Unit 1 Topic 2B Proteins and DNA Flashcards
(54 cards)
1
Q
Describe the function of carrier proteins in a cell surface membrane.
A
- Involved in facilitated diffusion
- Movement of {large / polar molecules}
- (Facilitated diffusion from a high concentration to a low concentration
- Involved in active transport
- Needs ATP to move molecules against concentration gradient
2
Q
Describe the roles of RNA in the synthesis of translation of haemoglobin
A
- a mRNA molecule(produced by transcription) codes for each of the polypeptide chains of haemoglobin
- mRNA leaves nucleus and moves to cytoplasm
- Each tRNA carries its own specific amino acid to ribosome
- anticodon on tRNA binds to codons on mRNA by complementary base pairing
- tRNA holds amino acid in place while peptide bonds are formed between amino acids
- Reference to start and stop codons on mRNA (start and stop transcription)
3
Q
Process of DNA replication
A
- The 2 strands of DNA molecule unzips along the line of hydrogen bonds by action of DNA helicase. (catalyzes the breakage of hydrogen bonding)
- 2 strands of unzipped DNA act as templates for synthesis of new DNA strands
- Exposed nitrogenous bases attract free DNA nucleotides; new hydrogen bonds are formed between matching complementary base pairs
- New DNA strands are formed along the template strands by DNA polymerase between adjacent free nucleotides (5’ to 3’ end)
- New and short sections of DNA are formed by DNA ligase between nucleotides between adjacent segments of new DNA
- Identical DNA molecules are formed
4
Q
Properties of triplet code
A
(DONUT)
- Degenerate - codes for limited amino acids
- Overlapping between codons
- Non (goes with above)
- Universal - codes for same amino acid in all organisms
- triplet code - 3 consecutive bases codes for 1 amino acid
5
Q
Describe how the induced-fit mechanism differs from the ‘lock-and-key’ model.
A
- The shape of the active site of enzyme is modified when the substrate binds to it according to induced-fit mechanism
- The shape of the active site is fixed and perfectly complementary to the shape of the substrate according to the lock-and-key model
6
Q
Describe the roles of transcription and translation in the synthesis of globular protein by a muscle cell.
A
- The {gene / sequence of DNA} for the (globular protein is transcribed
- Complementary base pairing between RNA nucleotides and DNA (to produce mRNA)
- mRNA leaves the nucleus and attaches to a ribosome
- Pairing between codons on mRNA and anticodons on tRNA
- tRNA provides specific amino acids
- The sequence of {bases / codons} determines the {sequence of amino acids / primary structure of the protein}
7
Q
Function of enzymes
A
- Can lower activation energy
- Increase rate of reactions, without itself being changed or used up
8
Q
Process of enzymes
A
- Substrates bind to the active site of an enzyme molecule to form an enzyme-substrate complex
- Enzymes lower the activation energy of the reaction to catalyse the reaction
- After the reaction, enzyme detaches from the product
9
Q
Conservative replication
A
- Original DNA remains intact,
- New identical double helix formed from entirely new material.
10
Q
Semi-conservative replication
A
- DNA unzips, new nucleotides align along each strand
- Each new double helix has one strand made of completely new material and the other strand of the original DNA
11
Q
State the steps of Meselson and Stahl’s experiment
A
- Bacteria was grown in a 15N (heavy) medium for many generations
- Preparation of DNA by adding CsCl to make the solution dense enough for centrifugation.
- The sample (gen 0) was centrifuged and only 15N (heavy) bands were observed.
- The bacteria was transferred to a 14N (light) medium and allowed to replicate.
- A sample was taken every 20 minutes and each sample was centrifuged
- Gen 1: All DNA were of intermediate density and only 15N-14N bands were observed
- Gen 2: ½ DNA were of intermediate density and ½ were of light density.
- Both 15N-14N bands and 14N bands were observed.
12
Q
State the steps of DNA replication
A
In semi-conservative DNA replication,
- DNA helicase unzips the two strands of DNA
- The nitrogenous bases are exposed, nucleotides line up along both strands.
