Chapter 4 Nucleic acids Flashcards

Question

Outline the role of RNA in plasma B-cells

Answer 1

DNA cannot leave the cell ∴ mRNA is a copy of a section of DNA (i.e. transcription) RNA passes through nuclear pores into the cytoplasm to a ribosome, which is made of rRNA RNA is required to produce proteins (antibodies) tRNA bring amino acids to the ribosomes to be synthesised into polypeptide chains

Answer 2

carries genetic information from nucleus to ribosomes for translation in protein synthesis

Answer 3

binds to specific amino acids for delivery to ribsosomes in translation

Answer 4

combined with proteins to produce ribosomal sub-units (which are then assembled in the nucleolus to form ribosomes)

Answer 5

DNA helicase unwinds a cistron by breaking H bonds RNA polymerase binds to a specific binding site at the start of the gene RNA polymerase moves along the template strand RNA nucleotides activated by addition of phosphate groups (from ATP) activate RNA nucleotides diffuse through the nucleus and align next to complement bases on the template strand creating mRNA RNA polymerase forms covalent phosphodiester bonds between adjoining nucleotides DNA strands rejoin behind the enzyme when stop codon reached, enzyme detaches

Answer 6

mRNA moves into cytoplasm through nuclear pores mRNA attaches to groove between ribosomal subunits ribosome binds at 5' end tRNA complexes with complement anticodons bind to the first mRNA codon via H bonding another tRNA does the same a peptide bond forms between two adjacent amino acids ribosome moves along to next codon empty tRNA molecule released a new amino acid-tRNA complex binds to vacant mRNA codon repeats until stop codon reached ribosome falls off mRNA molecule and methionine removed

Answer 7

* 2 condensation reactions 1. * one joins phosphate to pentose sugar = ester bond 2. one joins NCB to pentose sugar = glycosidic bond * Hence 2 water molecules produced for each nucleotide produced** **

Answer 8

1. NAD 2. FAD 3. Coenzyme A

Answer 9

They are both: * coenzymes which act as an electron (and hydrogen atom) carriers * have vital roles in the Krebs cycle and electron transport chain enabling the production of ATP in aerobic respiration

Answer 10

* linear structures * which consist of one molecule of DNA * usually are found in pairs (in somatic cells) * carry 100s – 1000s of genes each * are lengths of DNA that are wrapped around a protective layer of protein called the histone coat

Answer 11

* is the total collection of all the genes within an organism (or cell) * human genome consists of ~25 000 genes * ## Footnote Remember: some of these genes are also located in the mitochondria

Answer 12

1. * Stable 2. Carris large quantities of genetic information 3. Demonstrates complementary base pairing 4. Passes genetic information to mRNA 5. Genetic information is preserved (should one polynucleotide strand be damaged) 6. Forms double helix 7. 2 polynucleotide strands are anti-parrallel

Answer 13

1. The histone coat is removed (in eukaryote) 2. The DNA molecule is ¬unwound by the enzyme DNA helicase 3. H-bonds between the complementary nitrogenous bases break i.e. DNA unzips by the enzyme DNA helicase 4. Free DNA-nucleotides are activated by addition of 2 phosphate* groups (from ATP): ATP, GTP, CTP and TTP 5. Activated DNA-nucleotides randomly diffuse in close proximity to exposed bases on the template strand 6. H-bonds form between 2 complementary nitrogen-containing bases (A=T or GC) 7. Both polynucleotide stands act as templates & are copied 8. DNA polymerase catalyses synthesis of two new polynucleotide strands (enzyme moves in 3’5’ direction hence a continuous polynucleotide strand is made in the 5’3’ direction) 9. Two extra phosphate groups on the activated DNA-nucleotides are hydrolysed which release the energy for bonding adjacent DNA-nucleotides 10. The sugar-phosphate backbone is joined by phosphodiester bonds (covalent bonds) by DNA polymerase 11. The process continues until all the DNA has been copied 12. The DNA rewinds (controlled by the enzyme DNA helicase) 13. The histone coat is replaced

Answer 14

DNA ligase joins short Okazaki fragments together on the lagging strand while the polynucleotide is made in a continuous strand on the leading strand

Answer 15

* is a globular protein (consists of 10 ppc  40 structure) * it has an active site specific for a DNA nucleotide base

