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9.7: Structure of DNA - Biology

9.7: Structure of DNA - Biology


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Learning Outcomes

Diagram the structure of DNA

The building blocks of DNA are nucleotides. The important components of each nucleotide are a nitrogenous base, deoxyribose (5-carbon sugar), and a phosphate group (see Figure 1). Each nucleotide is named depending on its nitrogenous base. The nitrogenous base can be a purine, such as adenine (A) and guanine (G), or a pyrimidine, such as cytosine (C) and thymine (T). Uracil (U) is also a pyrimidine (as seen in Figure 1), but it only occurs in RNA, which we will talk more about later.

The nucleotides combine with each other by covalent bonds known as phosphodiester bonds or linkages. The phosphate residue is attached to the hydroxyl group of the 5′ carbon of one sugar of one nucleotide and the hydroxyl group of the 3′ carbon of the sugar of the next nucleotide, thereby forming a 5′-3′ phosphodiester bond.

In the 1950s, Francis Crick and James Watson worked together to determine the structure of DNA at the University of Cambridge, England. Other scientists like Linus Pauling and Maurice Wilkins were also actively exploring this field. Pauling had discovered the secondary structure of proteins using X-ray crystallography. In Wilkins’ lab, researcher Rosalind Franklin was using X-ray diffraction methods to understand the structure of DNA. Watson and Crick were able to piece together the puzzle of the DNA molecule on the basis of Franklin’s data because Crick had also studied X-ray diffraction (Figure 2). In 1962, James Watson, Francis Crick, and Maurice Wilkins were awarded the Nobel Prize in Medicine. Unfortunately, by then Franklin had died, and Nobel prizes are not awarded posthumously.

Watson and Crick proposed that DNA is made up of two strands that are twisted around each other to form a right-handed helix. Base pairing takes place between a purine and pyrimidine; namely, A pairs with T and G pairs with C. Adenine and thymine are complementary base pairs, and cytosine and guanine are also complementary base pairs. The base pairs are stabilized by hydrogen bonds; adenine and thymine form two hydrogen bonds and cytosine and guanine form three hydrogen bonds. The two strands are anti-parallel in nature; that is, the 3′ end of one strand faces the 5′ end of the other strand. The sugar and phosphate of the nucleotides form the backbone of the structure, whereas the nitrogenous bases are stacked inside. Each base pair is separated from the other base pair by a distance of 0.34 nm, and each turn of the helix measures 3.4 nm. Therefore, ten base pairs are present per turn of the helix. The diameter of the DNA double helix is 2 nm, and it is uniform throughout. Only the pairing between a purine and pyrimidine can explain the uniform diameter. The twisting of the two strands around each other results in the formation of uniformly spaced major and minor grooves (Figure 3).


Structure of an H1-Bound 6-Nucleosome Array Reveals an Untwisted Two-Start Chromatin Fiber Conformation

Chromatin adopts a diversity of regular and irregular fiber structures in vitro and in vivo. However, how an array of nucleosomes folds into and switches between different fiber conformations is poorly understood. We report the 9.7 Å resolution crystal structure of a 6-nucleosome array bound to linker histone H1 determined under ionic conditions that favor incomplete chromatin condensation. The structure reveals a flat two-start helix with uniform nucleosomal stacking interfaces and a nucleosome packing density that is only half that of a twisted 30-nm fiber. Hydroxyl radical footprinting indicates that H1 binds the array in an on-dyad configuration resembling that observed for mononucleosomes. Biophysical, cryo-EM, and crosslinking data validate the crystal structure and reveal that a minor change in ionic environment shifts the conformational landscape to a more compact, twisted form. These findings provide insights into the structural plasticity of chromatin and suggest a possible assembly pathway for a 30-nm fiber.

Keywords: 30-nm fiber chromatin cryo-EM crystallography histone H1 linker histone nucleosome array.


9.7: Structure of DNA - Biology

* Packing ratio = length of DNA or DNA/protein complex relative to original length of DNA.

E. Modifications of histones -- helps tighten up or loosen chromatin

1. Histones have tails that stick out (see handout 9B).

a. Many different modifications of amino acids in tail are possible: For example, addition/removal of acetate, methyl or phosphate groups to side chains of AA.

b. Regulatory Function: Modifications of histones affect folding of chromatin and therefore activity of genes.

a. Phosphorylation of H1 occurs in M changes in kinase and phosphatase activity affect state of histones and folding of chromatin in parallel with changes in lamins.

b. Acetylation of lys side chains of histones. Acetylation of histones --> more active, looser chromatin (see articles). Acetylation of H3 & H4 is higher in active chromatin.

c. Other modifications , such as methylation, occur and are thought to serve regulatory purposes as well. (Both DNA and histones can be methylated.)