- DNA polymerase facilitates the complementary base pairing between bases by catalysing the formation of covalent phosphodiester bonds between sugars and phosphate groups by condensation reactions.
- DNA ligase binds the DNA fragments together by catalysing the formation of covalent phosphodiester bonds between newly-bound nucleotides by condensation reactions.
- The completed DNA strands recoil as a double helix.
13
Q
State the process of transcription
A
- DNA helix unwinds, expose base to act as template, hydrogen bonds between complementary strands are broken. (catalysed by DNA helicase)
- Free RNA mononucleotides in nucleus line up against and attach to one coding strand of DNA. There is complementary base pairing between ribonucleotides by RNA polymerase.
- RNA nucleotides are joined together by the formation of phosphodiester bonds in a condensation reaction by RNA polymerase.
- The completed pre-mRNA strand detaches from the DNA strand
- Pre-mRNA modified to become mRNA (splicing, remove introns) and moves out of the nucleus to the cytoplasm for translation.
14
Q
Process of translation
A
- mRNA molecules bind to the small subunit of ribosomes
- tRNA molecules bind to a specific amino acid corresponding to the anticodon
- anticodon of tRNA line up against and attach to the codon of mRNA and hydrogen bonds are formed between the bases of tRNA and mRNA
- another tRNA molecule binds to the next codon
- peptide bond formed between two adjacent amino acids by condensation reaction
- polypeptide chain is formed as the process repeats
- when stop codon reached, polypeptide chain released, ribosome complex disassembles and is ready for reuse
15
Q
Stop codon properties
A
a) Occurs at the end of the gene
b) Recognised by the ribosomes and Indicates when polypeptide synthesis ends and
c) Indicates when tRNA molecules stop attaching on the mRNA strand.
d) The polypeptide chain would then be released and the ribosome complex disassembles and is ready for reuse.
16
Q
Past paper: tRNA vs mRNA (differences and similarities)
A
Differences:
- tRNA is folded/ is clover shaped but mRNA is straight and unfolded
- tRNA has a fixed length but mRNA does not and the length depends on the size of the gene
- tRNA has anticodon but mRNA has codon
- tRNA has an amino acid binding site but mRNA does not.
- tRNA has hydrogen bonds holding the structure together but mRNA does not
Similarities:
- Both contain RNA mononucleotides and have uracil, ( adenine, cytosine and guanine )
- Both are single stranded
17
Q
Explain how the three-dimensional structure of lactase affects the mechanism of action of this enzyme. (3)
A
- lactase is soluble because of its {globular shape / external polar R groups} (1)
- (and therefore) lactase collides with lactose (1)
- active site of lactase is complementary to the lactose (1)
- Formation of enzyme-substrate complex lowers the activation energy (1)
18
Q
What is RNA
A
- polymer of a nucleotide formed of a ribose, nitrogenous base, and phosphate group
19
Q
What are the nitrogenous base of RNA
A
- adenine
- guanine
- cytosine
- uracil
20
Q
What is the function of the RNA
A
- copy and transfer genetic code from DNA in the nucleus to ribosomes
21
Q
What is mRNA
A
- copy of gene from DNA
- created in the nucleus
- leaves the nucleus to carry the copy of genetic code of one gene to a ribosome in the cytoplasm
22
Q
Why don’t DNA just leave nucleus and we need mRNA
A
- mRNA is much shorter (length of one gene)
- DNA is too large to leave the nucleus and would be at risk of being damaged by enzymes