Answer 16

o to ensure there are no mutations o as this would lead to the production of altered proteins with potentially impaired / loss of function o as different primary structure may lead to different antigens so cells are rejected (by immune system) o as different surface antigens may mean cells cannot function together o as this may result in some cells becoming cancerous as a result of uncontrolled mitosis o as this may lead to no or unregulated apoptosis

Answer 17

**Definition of semi-conservative replication** * each polynucleotide strand of DNA acts as a template for the synthesis of a complementary strand * new DNA molecule has one original polynucleotide strand and one new polynucleotide strand ## Footnote Remember: semi” = half i.e. each new DNA molecule retains one polynucleotide from the original DNA molecule (i.e. half is retained) “conservative” = the genetic information (i.e. sequence of nucleotide bases) is copied exactly

Answer 18

Meselson and Stahl They grew the bacteria Escherichia coli with different isotopes of nitrogen (14N and 15N)  All the bases in DNA contain nitrogen.  Nitrogen has two forms:  light isotope 14N  heavy isotope 15N  The bacteria were exposed to 15N for several generations until all the DNA was formed from 15N  The bacteria were then exposed to 14N which is lighter  Scientists could then distinguish between the different DNA densities by centrifuging them  Bacteria will incorporate nitrogen from their growing medium into any new DNA they make  The 15N strand is heavier so makes a band low down the tube

Answer 19

1. Dissolve salt (which binds to the DNA and causes it to clump together) in distilled water. Add the washing-up liquid (to break down the cell membranes) and mix gently. 2. Breakup the kiwi cells using a hand blender and then add to a beaker with the salty- detergent solution. 3. Stand the beaker in a water bath at 60 degrees C for exactly 15 minutes to ensure the enzymes in the cells are denatured and prevent the DNA from being hydrolysed). 4. Cool the mixture by placing the beaker in an ice water bath for 5 minutes, stirring frequently. 5. Pour the mixture into a blender and blend it for no more than 5 seconds. 6. Filter the mixture into a second beaker. Ensure that any foam on top of the liquid does not contaminate the filtrate. 7. Add 2–3 drops of protease (to hydrolyse the proteins in the mixture e.g. proteins bound to the DNA) to about 10 cm 3 of the onion extract in a boiling tube and mix well. 8. If only DNA is required, add RNase enzymes (to break down any RN A in the mixture). Omit this step if DNA and RNA are to be extracted. 9. Very carefully pour ice cold ethanol down the side of the boiling tube, to form a layer on top of the DNA-detergent mixture. 10. Leave the tube, undisturbed, for a few minutes. DNA will form a white precipitate in the upper (ethanol) layer. ## Footnote Remember the same method will work with onion, strawberry etc cells

Answer 20

1. chromosome 2. point

Answer 21

= change to the structure (whole or part) of a chromosome can be  deletion(s),  insertion(s) or  translocation(s)

Answer 22

= change to the structure of a gene o due to a change in the DNA-nucleotide base sequence

Answer 23

* tar * mustard gas * benzene * radiation: UV light, X-rays, g-rays ## Footnote Only give one type of radiation in an exam and don't just say "radiation" give a named type

Answer 24

o mutation shift triplets by one base by either insertion or deletion o amino acids in the ppc sequence before the insertion/deletion remain unaffected o codons after the insertion/deletion will all be altered o may result in a codon for ‘STOP’ = truncated ppc (prematurely shortened ppc) o may result in completely new sequence of codons = new sequence of amino acids in the ppc o this is likely to affect the 20 and 30 structure of the protein -> probable loss of function

Answer 25

o each amino acid coded for by more than one codon o code is degenerate (do NOT say degenerative or degenerated) o hence a change of one base may not change the amino acid that is coded for part of protein may not be involved in its function o mutated may be recessive to normal allele o 2 alleles on homologous pair of chromosomes so may be masked by dominant allele o mutation may occur in an intron (so not have any effect on the primary structure of the ppc)

Answer 26

* mutations occur randomly & spontaneously * risk of mutation increases with exposure to chemical mutagens or radiation * environmental factors influence rate of mutations * mutations contribute to both normal and abnormal biological processes e.g. – evolution – cancer – immune system development

Answer 27

* Production of melanin in early humans in Africa a provided protection from UV-light but were still able to synthesise Vit. D * Those in Africa retained the advantage * Those humans who migrated to cooler climates did not need the dark melanin but did need a greater ability to produce Vit. D \ those with paler skin (mutation) had an advantage (lack of vitamin D à rickets & a narrow pelvis à difficulties in child-birth for mothers potentially fatal for mother and baby).