To review nucleosome structure, try problems 4-1, 4-3 A, 4-4 A & 4-8 A.

IV. Does Genetic Activity Correlates with States of Folding of Euk. Chromatin? How to test?

A. Polytene chromosomes (See Becker Ch. 21 esp. fig. 21-15 & 21-16) Special Interesting Case. Will not be covered in lecture is here FYI.

1. Advantages:

a. Special case where many chromatids -- about 1000 -- are lined up together with genes and loops of chromatin aligned therefore can see differences in state of folding in light microscope.

b. Can compare changes in folding of chromatin and level of transcription (by incorporation of labeled U) in a single tissue as it responds to hormones that alter gene expression.

2. Disadvantage : Can't compare regions of active/inactive chromatin from many dif. tissues.

3. Results: Active (transcribed) regions are clearly looser -- individual loops stretched out more. (See Becker )

B. Results of treatment with DNase I (of ordinary chromosomes/chromatin) -- How folded is Euchromatin?

1. The problem -- Not all euchromatin is transcribed in any one cell -- is it all folded to the same degree or not?

a. All (eu)chromatin looks about the same in the light microscope. Therefore indirect methods (such as DNase treatment) are necessary to test state of folding.

b. Use of DNases. States of folding of (eu)chromatin are often distinguished by effects of treatment with various types of DNase. State of (eu)chromatin will determine relative sensitivity of DNA (while still in chromatin) to degradation by various DNases. DNA that is in tighter areas of chromatin will be more protected from degradation. (Some examples of this will be discussed below and are in problem sets.)

2 . Method -- see Becker fig. 21-17. Want to examine chromatin region containing a particular gene, say globin genes, from dif. tissues. (Note: this region is euchromatic in interphase in all these tissues, even though it is not expressed in all of them. No gross difference is visible that's why indirect methods are necessary.)

a. What you want to do: Fish out section of chromatin containing globin genes (from cells that make globin and cells that don't) and test state of chromatin folding.

b. What you have to do: Need to test for folding indirectly, and to do experiment in reverse. Test state of folding first (indirectly) and then test state of genes.

c. General procedure: Treat chromatin with DNase, differentially degrading DNA in active and inactive chromatin. Then use probe to test for state of known genes and see if genes were degraded or not.

d. Details of DNase treatment: Treat chromatin with DNase (Usually DNase I, a different enzyme from one used previously to get a "ladder"). Then isolate DNA and see what state it's in.

(1). Can vary conditions -- amount of enzyme, time, etc. to distinguish various degrees of sensitivity to the enzyme.

(2). Can test chromatin from many different tissues, say erythrocytes (in chickens -- still have nuclei) vs. brain

(3). Can test state of many different genesusing diff. probes (in step e below).

(4). DNase I cuts the DNA differently than micrococcal nuclease. DNA that is simply wound around nucleosomes is not totally protected from DNase I -- further folding up is required. So resistance to DNase I is a measure of how tightly the nucleosomes are folded in on themselves. DNase I does not cut preferentially at any particular sequence or at any particular place relative to the position of the nucleosome. Does not give "ladder."

e. Expected result: If chromatin is relatively loose, DNA will be unprotected and readily degraded by DNase I. (There will be nothing left to hybridize to the probe for, say, globin.) If chromatin is relatively tight, DNA in that region will NOT be easily degraded. (And there will be DNA to hybridize to probe.)

f. Details of how to measure state of DNA (after treatment of chromatin with DNase I):

(1). Prepare DNA. Extract DNA (remove histones) treat naked DNA with restriction enzyme (to give pieces of reasonable size IF DNA was protected from DNase I). If DNA was in loose region of chromatin, it will already have been degraded by DNase I.

(2). Find regions corresponding to known genes. Run restriction fragments on gel do blot, identify regions of interest (say, globin genes) with probe. Only regions from areas with relatively tight chromatin will give clear, undegraded bands on the gel.

3. Results of treatment of euchromatin with DNase I and other nucleases.

a. Almost all eukaryotic DNA is in nucleosomes. (See (c) for exceptions.) How do we know?

(1). Almost all DNA is much more resistant to digestion by DNase I than naked DNA

(2). Almost all DNA forms ladder when treated with micrococcal DNase.

b. Actually transcribed (coding) DNA is more sensitive to DNase I than ordinary euchromatic DNA. But transcribed DNA is much more resistant than naked DNA -- it is still in nucleosomes.

c. Hypersensitive sites exist -- these are the only sections not in nucleosomes.

(1). Some hypersensitive sites found= very sensitive regions (100X more sensitive to degradation by DNase I than heterochromatin 10X more sensitive than active euchromatin.)