23
Q
What are the properties of mRNA
A
- short lived: only for temporarily helping to create a protein, would have carried out its function before enzyme break it down
- single-stranded
24
Q
What is a codon
A
every 3 bases in the sequence code for a specific amino acid
25
Where is tRNA found at
cytoplasm only
26
What are the shape and properties of tRNA
- single stranded
- cloverleaf shape (folded)
- held in place by hydrogen bonds
27
What is the function of tRNA
- attach to one of the 20 amino acids
- transfer the amino acid to ribosome to create peptide chain
28
What is the mechanism behind amino acid attaching to tRNA
- specific amino acids attach to specific tRNA molecules and brought to the mRNA (at ribosome)
- determined by the 3 bases (exposed on top) found on the tRNA
29
anticodon
tRNA
- complementary to the codon on mRNA
30
What is rRNA
- combines with protein to make up the bulk of ribosomes
31
Differences between DNA and RNA **monomers**
DNA
- contains base thymine
- pentose sugar: deoxyribose
RNA
- contains uracil
- pentose sugar: ribose
32
Differences between DNA and RNA **polymers**
DNA
* larger
* 23000 genes
* double-stranded
RNA
* shorter
* only the length of one gene
* single-stranded
33
(past paper: spec) Suggest one way in which the cells use uridine monophosphate
- synthesis of RNA
34
(past paper: spec) Explain why each codon for the DNA genetic code must contain at least three bases? (3)
- enough codons needed for 20 different amino acids
- four bases used in genetic code
- triple codes provide enough possible (43 / 64) codons for amino acids
35
(past paper: spec) Collagen is an insoluble, fibrous protein. Describe the roles of RNA in the synthesis of collagen. (5)
- mRNA molecule **codes for each of the polypeptide chains** in collagen
- mRNA carries copy of genetic code for collagen out of the **nucleus to the ribosomes**
- each tRNA carries its own **specific amino acid to the ribosome**
- **anticodon** on tRNA binds to **codon** on mRNA
- tRNA **holds the amino acid** in place while **peptide bonds** form
- start and stop codons (signal start and end of translation)
36
(past paper: spec) Explain how the primary structure of collagen determines its properties. (4)
insoluble
- amino acids
- many hydrophobic R groups
strong
- triple helix
- many repeating amino acid sequences
- many small R groups so the triple helix can form
37
DNA replication vs transcription
1. DNA polymerase is used in DNA replication while RNA polymerase is used in transcription.
2. DNA replication is semi-conservative while transcription is not semi-conservsative.
3. In DNA replication, both DNA strands are copied in DNA replication but only one DNA strand is
copied in transcription.
4. In DNA replication, Thymine is used while Uracil is used in transcription.
38
Transcription vs translation
1. Transcription occurs in the nucleus while translation occurs on ribosomes in the cytoplasm
2. In transcription, the antisense DNA strand is used as a template while mRNA is used as a
template in translation.
3. mRNA is the product of transcription while a polypeptide chain is the product of translation.
4. RNA mononucleotides are used as the raw material in transcription while amino acids are used
as the raw material in translation.
39
(pastpaper: May19) Explain why some amino acids, such as alanine, have more than one genetic code. (3)
- Degenerate
- More codes than needed
- 64 possible arrangements, 20 amino acids to code for
- Minimise risk of mutations!!
