Open ended Genetics CH 3 & 4

How will DNA replication be affected if DNA polymerase I has a mutation that inactivates 5ʹ-to-3ʹ exonuclease activity?

The 5ʹ-to-3ʹ exonuclease activity of DNA polymerase I removes the RNA primers from the ends of Okazaki fragments. If this activity is missing, the primers may not be removed from the growing DNA strands.

A person with PKU controls the buildup of phenylpyruvic acid in his or her blood by restricting dietary intake of phenylalanine, thus avoiding the severe symptoms of mental retardation, slow growth rate, and early death. Yet he or she still exhibits the symptoms of fair skin and low adrenaline levels. Why?

The PKU mutation is in the gene for phenylalanine hydroxylase, the enzyme that converts phenylalanine to tyrosine. Tyrosine is essential for melanin and adrenaline production. A PKU individual cannot make tyrosine, and it is difficult to obtain enough tyrosine from food intake to avoid any symptoms of the disease.

While the defective gene product in patients with PKU was identified using biochemical analysis, how was the cystic fibrosis gene identified?

The cystic fibrosis gene was identified by a combination of genetic and modern molecular biology techniques. It was first localized to chromosome 7, and then molecularly cloned and sequenced. The sequence of the mutant allele was compared with the allele of a normal person to discover a three base‐pair deletion in the cystic fibrosis gene. To determine the geneʹs function, its amino acid sequence and three‐dimensional structure was deduced. This was compared with a database of protein sequences to determine the geneʹs function as an active transport protein.

The following is a representation of a metabolic pathway that involves genes for enzymes A, B, C, and D: A → B → C → D → product A mutation in the B gene would result in an accumulation of which gene product?

A

Explain the difference between the one‐geneone‐enzyme hypothesis and the one‐gene one‐polypeptide hypothesis.

A relationship is found between a mutation in a single gene and the loss of production of a single enzyme in a metabolic pathway. However, more than one gene may control each step in a pathway because some enzymes (and some non‐enzyme proteins such as hemoglobin) consist of two or more polypeptide chains, each of which is coded for by a different gene. Thus, it is more accurate to use the phrase ʺone‐geneone‐polypeptide.ʺ

Why is an AT-rich sequence characteristic of DNA replicators in all organisms?

AT-rich regions of DNA are relatively easy to denature to single strands. AT base pairs are held together by only two hydrogen bonds, while GC pairs are held together by three.

What is an essential amino acid? Give an example.

An essential amino acid is one that humans must consume as part of their diet, because they are unable to synthesize it. An example is phenylalanine.

The diploid set of chromosomes in Drosophila embryos replicates six times faster than the single E. coli chromosome, even though there is about 100 times more DNA in Drosophila than in E. coli and the rate of movement of the replication fork in Drosophila is much slower. How is this so?

Eukaryotic chromosomes duplicate rapidly because DNA replication initiates at many origins of replication throughout the genome. In E. coli, there is only one replicon, while in eukaryotes there are multiple, smaller replicons.

How did Meselson and Stahl rule out the conservative model of DNA replication using equilibrium density gradient centrifugation?

First, they grew E. coli cells in a medium containing only high-density nitrogen (15N). Then they allowed the cells to undergo successive rounds of replication in the presence of normal density nitrogen (14N). They used equilibrium density gradient centrifugation after each round of replication to separate the DNA produced by density. If the conservative model was correct, they would have found no intermediate-density DNA after a round of replication. However, after one round of replication, they found that the entire amount of DNA had a density exactly intermediate between 14N and 15N. Subsequent rounds of the experiment showed that semiconservative replication was the correct model.

A cross is made between yeast cells with different alleles for a set of linked genes: pr+q × p+rq+. The resulting tetrads show a 3:1 ratio for r+ to r instead of the expected 2:2 ratio. Can you explain how this could have occurred?

