Reading...
Play button
Play button
Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
34 Cards in this Set
- Front
- Back
|
Describe the structure of DNA,including the antiparallel strands, 3’–5’ linkages and hydrogen bonding between purines and pyrimidines
|
- DNA is made up of two strands
- At one end of each strand there is a phosphate group attached carbon 5 of the sugar (5' end) and at the other end of each strand is a hydroxyl group attached to the carbon atom number 3 of the deoxyribose (3' end) - The strands run in opposite directions: One strand runs in a 5'-3' direction and the other runs in a 3'-5' direction. They are therefore said to be antiparallel. - The sugars are linked by phosphate groups attached to carbons 3 and 5 of the sugar. This is called a 3'-5' linkage. - The bases of each strand are linked together by hydrogen bonds. Cytosine and thymine are pyrimidines and have one ring in their molecular structure. Adenine and guanine are purines, and they have two rings in their molecular structure. - A purine and a pyrimidine are always linked: A and T by two hydrogen bonds and G and C by three. |
|
|
Outline the structure of nucleosomes
|
- Consist of a group of 8 histone proteins around which the DNA is wrapped
- The DNA is locked in place by another histone protein |
|
|
State that nucleosomes help to supercoil chromosomes and help to regulate transcription
|
Nucleosomes help to supercoil chromosomes and help to regulate transcription
|
|
|
Distinguish between unique or single-copy genes and highly repetitive sequences in nuclear DNA
|
Unique/Single-copy genes:
- 55 - 95% of the genome consists of genes that have only a single copy - These code for the functional polypeptides used by the cell or body e.g structural proteins, transport proteins, hormones, enzymes. Highly repetitive sequences: - 5 - 45% of the genome - The sequences are normally 5-300 bases per repeated - May be duplicated as many as 10^5 times in a genome - Used in DNA profiling |
|
|
State that eukaryotic genes can contain exons and introns
|
Eukaryotic genes can contain exons and introns
|
|
|
State that DNA replication occurs in a 5’→ 3’ direction
|
DNA replication occurs in a 5’→ 3’ direction
|
|
|
State that DNA replication is initiated at many points in eukaryotic chromosomes
|
DNA replication is initiated at many points in eukaryotic chromosomes
|
|
|
State that transcription is carried out in a 5’→ 3’ direction
|
Transcription is carried out in a 5’→ 3’ direction
|
|
|
Distinguish between the sense and antisense strands of DNA
|
The sense strand (coding strand) has the same base sequence as mRNA with uracil instead of thymine. The antisense (template) strand is transcribed.
|
|
|
Explain the process of DNA replication in prokaryotes, including the role of enzymes (helicase, DNA polymerase, RNA primase and DNA ligase), Okazaki fragments and deoxynucleoside triphosphates
|
1.Helicase uncoils the DNA.
2.RNA primase adds short sequences of RNA to both strands (the primer). 3.The primer allows DNA polymerase III to bind and start replication. 4.DNA polymerase III adds nucleotides to each template strand in a 5'→3' direction. 5.These nucleotides are initially deoxyribonucleoside triphosphates but they lose two phosphate groups during the replication process to release energy. 6.One strand is replicated in a continuous manner in the same direction as the replication fork (leading strand). 7.The other strand is replicated in fragments (Okazaki fragments) in the opposite direction (lagging strand). 8.DNA polymerase I removes the RNA primers and replaces them with DNA. 9.DNA ligase then joins the Okazaki fragments together to form a continuous strand |
|
|
Outline the roles of the enzymes involved in DNA replication
|
Helicase: unwinds the DNA at the replication fork, breaks the hydrogen bonds between the bases.
DNA Polymerase lll: Adds deoxynucleoside triphosphates to the 3' end. RNA primase: Adds nucleoside triphosphates on the lagging strand to form an RNA primer. DNA polymerase l: removes the RNA primer, replaces it using deoxynucleoside triphosphates. DNA ligase: joins the okazaki fragments together. |
|
|
Explain the process of transcription in prokaryotes, including the role of the promoter region, RNA polymerase, nucleoside triphosphates and the terminator
|
1.RNA polymerase binds to the promoter region. This initiates transcription.
2.RNA polymerase uncoils the DNA. 3.Only one strand is used, the template (antisense) strand. 4.Free nucleoside triphosphates bond to their complementary bases on the template strand. 5.Adenine binds to uracil instead of thymine. 6.As the nucleoside triphosphates bind they become nucleotides and release energy by losing two phosphate groups. 7.The mRNA is built in a 5'→3' direction. 8.RNA polymerase forms covalent bonds between the nucleotides and keeps moving along the DNA until it reaches the terminator. 9.The terminator signals the RNA polymerase to stop transcription. 10.RNA polymerase is released and mRNA separates from the DNA. 11.The DNA rewinds. |
|
|
State that eukaryotic RNA needs the removal of introns to form mature mRNA
|
Eukaryotic RNA needs the removal of introns to form mature mRNA. Much Eukaryotic DNA is transcribed into primary RNA. Primary RNA contains sequences called introns which are not translated. Only exons make up the mature mRNA, therefore introns are removed. This takes place in the nucleus.
|
|
|
Explain the process of DNA replication in prokaryotes, including the role of enzymes (helicase, DNA polymerase, RNA primase and DNA ligase), Okazaki fragments and deoxynucleoside triphosphates
|
1.Helicase uncoils the DNA.