39
(pastpaper: May19) Explain the role of the other three genetic codes. (stop codons) (2)
- Stop codons
- End translation / no more amino acids can add to the polypeptide chain
40
(pastpaper: May19) State what is meant by the term semi-conservative replication. (2)
- Increase in number of DNA molecules
- Each new molecule consists of old / parent strand and one new strand
41
(pastpaper: May19) Explain the importance of semi-conservative replication in the production of new cells. (2)
- Genetically identical cells
- Accept genetic information / genotype / alleles / DNA sequence
- Same structure / function as the parent cells
42
(pastpaper: May19) Transfer RNA is involved in translation. The amino acid AzPhe requires a special tRNA molecule during synthesis of silk. Suggest why AzPhe is not inserted into proteins in silkworms that have not been selectively bred. (3)
- Not naturally-occurring amino acids
- No DNA / mRNA that codes for AzPhe
- No tRNA that can bind to the codes
- Not held in position for peptide bonds to form between amino acids
43
(pastpaper: May19) The R group of phenylalanine is smaller than the R group of AzPhe. Suggest how inserting an amino acid with larger R group could affect the properties of silk fibres (4)
- Properties of protein, depends on structure
- AzPhe ⇒ affects secondary / folding / 3D of silk
- Because different bonds are formed between R groups
- New bonds make silk stronger ⇒ larger ⇒ insoluble
Or
- Larger R group ⇒ makes the silk weak (because fibres are not held as closely)
44
Explain why several enzymes are involved in the urea cycle. (3)
because (the urea cycle has) many {stages / steps / reactions} (1)
therefore, the product of one stage is the substrate of the next stage (1)
and enzymes are specific to one substrate (1)
ACCEPT each stage has {new / different} substrate / different substrates (in the process) (3)
substrates {bind / attach / fit} to (specific) active site / credit reference to lock and key theory / credit reference to induced fit (1)
45
Describe how a chain of amino acids can form an enzyme molecule (4)
- Chain of amino acids form secondary structure / alpha helix / beta pleated sheets (hydrogen bonds)
- Then forms tertiary / quaternary / 3D / globular structure (hydrogen bonds, disulfide / ionic bonds)
- With active site
- With hydrophilic R groups arranged on the outside
46
Describe the role of RNA polymerase in protein synthesis. (2)
transcription / transcribe the DNA / synthesis of RNA (1)
credit detail of role of RNA polymerase in transcription (1)
binds to the (DNA) promotor (region),
unwinds the DNA (helix),
separates the DNA strands,
lines up the (RNA) nucleotides on the {DNA / template strand},
forms phosphodiester bonds between (RNA) nucleotides
repairs mistakes in the (newly-synthesised) RNA (1)
47
Describe the roles of the ribosomes in protein synthesis. (2)
Translation / synthesis of polypeptide chain
Holds two tRN Atogether on mRNA
Until peptide bonds joins amino acids together
48
Explain why this amylase is able to break only 1-4-glycosidic bonds. (2)
- (amylase) {is specific / is complementary / only binds / only fits} {substrates / 1 – 4 glycosidic bonds} (1)
- {1 – 4 glycosidic bonds / substrate} can {fit into / bind with} the active site (1)
- ACCEPT (the parts of the glucoses joined by) 1 – 6 glycosidic bonds do not {fit into / bind with} the active site
49
The digestion of carbohydrates is completed in the small intestine. Glucose is then absorbed into the bloodstream. Suggest how disaccharides and oligosaccharides are broken down in the small intestine. Give reasons for your answer. (3)
By hydrolysis of glycosidic bonds
Different enzymes needed to break each type of glycosidic bond
Different enzymes needed for disaccharides and oligosaccharides / each type of oligosaccharides / each type of sugars
50
Explain why amoeba can rely on diffusion to supply the oxygen it needs. (2)
Higher concentration of oxygen outside / in water than in the cytoplasm
All parts of amoeba are close to cell membrane = small volume = high surface area to volume ratio
Low energy requirements / metabolism
ACCEPT amoeba is small / diffusion distance is small (to all parts of cell)
51
Describe the role of mRNA in the production of enzymes. (2)
- Carry a copy of gene out of the nucleus / into the cytoplasm / to ribosomes
- Used for connecting amino acid / producing amino acid sequence / formation of polypeptide chain
52
The extent to which soybeans reduce cholesterol depends on the variety of soybeans. This reduction is influenced by two proteins found in soybeans, glycinin and beta-conglycinin. Give two ways that the primary structure of two proteins could differ from each other. (1)
- The sequence of amino acid
- Number of amino acids / length of polypeptide chain
- Proportion / number of each type of amino acid
53
Explain why eating soybeans may reduce cholesterol levels in the blood. Use the information in the graph to support your answer. (3)
- As the concentration of soybean peptides increase, the inhibition of HMGCR increases AND the synthesis of cholesterol decreases
- Because soybean peptides inhibit HMGCR
- Because the peptides bind to the HMGCR / compete with the substrate
- As the cholesterol precursor cannot be converted into cholesterol by HMGCR (ACCEPT cholesterol precursor cannot bind to HMGCR)