Gene conversion by a mismatch repair mechanism could have caused this deviation from expected ratios. During pairing of homologous chromosomes in meiosis, recombination between the two inner chromatids occurred, resulting in heteroduplex (mismatched) DNA strands. Both mismatches were repaired by excision and DNA synthesis to match the parent DNA with the r+ allele.

How do scientists, lacking the ability to make controlled genetic crosses in humans, study the inheritance of and give genetic counseling on human genetic disorders?

Pedigree analysis can be done by compiling phenotypic records from both families over several generations. This can determine the likelihood that a particular allele is present in either family.

You are a doctor who specializes in genetic diseases. A married couple comes to see you for genetic counseling. The woman has PKU, the symptoms of which she manages by diet. The man and woman are first cousins and would like to know the probability that any of their children will have the disease. What would you tell them?

Since the woman has the disease, she must be homozygous recessive for the defective autosomal gene. If her husband is not a carrier, all their children will be heterozygous carriers of the disease. However, since the man and woman are closely related, there is a high probability that the man is a carrier of the PKU gene. If he is a carrier, there is a 50% chance that any of their children will have the disease. You would perform a detailed pedigree analysis for the couple to better determine the probability that the husband carries the recessive allele.

How do the DNA polymerase repair mechanisms work?

Both DNA Pol I and DNA Pol III have 3ʹ-to-5ʹ exonuclease activity and can remove nucleotides from the end of a DNA chain as part of an error correction mechanism. If an incorrect base is inserted by DNA polymerase, and the error is recognized immediately, the exonuclease activity excises the erroneous nucleotide from the new strand. After excision, the DNA polymerase resumes motion in the forward direction and inserts the correct nucleotide.

What are some key differences in replication between E. coli DNA and λ phage DNA?

Both start replication as a circular molecule, and in E. coli, the parental DNA strands remain in a circular form throughout the replication cycle. A replication bubble opens to form two replication forks, and replication proceeds bidirectionally. Lambda phage DNA is replicated by a rolling circle mechanism, in which a nick is made in one of the two strands of the circle, and the 5ʹ end of the cut DNA strand is rolled out as a free ʺtongueʺof increasing length as replication proceeds . The parental DNA circle is replicated continuously, while the linear displaced strand is replicated discontinuously. As long as the circle continues to roll, concatamers of phage DNA can be produced. This is later cut up into linear chromosomes and packaged into new phage heads.

What advantages does chorionic villus sampling have over amniocentesis for fetal analysis of genetic defects?

Chorionic villus sampling is generally performed up to four weeks earlier than amniocentesis, allowing parents to make an earlier decision regarding potential termination of pregnancies. It is also a faster test that does not require cell culture to do biochemical assays.

What is an inborn error of metabolism? Give an example.

IEM are a large group of rare genetic diseases that generally result from a defect in an enzyme or transport protein which results in a block in a metabolic pathway.An inborn error of metabolism is a genetic disease caused by the absence of a particular enzyme necessary to fulfill a biochemical pathway. An example is alkaptonuria, in which an enzyme necessary for homogentisic acid (HA) metabolism is not made. Therefore, people with this disease cannot metabolize HA and it is excreted in their urine.

How can enzyme assay be used to detect carriage of a recessive gene mutation in a person of normal phenotype?

If they were a carrier, the person, as a heterozygote, would be expected to produce only half the normal enzyme or protein product for the gene in question.

What experiment demonstrated that the 5ʹ-to-3ʹ exonuclease activity of DNA polymerase I was essential for cell viability?

In E. coli, the DNA Pol I polAex1 mutant strain survives at 37 ̊C but dies at 42 ̊C. This was shown to be because the mutant DNA Pol I enzyme had normal activity at 37 ̊C but a defective 5ʹ-to-3ʹ activity at 42 ̊C. This demonstrated that the 5ʹ-to-3ʹ exonuclease activity of DNA Pol I was essential for cell survival.

Describe the method devised by Arthur Kornberg which first successfully achieved DNA synthesis in vitro, including its components and their uses.

