DIFFERENCES IN SIZE BETWEEN EUKARYOTIC AND PROKARYOTIC DNA

Một phần của tài liệu Ebook Marks'' essentials of medical biochemistry a clinical approach (2nd edition): Part 1, 2E 1 (Trang 184 - 187)

A. Diploid versus Haploid

Except for the germ cells, most normal human cells are diploid. Therefore, they contain two copies of each chromosome, and each chromosome contains genes that are alleles of the genes on the homologous chromosome. Since one chromosome in each set of homologous chromosomes is obtained from each parent, the alleles can be identical, containing the same DNA sequence, or they can differ. A diploid human cell contains 2,000 times more DNA than the genome of the bacterium in the haploid E. coli cell (about 4 ⫻ 106 base pairs).

B. Introns

Eukaryotic introns contribute to the DNA size difference between bacteria and human cells. In eukaryotic genes, introns (noncoding regions) occur within se- quences that code for proteins. Consequently, the primary transcript (heterogeneous nuclear RNA or hnRNA) averages about 10 times longer than the mature mRNA produced by removal of the introns. In contrast, bacterial genes do not contain introns.

C. Repetitive Sequences in Eukaryotic DNA

Although being diploid, and containing introns account for some of the difference between the DNA content of humans and bacteria, a large difference remains that is related to the greater complexity of the human organism. Bacterial cells have a single copy of each gene, called unique DNA, and they contain very little DNA that does not produce functional products. Eukaryotic cells contain substantial amounts of DNA that does not code for functional products (i.e., proteins or rRNA and

T T

DNA

Transcription

Removal of intron tRNA

precursor Intron

Nucleus

Cytoplasm

Nuclear pore

5' 3'

5' 3'

5' CCA3'

OH

tRNA p

D

D T

5' CCA3'

tRNA D

D T

Modification of bases and addition

of CAA 1

2

3

4

FIG. 11.13. Overview of tRNA synthesis. D, T, Ψ, and 䊏 indicate modifi ed bases. D, dihy- drouracil; T, ribothymine; ψ, pseudouridine; 䊏, other modifi ed bases.

Calculate the number of different proteins, 300 amino acids in length, which could be produced from the E. coli genome (4 ⫻ 106 base pairs of DNA).

Lieberman_Ch11.indd 172

Lieberman_Ch11.indd 172 9/16/14 1:31 AM9/16/14 1:31 AM

CHAPTER 11 TRANSCRIPTION: SYNTHESIS OF RNA 173

tRNA). In addition, some genes that encode functional products are present in mul- tiple copies, called highly repetitive or moderately repetitive DNA. About 64% of the DNA in the human genome is unique, consisting of DNA sequences present in one or a very few copies in the genome (Fig. 11.14). Some of the unique DNA se- quences are transcribed to generate mRNA, which is translated to produce proteins.

Highly repetitive DNA consists of sequences about 6 to 100 base pairs in length that are present in hundreds of thousands to millions of copies, clustered within a few locations in the genome (see Fig. 11.14). It occurs in centromeres (which join sister chromatids during mitosis) and in telomeres (the ends of chromosomes). This DNA represents about 10% of the human genome. It is not transcribed.

Moderately repetitive DNA is present in a few to tens of thousands of copies in the genome (see Fig. 11.14). This fraction constitutes about 25% of the human ge- nome. It contains DNA that is functional and transcribed to produce rRNA, tRNA, and also some mRNA. The histone genes, present in a few hundred copies in the ge- nome, belong to this class. Moderately repetitive DNA also includes some gene se- quences that are functional but not transcribed. Promoters and enhancers (which are involved in regulating gene expression) are examples of gene sequences in this cate- gory. Other groups of moderately repetitive gene sequences that have been found in human are called the Alu sequences (about 300 base pairs in length). Alu sequences are also examples of short interspersed elements (SINEs). The long interspersed elements (LINE) sequences are 6,000 to 7,000 base pairs in length. The function of the Alu and LINE sequences has not been determined.

Major differences between prokaryotic and eukaryotic DNA and RNA are sum- marized in Table 11.1.

Four million base pairs contain 4 ⫻ 106/3 or 1.33 million codons. If each protein contained about 300 amino acids, E. coli could produce about 4,000 differ- ent proteins (1.33 ⫻ 106/300).