(2). Hypersensitive sites correspond to sites without nucleosomes

(3). DNA in hypersensitive sites is not naked -- it has other proteins, but no histones. See Becker fig. 21-18.

(4). Hypersensitive sites correspond to regulatory, not coding, regions(in areas of active transcription).

(5). Why are hypersensitive sites so sensitive to DNase? Transcription factors (regulatory proteins) have replaced histones and the other proteins don't protect the DNA as well as histones do.

C. Possible States of Interphase Chromatin -- what is inferred from all the results, and the implications. (See table on handout. )

a. Super Tight = heterochromatin = essentially not uncoiled from mitosis. Genetically inactive.

b. Euchromatin -- 3 major states?

(1). Loose. Looser than heterochromatin, but not being transcribed (genetically inactive) at the moment. 30 nm fiber?

(2). Looser -- but still has nucleosomes. Beads-on-a-string? Stretched out loops? Example: coding region that are active -- they are actually being transcribed. (Also includes genes that were transcribed recently, or are next to transcribed regions, etc.)

(3). Loosest or hypersensitive region (missing some histones has regulatory proteins instead). Example: active promotor or enhancer.

c. A caution: There are probably intermediate states, and the ones above are simply the only ones that have been clearly distinguished using the methods currently available. For example, all inactive euchromatin is probably not the same. See ** below.

To review the differences in states of chromatin, uses of DNase, etc., try the rest of 4-3 to 4-8.

V. Introduction to Regulation of Eukaryotic Gene Expression

A. What has to be done to turn a eukaryotic gene on/off? What steps can be regulated?

1. In prokaryote (for comparison) -- process relatively simple.

a. Most regulation at transcription.

b. Translation in same compartme nt as transcription translation follows automatically.

c. Most mRNA has short half-life.

2. In eukaryote -- Most regulation is at transcription, but you have more steps & complications -- more additional points of regulation. See Becker fig. 21-11.

a. Need to unfold/loosen chromatin before transcription possible. We know that modification of histones, binding of certain non-histone proteins, and/or methylation of DNA is correlated with state of folding. Not clear what is primary cause and what is secondary effect of unfolding. Two possible models (most current data favors the first):

(1) Two step model for regulation. See Becker fig. 21-10.

(a). First must de-condense (loosen up) euchromatin to a transcribable state = loose (compared to heterochromatin and compared to inactive euchromatin). Pull out 30nm fiber to beads-on-a-string stage?

(b). Then add transcription factors (more or less as for prokaryotes) --> actual transcription

(1). Regions with transcription factors = nucleosomes removed = hypersensitive sites

(2). Regions being transcribed = nucleosomes are somehow "loosened up" or "remodeled" but not removed.

(c). What changes state of chromatin? (To tighten or loosen.)

(1). Remodeling proteins: these are responsible for moving and/or loosening up nucleosomes. See Purves 14.16 (14.17).

(2). Enzymes that modify histone tails. Changes in modification may have a direct effect and/or affect binding of regulatory proteins.

(2). One Step Model -- addition of TF's is primarily what opens up the chromatin -- don't need a separate remodeling step first. Alternatively, both unfolding or decondensation and addition of TF's occur simultaneously.

b. Transcript must be processed (capped, spliced, polyadenylated, etc.) -- any of these steps can be regulated, and there is more than one way to process most primary transcripts. (An example will be discussed next time.)

c. Transcription & translation occur in separate compartments .

(1). mRNA must be transported to cytoplasm.

(2). Translation can be regulated (independently of transcription) -- can control usage of mRNA, not just supply of mRNA. (An example will be discussed next time.) For any particular mRNA, can regulate 1 or both of following:

(a). rate of initiation of mRNA translation (how often ribosomes attach and start translation)

(b). rate of degradation of mRNA

IV. Major features of gene regulation in Eukaryotes

A. How can amount of protein be controlled? If cell makes more or less protein, which step(s) are regulated? Many possible points of regulation in eukaryotes. See Becker fig. 21-11, or Purves 14.11 (14.13), and list of steps above.

1. Most common point of regulation is at transcription (in both euk. & prok.) If you need more protein, usually make more mRNA.

2. Transcription is not the only step controlled, especially in euk. (Some examples will be discussed later.)

B. If cells make different proteins, how is that controlled? If two eukaryotic cells (from a multicellular organism) make different proteins, what is (usually) different between them?

1. Is DNA different? (No, except in immune system.)

2. Is state of chromatin usually different? (Ans: yes) How is this tested? Method & result described above. See figure 21-17 in Becker. What causes the difference in states of chromatin? Not clear what is cause and what is effect.

3. Is mRNA usually different? (Ans: yes). This means you can get tissue specific sequences from a cDNA library. (cDNA library = collection of all cDNA's from a particular cell type.) DNA from each cell type is the same mRNA and therefore cDNA is not. See Becker fig. 21-20.