2.RNA primase adds short sequences of RNA to both strands (the primer). 3.The primer allows DNA polymerase III to bind and start replication. 4.DNA polymerase III adds nucleotides to each template strand in a 5'→3' direction. 5.These nucleotides are initially deoxyribonucleoside triphosphates but they lose two phosphate groups during the replication process to release energy. 6.One strand is replicated in a continuous manner in the same direction as the replication fork (leading strand). 7.The other strand is replicated in fragments (Okazaki fragments) in the opposite direction (lagging strand). 8.DNA polymerase I removes the RNA primers and replaces them with DNA. 9.DNA ligase then joins the Okazaki fragments together to form a continuous strand |
|
|
Outline the roles of the enzymes involved in DNA replication
|
Helicase: unwinds the DNA at the replication fork, breaks the hydrogen bonds between the bases.
DNA Polymerase lll: Adds deoxynucleoside triphosphates to the 3' end. RNA primase: Adds nucleoside triphosphates on the lagging strand to form an RNA primer. DNA polymerase l: removes the RNA primer, replaces it using deoxynucleoside triphosphates. DNA ligase: joins the okazaki fragments together. |
|
|
Explain the process of transcription in prokaryotes, including the role of the promoter region, RNA polymerase, nucleoside triphosphates and the terminator
|
1.RNA polymerase binds to the promoter region. This initiates transcription.
2.RNA polymerase uncoils the DNA. 3.Only one strand is used, the template (antisense) strand. 4.Free nucleoside triphosphates bond to their complementary bases on the template strand. 5.Adenine binds to uracil instead of thymine. 6.As the nucleoside triphosphates bind they become nucleotides and release energy by losing two phosphate groups. 7.The mRNA is built in a 5'→3' direction. 8.RNA polymerase forms covalent bonds between the nucleotides and keeps moving along the DNA until it reaches the terminator. 9.The terminator signals the RNA polymerase to stop transcription. 10.RNA polymerase is released and mRNA separates from the DNA. 11.The DNA rewinds. |
|
|
State that eukaryotic RNA needs the removal of introns to form mature mRNA
|
Eukaryotic RNA needs the removal of introns to form mature mRNA. Much Eukaryotic DNA is transcribed into primary RNA. Primary RNA contains sequences called introns which are not translated. Only exons make up the mature mRNA, therefore introns are removed. This takes place in the nucleus.
|
|
|
Explain that each tRNA molecule is recognized by a tRNA-activating enzyme that binds a specific amino acid to the tRNA, using ATP for energy
|
.
|
|
|
Outline the structure of ribosomes, including protein and RNA composition, large and small subunits, three tRNA binding sites and mRNA binding sites
|
.
|
|
|
State that translation consists of initiation, elongation, translocation and termination
|
Translation consists of initiation, elongation, translocation and termination
|
|
|
State that translation occurs in a 5’→ 3’ direction
|
Translation occurs in a 5’→ 3’ direction. During translation, the ribosome moves along the mRNA towards the 3’ end. The start codon is nearer to the 5’ end
|
|
|
Draw and label a diagram showing the structure of a peptide bond between two amino acids
|
.
|
|
|
Explain the process of translation, including ribosomes, polysomes, start codons and stop codons
|
1. tRNA with the anticodon complementary to the start codon binds to the small subunit of the ribosome.
2. The small subunit, carrying the tRNA binds to the 5' end of the the messenger RNA. 3. The small subunit slides along the mRNA until it reaches the start codon, which shows where translation should begin. 4. The large subunit of the ribosome binds to the small subunit. 5. Another tRNA, with the anticodon complementary to the next codon on the mRNA binds to the ribosome. Elongation of the polypeptide can now begin. 6. The ribosome moves along the mRNA in a 5' to 3' direction, translating each codon into an amino acid on the elongating polypeptide. 7. The ribosome reaches a stop codon. No tRNA has the complementary anticodon. 8. The large subunit advances over the small subunit. The polypeptide is released from the tRNA. 9. The tRNA detaches and the large subunit, small subunit and mRNA all separate. |
|
|
State that free ribosomes synthesize proteins for use primarily within the cell, and that bound ribosomes synthesize proteins primarily for secretion or for lysosomes
|
Free ribosomes synthesize proteins for use primarily within the cell (cytoplasm), and bound ribosomes (attached to ER) synthesize proteins primarily for secretion or for lysosomes.
|
|
|
Explain the four levels of protein structure, indicating the significance of each level
|
Primary: Linear sequence of amino acids, peptide bonds (covalent), a change in just one amino acid may completely alter the properties of the polypeptide or protein.