Kornberg mixed together DNA fragments, all four dNTPs (DNA precursors), and an E. coli lysate to achieve DNA synthesis in vitro. The DNA fragments acted as a template for the synthesis of new DNA. The dNTPs were precursors of the new DNA strand, and the cell lysate contained the enzyme necessary to catalyze DNA synthesis (DNA Pol I).

When the RNA primers are removed from the 5ʹ ends of eukaryotic chromosomes after replication, DNA polymerase is unable to fill in the gaps. What prevents the chromosomes from getting shorter and shorter with each round of replication?

The enzyme telomerase maintains chromosome lengths by adding telomere repeats to the chromosome ends. Telomerase contains an RNA component that is complementary to the telomere repeat unit of the chromosome. After binding to the overhanging repeat, the telomerase RNA is used as a template to synthesize new chromosomal telomere repeats.

What are the key replication enzymes at the replisome, and how is DNA replication on both leading and lagging strands made efficient through the conformation of the DNA at the replisome in prokaryotes?

The key replication enzymes at the replisome are helicase, primase, and DNA polymerase III. To make replication more efficient, the lagging-strand DNA is folded so that its DNA polymerase III is complexed with the DNA polymerase III on the leading strand (forming the DNA Pol III holoenzyme). The folding of the lagging-strand template also makes production of sequential Okazaki fragments more efficient by bringing the 3ʹ end of each completed Okazaki fragments near the site where the next Okazaki fragment will start.

What are the differences in replication between leading and lagging strands in terms of continuity and directionality in relation to the replication fork?

The leading strand is copied continuously from the 3ʹ end toward the replication fork, while the lagging strand is copied in fragments away from the replication fork.

Why is DNA replication referred to as semiconservative?

The progeny double helices consist of one parental DNA strand and one newly synthesized strand.

In the laboratory, how could you distinguish between normal hemoglobin A and hemoglobin S?

The two forms of hemoglobin have different electrical charges. Therefore, you could distinguish between them by their migration patterns on an electrophoresis gel.

Classic albinism is an autosomal recessive mutation. However, two parents with albinism may produce a child with normal pigment. How can this be so?

There are a number of biochemical steps in melanin biosynthesis from tyrosine. For this reason, albinism can be caused by different enzyme deficiencies at different steps of the pathway. If each of the parentsʹ albinism stems from a different enzyme deficiency, they can therefore have a child with normal pigment.

What factor do the diseases PKU, albinism, and alkaptonuria all have in common?

They are all marked by a block in the phenylalanine‐tyrosine metabolic pathways.

Why are genetic diseases much more common among children of marriages involving first cousins than among children of marriages between unrelated partners?

This is because first cousins have many alleles in common, so the chances are greater for recessive alleles to be homozygous in children of first‐cousin marriages.

Why do tumor cells in mammals have telomerase activity?

Tumor cells are ʺimmortalʺ cells, and the enzyme telomerase is required for long-term cell viability. It has been demonstrated that mutations in genes such as TLC1 and EST1 decrease cell longevity by continuous shortening of telomere length.

How were the sequences that compose the replication origins in yeast discovered?

Yeast cells were grown in heavy isotope media to produce denser DNA and then shifted to media with normal, light isotopes. After a few minutes, DNA from these cells was extracted and cut into small pieces. The lighter DNA fragments (which should contain the origins of replication) were collected, fluorescently labeled, and used to hybridize a microarray of yeast sequences. Determination of the sequences to which this light, fluorescently labeled DNA hybridized led to the identification of a number of yeast replication origins.

You discover a mutant strain of Neurospora crassa that will not grow on a minimal medium but that will grow on a medium supplemented by the amino acid methionine. How can you determine which step in the methionine synthesis pathway is affected by the mutation?

You would grow the strain on a series of media each supplemented with a chemical intermediate in the methionine biosynthetic pathway and observe for growth. The mutant will be able to grow on any medium that contains an intermediate substance in the pathway that comes after the step controlled by its mutant gene. It will also accumulate the intermediate compound used in the step where it is blocked.