Answer 28

* **Triplet code** – a sequence of 3 nucleotide bases (on the DNA or mRNA) is called a codon. Each codon codes for one specific amino acid. There are 4 nitrogen-containing bases \ there are 43 = 64 different triplets. 20 amino acids exist naturally so some have more than one codon. * **Degenerate code** *(do NOT say degenerative or degenerated)* – all amino acids (except methionine) have more than one codon. * **Widespread** (but not universal) e.g. TCT codes for serine in all organisms; but some variations do exist * ‘**Incomplete’**- some codons do not code for an amino acid but instead mean ‘stop’ i.e. This stops the synthesis of the ppc. * **Non-overlapping** – sequence of bases is read so that each base is only part of one codon i.e. each triplet is read separately ## Footnote Think "TWIND"

Answer 29

**a) Transcription:** * Occurs in nucleoplasm * Produces single stranded mRNA **b) Translation** * Occurs in cytosol * Produces polypeptide chain

Answer 30

Introns = genetic rubbish i.e. sections of DNA that do not code for a sequence of amino acids within a ppc

Answer 31

Exons = sections of DNA that code for specific sequence of amino acids in a ppc

Answer 32

* Consist of 3 exposed RNA-nucleotides * Only found at base of tRNA molecules * Each anticodon is specific for one amino acid * Each tRNA will have a specific anticodon which determines the amino acid is carries

Answer 33

* Sub-ceullular structure found in eukaryotes (80s) & prokaryotes (70s) * Assembled in nucleolus of eukaryotes * Assembled in cytosol of prokaryotes * Made from rRNA and protein * Consist of 2 subunits: one large & one small * Contains grove between the 2 subunits into which mRNA fits * Can be found free in cytosol or attached to RER * Site of translation * Many ribosomes can read one mRNA at a time = polysome (increases efficiency) ## Footnote Remember: ribosomes are not true organelles (as they have no membrane surrounding their structure)

Answer 34

The addition of each **specific** amino acid is to its **specific** tRNA molecule catalysed by a **specific** enzyme & ATP remember: this must occur before translation can occur

Answer 35

1. mRNA moves into the cytosol through the nuclear pore 2. mRNA attaches to the groove between the 2 subunits of the ribosome 3. Ribosome binds at 5’ end of mRNA (translation occurs in the 5’3’ direction) 4. First αα -tRNA complex with correct complimentary anticodon pairs to first mRNA codon by hydrogen bonding between complementary bases (occurs in the peptidyl binding site) Remember: the first mRNA codon is always AUG=initiation codon and codes for amino acid methionine 5. In the vacant amino-acyl site a second αα -tRNA complex pairs with the exposed mRNA 6. Peptide bond is formed between the two adjacent amino acids (by condensation reaction); energy to form the peptide bond is provided by breaking of the bond between amino acid and the αα -tRNA complex. This is catalysed by the enzyme peptidyl transferase which is found in the large subunit of the ribosome 7. Ribosome moves along one mRNA codon to next exposed mRNA codon 8. Empty tRNA molecule is released and returns to cytosol to become re-activated (i.e. reused) 9. New αα-tRNA complex binds to vacant mRNA codon as in step 5 10. Steps 5 to 8 continue until the ribosome reaches a ‘stop’ codon (UAA, UAC, UGA). There are no tRNA molecules with complementary anticodons for these codons 11. This causes the ribosome to fall off mRNA molecule 12. PPC released

Answer 36

1. The polypeptide chain is released and the initial amino acid, methionine, is removed 2. The polypeptide chain folds spontaneously to take up its secondary &tertiary (and if necessary quaternary by the addition of other ppc) structure. 3. The (now folded) protein is then modified in the golgi apparatus e.g. * the addition of a glycocalyx to the protein to form a glycoprotein or * the addition of any prosthetic group(s).

Answer 37

DNA synthesis in prokaryotes: * Faster as shorter DNA & no histone coat to remove; * occurs in cytosol (as no nuclear envelope) Protein synthesis in prokaryotes also tends to be faster: as no delay in moving mRNA to cytosol & again ppc tend to be shorter