Secondary: the amino acids in a polypeptide chain interact with each other, it may coil into an alpha helix (held in shape by hydrogen bonding some between amino acids some distance apart), sometimes a looser, straighter shape is formed called a beta pleated sheet, other secondary proteins do not show regular arrangement at all (depends on which R groups are present). Tertiary: folded into 3d shapes, the folding creates pockets which form the active site in enzymes or binding sites for other molecules, main bonding is disulfide bridges (covalent), weak bonds include hydrogen and ionic, the shape is determined by the sequence of amino acids, each protein has an individual and precise shape (important for precise active/binding sites) Quaternary: two or more polypeptide chains linked together, can also create pockets, bonding is the same as in tertiary. |
|
|
Outline the difference between fibrous and globular proteins, with reference to two examples of each protein type
|
.
|
|
|
Explain the significance of polar and non-polar amino acids
|
Structure of amino acids:
. . . . Some of these side chains are polar and some are non-polar. These play an essential role in protein function. Integral membrane proteins - these are anchored in place by having non-polar (hydrophobic) R groups positioned in contact with the hydrophobic core of the membrane. Hydrophilic ion channels in membranes - Polar amino acids are found inside membrane proteins and create a channel through which hydrophilic molecules can pass. The active site of an enzyme- If the amino acids in the active site of an enzyme are non-polar then it makes this active site specific to a non-polar (water hating) substance. On the other hand, if the active site is made up of polar amino acids then the active site is specific to a polar (water-loving) substance. |
|
|
State four functions of proteins, giving a named example of each
|
Transport e.g Haemoglobin
Defence e.g Immunoglobulin Protection e.g Keratin Structural e.g Collagen Enzymes e.g Catalase |
|
|
State that metabolic pathways consist of chains and cycles of enzyme catalysed reactions
|
Metabolic pathways consist of chains and cycles of enzyme catalysed reactions.
|
|
|
Describe the induced-fit model
|
The lock and key model indicates the exact fit between enzymes and the substrate in the active site. The induced fit hypothesis states that when the substrate and the enzyme interact at the active site a slight change occurs in the shape of the enzyme molecule. This hypothesis explains how enzymes reduce the amount of activation energy required for a reaction. By changing shape, the substrate becomes distorted, and this puts strain on bonds within the substrate. It therefore takes less energy to break these bonds. This hypothesis also accounts for the ability of some enzymes to bind to several substrates.
|
|
|
Explain that enzymes lower the activation energy of the chemical reactions that they catalyse
|
- In order for a chemical reaction to start, it usually requres a certain input of energy.
- This is called the activation energy. - The function of enzymes is to lower the activation energy so that reactions can take place at physiological temperatures. . . . . . . |
|
|
Explain competitive inhibition, with reference to one example
|
Competitive inhibitors:
• They have a shape similar to the enzyme’s normal substrate molecules. This means that they can temporarily fit into the active site, therefore the substrate cannot enter. • The inhibitor and the substrate compete for the active site, whichever gets there first is successful. • Concentration of the inhibitor and substrate are important to which molecule is more likely to enter the active site. If the substrate concentration is decreased, the degree of inhibition is lessened. • E.g Treatment of ethanol glycol (antifreeze) ingestion. When ingested it is converted into oxalic acid, which can cause irreversible kidney damage. However, the active site of the enzyme that converts ethylene glycol will also accept ethanol. Therefore, a large dose of ethanol acts as a competitive inhibitor for the ingested ethylene glycol, giving the body time to excrete it. |
|
|
Explain non-competitive inhibition, with reference to one example
|
Non-competitive inhibitors:
• These molecules bind to the enzyme molecule, although not at the active site. This distorts the enzyme and alters the shape of the active site. The substrate can no longer bind to the active site and the enzyme cannot convert the substrate to the product. • E.g Nitric oxide synthase catalyses arginine into citrulline and nitric oxide. Opiods bind to the enzyme at a different spot, therefore inhibiting the reaction. |
|
|
Explain the control of metabolic pathways by end-product inhibition, including the role of allosteric sites
|
- In many metabolic pathways, the product of the last reaction in the pathway inhibits the enzyme that catalyses the first reaction.
- This is called end-product inhibition. - The enzyme that is inhibited by the end-product is an allosteric enzyme. This means it has two non-overlapping binding sites. One is the active site, the other the alloster site. - The allosteric site is the binfing site for the end product, when it binds, the structure of the enzyme is altered so that the substrate is less likely to bind. - It is reversible, and if the end product detatches, the active site is restored. - An advantage of this method for controlling metabolic pathways is that if there is an excess of the end-product, the whole pathway switches off, so intermediates do not build up. |