5S pre-rRNA genes

Telomere

mRNA gene

mRNA gene

mRNA gene mRNA gene

tRNA genes

Moderately repetitive sequences

Highly repetitive sequences

Highly repetitive sequences

mRNA gene mRNA gene mRNA gene

mRNA gene mRNA gene

mRNA gene

tRNA genes

sn/scRNA genes Unique sequences

Centromere

mRNA gene

mRNA gene mRNA gene

Telomere

Large pre-rRNA genes

Nucleolar organizer

mRNA gene

FIG. 11.14. Distribution of unique, moderately repetitive, and highly repetitive sequences in a hypothetical human chromosome. Unique genes encode mRNA. These genes occur in single copies. The genes for the large rRNA and the tRNA precursors occur in multiple copies that are clustered in the genome. The large rRNA genes form the nucleolar organizer. Moderately repetitive sequences are dispersed throughout the genome, and highly repetitive sequences are clustered around the centromere and at the ends of the chromosome (the telomeres). Small nuclear RNA (snRNA) and small cytoplasmic RNA (scRNA) are usually found in ribonucleoprotein particles. (From Wolfe SL. Mol Cell Biol 1993:761.)

Lieberman_Ch11.indd 173

Lieberman_Ch11.indd 173 9/16/14 1:31 AM9/16/14 1:31 AM

Table 11.1 Differences Between Eukaryotes and Prokaryotes

Eukaryotes (Human) Prokaryotes (E. coli)

Nucleus Yes No

Chromosomes

Number 23 per haploid cell 1 per haploid cell

DNA Linear Circular

Histones Yes No

Genome

Diploid Somatic cells No

Haploid Germ cells All cells

Size 3 ⫻ 109 base pairs per haploid cell 4 ⫻ 106 base pairs Genes

Unique 64% 100%

Repetitive

Moderately 25% None

Highly 10% None

Operons No Yes

mRNA

Polycistronic No Yes

Introns (hnRNA) Yes No

Translation Separate from transcription Coupled with transcription

Table 11.2 Diseases Discussed in Chapter 11 Disorder or Condition

Genetic or

Environmental Comments

β-Thalassemia Genetic An anemia due to an imbalance in β- and α-globin chain synthesis. For a β-thalassemia, more α-chain is synthesized than functional β-chain.

Tuberculosis, a compli- cation of AIDS

Environmental The drug rifampin, amongst others, is used to treat tuberculosis via inhibition of bacterial RNA polymerase.

Mushroom poisoning (α-amanitin poisoning)

Environmental Inhibition of RNA polymerase II by α-amanitin.

There is no effective antidote for this poison.

Systemic lupus erythe- matosus (SLE)

Both The development of autoantibodies directed against various cellular proteins, including those involved in RNA processing (such as complexes involved in RNA splicing, the snRNPs).

C L I N I CA L CO M M E N T S

The diseases discussed in this chapter are summarized in Table 11.2.

Lisa N. Patients with β⫹-thalassemia who maintain their hemoglobin levels above 6.0 to 7.0 g/dL are usually classifi ed as having thalassemia in- termedia. In the β-thalassemias, the α-chains of adult hemoglobin A (α2β2) continue to be synthesized at a normal rate. These chains accumulate in the bone marrow in which the red blood cells are synthesized during the process of erythro- poiesis (generation of red blood cells). The accumulation of α-chains diminishes erythropoiesis, resulting in an anemia. Individuals who are homozygous for a severe mutation require constant transfusions.

Individuals with thalassemia intermedia, such as Lisa N., could have inherited two different defective alleles, one from each parent. One parent may be a “silent”

carrier, with one normal allele and one mildly affected allele. This parent produces enough functional β-globin so few or no clinical symptoms of thalassemia ap- pear. (However, they generally have a somewhat decreased amount of hemoglobin, resulting in microcytic hypochromic red blood cells.) When this parent contributes the mildly defective allele and the other heterozygous parent contributes a more se- verely defective allele, thalassemia intermedia occurs in the child. The child is thus heterozygous for two different defective alleles.

The mutations that cause the thalas- semias affect the synthesis of either the α- or the β-chain of adult hemo- globin, causing an anemia. They are classifi ed by the chain affected (α- or β-chain) and by the amount of chain synthesized (0 for no synthesis and ⫹ for synthesis of some functional chains).

They are also classifi ed as major, intermediate, or minor, according to the severity of the clini- cal disorder. β-Thalassemia major (also called homozygous β-thalassemia) is a clinically se- vere disorder requiring frequent blood trans- fusions. It is caused by the inheritance of two alleles for a severe mutation. In β-thalassemia intermedia, the patient exhibits a less severe clinical phenotype and is able to maintain he- moglobin levels above 6 g/dL. It is usually the result of two different mild mutations or ho- mozygosity for a mild mutation. β-Thalassemia minor (also known as β-thalassemia trait) is a heterozygous disorder involving a single muta- tion that is often clinically asymptomatic.

During embryonic and fetal life, the β-chain is replaced by the ε-and γ-chains, respectively.

As a result, patients with severe mutations in the α-chain tend to die in utero, while those with mutations in the β-chains exhibit symp- toms postnatally as hemoglobin F (α2γ2) is nor- mally replaced with adult hemoglobin A (α2β2) after birth.

