than 4kb and the generated fragments are only a small part of the complete gene. Hence, this procedure is not suitable for cloning large genes.

Look at 17

This particular sample comes from a person who is heterozygous for a point mutation in the gene. One allele is wild type, and the other contains the point mutation. Because the alleles are amplified equally by polymerase chain reaction (PCR), we see a double peak at the relevant position.

A successful PCR experiment often depends on deigning the correct primers. In particular, the Tm, the melting temperature of a double helix, for each primer should be approximately the same. What is the basis of this requirement?

Tm means the melting temperature of the double stranded nucleic acid. In PCR technique, the Tm of two primers should be the same. If there is too much of a difference in the melting temperature of the primers, the extent of hybridization with the target dna would differ during the annealing phase. This results in differential replication of the strands.

Propose a method for isolating a DNA fragment that is adjacent in the genome to a previously isolated DNA fragment. Assume that you have access to a complete library of DNA fragments in a vector but that the sequence of the genome under study has not yet been determined.

Using chemical synthesis or polymerase chain reaction to prepare hybridization probes that are complementary to both ends of the known DNA fragment. Add the hybridization probes to the clones representing the library of DNA fragment, select the clones that hybridize to one of the probes but not the other. Such clones are likely to represent DNA fragments that conain one end of the known fragments region with adjacent region of the particular chromosome

Why might the genomic analyses of dogs be particularly useful for investigating genes responsible for body size and other physical characteristics?

Within a single gene species, individual dogs show enormous variations in body size and other physical characteristics. The genomic analysis of the individual dogs would provide valuable clues which concern to the genes that are responsible for the diversity within the species.

Which of the following amino acid sequences would yield the most optimal oligonucleotide probe? Ala-Met-Ser-Leu-Pro-Trp Gly-Trp-Asp-Met-His-Lys Cys-Val-Trp-Asn-Lys-Ile Arg-Ser-Met-Leu-Gln-Asn

You use the amount of codons each amino acid codes for to figure this out. a. 4x1x6x6x4x1=576 possible coding sequences b. 4x1x2x1x2x2=32 possible coding sequences c. 2x4x1x2x2x3=96 possible codon sequences d. 6x6x1x6x2x2=864 possible codon sequences --> so fragment 2 is the most ideal probe with 32 possible coding combinations

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.

[1/4]^4 = 1/256 There are 4 bases and these 4 bases can be in any of the 4 positions. [1/4]^8 = 1/65536 There are 4 bases and they can each be in any of the 8 positions. Thus the average product of digestion by Alul would be 250bp in length and that of Notl would be 66,000bp in length.

sticky ends

complementary single strands DNA

Do number 9

in notebook

Sanger dideoxy method

sequencing by strand termination