4. If mRNA's are different, why is that? Is the difference due entirely to differences in transcription?

a. Transcription is different in different cells . It could be that all cells transcribe all genes, but only some RNA's are exported to the cytoplasm and the remaining nuclear RNAs are degraded. This is not the case. Only selected genes are transcribed in each cell type, and RNA's from those genes are processed to make mRNA.

b. Splicing and processing of same primary transcripts can be different ( in different cells or at different times). Different mRNA's (& therefore proteins) can be produced from the same transcript by alternative splicing and/or poly A addition.

To review possible steps in regulation, try problems 4-9 to 4-11.


Forms of DNA

DNA is a very flexible molecule and has the ability to exist in various forms based on the environmental conditions. Naturally occurring DNA double helices are classified into A, B and Z-types. A and B-forms of DNA are the right handed forms whereas Z-DNA is the left handed form. When hydrated the DNA generally assumes B-form. The A conformation is found when there is little water to interact with the helix and is also the conformation adopted by the RNA. The formation of Z-DNA occurs with the methylation of deoxycytosine residues and also during transcription where negative supercoiling stabilizes it.

Parameter A-DNA B-DNA Z-DNA
Helix sense right-handed right-handed left-handed
Residues per turn 11 10.5 12
Axial rise [Å] 2.55 3.4 3.7
Helix pitch(°) 28 34 45
Base pair tilt(°) 20 −6 7
Rotation per residue (°) 33 36-30
Diameter of helix [Å] 23 20 18
Glycosidic bond configuration
dA,dT,dC
dG

anti
anti

anti
anti

anti
syn
Sugar pucker
dA,dT,dC
dG

C3'-endo
C3'-endo

C2'-endo
C2'-endo

C2'-endo
C3'-endo
Intrastrand phosphate-phosphate distance [Å]
dA,dT,dC
dG

5.9
5.9

7.0
7.0

7.0
5.9
Sources: [2] [3] [4]


Cell Lysis (Breaking Open the Cell Wall and Membranes)

Plant cells have a very rigid external structure — the cell wall — which protects it. To get to the DNA, the very first step would be to break open that wall.

The cell wall is the first barrier in that must be broken to extract the DNA molecule inside the cell. It is very rigid and acts as a protector and filter. It is made of cellulose, and is responsible for making wood hard and durable. To destroy the cell wall, a mechanical method is used to break apart the cellulose molecules. In this experiment, the fruit sample is mashed manually.


DNA, Chromosomes and Genes

DNA is called the blueprint of life because it contains the instructions needed for an organism to grow, develop, survive and reproduce. DNA does this by controlling protein synthesis. Proteins do most of the work in cells, and are the basic unit of structure and function in the cells of organisms.

Genes are the basic units of heredity and are located on chromosomes.

Genes are sections of DNA, whereas chromosomes are the structures that DNA folds into before cell division. Each human somatic cell contains 23 pairs of chromosomes. All of the genes that code for the creation, growth, and development of a human person are found in these chromosomes. In addition to DNA, these chromosomes contain histone proteins that help in the packaging of the DNA into chromosomes.

In eukaryotic cells, chromosomes are found in the nucleus but in prokaryotic cells they are free to move about.

Answer:

Deoxyribonucleic Acid is the meaning of it. It is a nucleic acid that is the carrier of genetic information.

Explanation:

DNA are the letters of deoxyribonucleic acid.

All life on earth uses this nucleic acid as the genetic code.

A nucleic acid is a polynucleotide. A polynucleotide consists of three basic units: a phosphate group, a 5 carbon sugar (pentose), and a nitrogenous base. The five carbon sugar is deoxyribose. Since a polynucleotide chain, the phosphate and deoxyribose units are repetitive, the variation is provided by the nitrogenous bases.

There are four bases: adenine, guanine, cytosine, and thymine.
Both adenine and guanine are purines which have a double ring structure. Cytosine and thymine are pyramidines which consist of a single ring structure.

The DNA molecule is double helix, a spiral shaped ladder. The upright or backbone of the ladder is made of alternating pentose and phosphate groups held together by covalent bonds. The rungs or steps of the ladder consist of the bases. These bases are joined to the pentose sugars with covalent bonds. Adenine pairs with thymine using two hydrogen bonds and cytosine pairs with guanine using three hydrogen.

The genetic code is determined by the linear sequence of the bases.

For example the sequence of adenine guanine thymine does not carry the same message as guanine thymine adenine.

The code is arranged in triplet form which codes for RNA which in turn codes for amino acids which form the basis of proteins.


Watch the video: Πώς αντιγράφεται το DNA? Και βασικά. Τι είναι το DNA? (July 2022).


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