Lieberman_Ch11.indd 174

Lieberman_Ch11.indd 174 9/16/14 1:31 AM9/16/14 1:31 AM

CHAPTER 11 TRANSCRIPTION: SYNTHESIS OF RNA 175

Isabel S. Isabel S. was treated with a multidrug regimen for tuberculosis because the microbes that cause the disease frequently become resistant to the individual drugs. As with patients with a normal immune system, the current approach in patients with AIDS who develop opportunistic infection with Mycobacterium tuberculosis is to initiate antimycobacterial therapy with four agents because the mycobacteria frequently become resistant to one or more of the particu- lar antitubercular drugs. Isabel S. was started on isoniazid (INH), rifampin, pyrazin- amide, and ethambutol. INH inhibits the biosynthesis of mycolic acids, which are important constituents of the mycobacterial cell wall. Rifampin binds to and inhibits bacterial RNA polymerase, which selectively kills the bacteria that cause the infec- tion. Pyrazinamide, a synthetic analogue of nicotinamide, targets the mycobacterial fatty acid synthase I gene involved in mycolic acid biosynthesis in M. tuberculosis.

Ethambutol blocks arabinosyl transferases that are involved in cell wall biosynthesis.

Just as bacteria can become resistant to drugs, so can HIV. Because of this con- cern, patients with HIV are no longer treated with a single-drug regimen. Multidrug regimens usually include two nucleoside reverse transcriptase inhibitors (NRTIs), such as dideoxyinosine (didanosine, formerly called ddI) and dideoxycytidine (zal- citabine, formerly called ddC), as well as nonnucleoside reverse transcriptase in- hibitors (NNRTIs) (e.g., efavirenz) or protease inhibitors (PIs) (e.g., indinavir). PIs prevent the HIV polyprotein from being cleaved into its mature products. (When the HIV mRNA is translated, a polyprotein is formed, which must be cleaved by HIV protease in order to form the mature proteins needed for the assembly of the virus. An inability to cleave the proprotein will block viral maturation and further infection of other cells.) There are also integrase strand transfer inhibitors (INSTIs), which block integration of the viral genome into the host genome, and entry inhibi- tors (EIs), which block viral entry into target cells.

Isabel S. was started on efavirenz (an NNRTI) as her third drug and was coun- seled on not getting pregnant because the drug is teratogenic.

Sarah L. Systemic lupus erythematosus (SLE) is a multisystem disease of unknown origin characterized by infl ammation related to the presence of autoantibodies in the blood. These autoantibodies react with antigens nor- mally found in the nucleus, cytoplasm, and plasma membrane of the cell. Such “self”

antigen–antibody (autoimmune) interactions initiate an infl ammatory cascade that produces the broad symptom profi le of multiorgan dysfunction found in Sarah L.

Pharmacological therapy for SLE involves anti-infl ammatory drugs and immuno- suppressive agents. It can include nonsteroidal anti-infl ammatory drugs (NSAIDs), corticosteroids, antimalarials, or several newer agents that cause immunosuppres- sion. Plaquenil is an antimalarial drug used to treat skin and joint symptoms in SLE, although its exact mechanism of action in these patients is not fully understood.

Sarah was placed on such a drug regimen.

Isoniazid is often prescribed with vitamin B6 (pyridoxine), as isoniazid can interfere with the activation of this vitamin (to pyridoxal phosphate), which can lead to an alteration in normal cellular me- tabolism and result in a clinical neuropathy. In patients susceptible to low levels of vitamin B6, taking additional pyridoxine overcomes this interference.

Studies have indicated that a failure to properly dispose of cellular debris, a normal byproduct of cell death, may lead to the induction of autoantibodies di- rected against chromatin in patients with SLE.

Normal cells have a fi nite lifetime and are pro- grammed to die (apoptosis) through a distinct biochemical mechanism. One of the steps in this mechanism is the stepwise degradation of cellular DNA (and other cellular components).

If the normal intracellular components are ex- posed to the immune system, autoantibodies against them may be generated. The enzyme in cells that degrades DNA is deoxyribonucle- ase I (DNase I), and individuals with SLE have reduced serum activity levels of DNase I, compared to individuals who do not have the disease. Through an understanding of the mo- lecular mechanism whereby autoantibodies are generated, it may be possible to develop therapies to combat this disorder.

1. The short transcript AUCCGUACG would be derived from which one of the following template DNA sequences?

Một phần của tài liệu Ebook Marks'' essentials of medical biochemistry a clinical approach (2nd edition): Part 1, 2E 1 (Trang 184 - 187)

Tải bản đầy đủ (PDF)

(340 trang)