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A Level Biology: DNA, genes and protein synthesis > SUMMARY QUESTIONS > Flashcards

SUMMARY QUESTIONS Flashcards

1
Q

Describe what a gene is

A

A base equine of DNA that codes for the amino acid sequence of a polypeptide or functional RNA

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2
Q

Calculate how many bases are required to code for a chain of 6 consecutive amino acids

A

18

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3
Q

Explain how a change in one base along a DNA molecule may result in an enzyme becoming non-functional

A
  • A different base might code for a different amino acid
  • the sequence of amino acids in the polypeptide produced will be different
  • this change to the primary structure of the protein might result in a different shaped tertiary structure
  • the enzyme shape will be different and may not fit the substrate
  • the enzyme-substrate complex can’t be formed and so the enzyme is non-functional
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4
Q

A section of DNA has the following sequence of bases along it: TAC GCT CCG CTG TAC. All of the bases are part of the code for amino acids. The first base in the sequence is the start of the code.
—> calculate the number of amino acids that the section of DNA codes for

A

5

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5
Q

A section of DNA has the following sequence of bases along it: TAC GCT CCG CTG TAC. All of the bases are part of the code for amino acids. The first base in the sequence is the start of the code.
—> determine which two sequences code for the same amino acid

A
  • the first and last (5th)
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6
Q

A section of DNA has the following sequence of bases along it: TAC GCT CCG CTG TAC. All of the bases are part of the code for amino acids. The first base in the sequence is the start of the code.
—> It is possible that this sequence codes for many different amino acids or many copies of the same amino acid. From your knowledge of the genetic code explain how this can happen

A

Because some amino acids have up to six different codes, while others have just one triplet

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7
Q

Using this table, state the 2 amino acids that have only one codon and state what it is in each case

A

Trp — UGG and met — AUG

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8
Q

Using this table, name the amino acids that have each of the following codon
A - CUC
B - AAA
C - GAU

A

A - leu
B - lys
C - asp

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9
Q

Using this table, For each of the following base quenches on a DNA molecule, deduce the sequences of amino acids in the order in which they would occur in the resultant polypeptide
A - ATGCGTTAAGGCAGT
B - GCTAAGTTTCCAGAT

A

A - try-ala-ile-pro-ser
B - arg-phe-lys-gly-leu

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10
Q

Contrast the DNA of a prokaryotic cell with that of a eukaryotic cell

A
  • prokaryotic: DNA is smaller, circular and not associated with proteins (i.e. doesn’t have chromosomes)
  • eukaryotic: DNA is larger, linear and associated with proteins/histones to form chromosomes
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11
Q

State the function of the protein found in chromosomes

A

It fixes the DNA into position

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12
Q

Explain how the considerable length of a DNA molecule is compacted into a chromosome

A

It is looped and coiled a number of times

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13
Q

Suppose the total length of all the DNA in a single human muscle cell is 2.3m
A - if all the DNA were distributed equally between the chromosomes, calculate the mean length of DNA in each one
B - Calculate in nm the length of DNA in a human brain cell

A

A - 50nm (46 chromosomes in every cell)
B - 2.3m (all diploid cells have same quantity of DNA)

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14
Q

Distinguish between the structure of mRNA and the structure of tRNA

A
  • mRNA is larger, has a greater variety of types and is shaped as a long single helix
  • tRNA is smaller, has fewer types and is clover-leaf in shape
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15
Q

State the 3 ways in which the molecular structure of RNA differs from DNA

A

ANY 3 FROM
- RNA is smaller than DNA
- RNA is usually a single strand and DNA a double helix
- the sugar in RNA is ribose and the sugar in DNA is deoxyribose
- in RNA the base uracil replaces the base thymine found in DNA

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16
Q

Distinguish between a codon and an anticodon

A
  • A codon is the triplet of bases on messenger RNA that codes for an amino acid
  • an anticodon is the triplet of bases on a transfer RNA molecule that is complementary to a codon
17
Q

This table states that, for DNA, the ‘quantity is constant for all cells of a species (except gametes)
A - state how the quantity in a gamete differs from that in a body cell
B - Explain the significance of the difference you have described

A

A - the amount of DNA in a gamete is half that in a body cell
B - allows gametes to fuse during sexual reproduction without doubling the total amount of DNA at each generation. In doing so it increases genetic variety by allowing the genetic info of two parents to be combined in the offspring

18
Q

Explain the advantage of:
A - DNA being a chemically stable molecule
B - mRNA being broken down relatively quickly

A

A - DNA needs to be stable to enable it to be passed from generation to generation unchanged and thereby allow offspring to be very similar to their parents. Any change to the DNA is a mutation and is normally harmful
B - mRNA is produced to help manufacture a protein, e.g. an enzyme. It would be wasteful to produce the protein continuously when it’s only needed periodically. mRNA therefore breaks down once it has been used and is produced again only when the protein is next required

19
Q

describe the role of RNA polymerase in transcription

A
  • moves along template DNA strand, causing the bases on this strand to join with the individual complementary nucleotides from the pool that is present in the nucleus
  • it adds the nucleotides one at a time, to build a strand of pre-mRNA until it reaches a particular sequence of bases on the DNA that it recognises as a ‘stop’ code
20
Q

state which other enzyme is involved in transcription and describe its role

A

DNA helicase — acts on a specific region of the DNA molecule to break the hydrogen bonds between the bases
—> causing the 2 strands to separate and expose the nucleotide bases in that region

21
Q

explain why splicing of pre-mRNA is necessary

A
  • because: pre-mRNA has nucleotide sequences derived from introns in DNA
    —> introns = non-functional + if left on the mRNA would lead to production of non-functional polypeptides/no polypeptides at all
  • splicing would remove these introns, preventing this problem
22
Q

A sequence of bases along the template strand of DNA is ATGCAAGTCCAG:
A - deduce the sequence of bases on a pre-messenger RNA molecule that has been transcribed from this part of the DNA molecule
B - calculate how many amino acids the sequence codes for

A

A - UACGUUCAGGUC
B - 4 amino acids (1 is coded for by 3 bases, so 12 bases would = 4 amino acids)

23
Q

A gene is made up of 756 base pairs. The mRNA that is transcribed from this gene is only 524 nucleotides long, explain why there is this difference

A
  • Some of the base pairs in the genes are introns
  • these introns are spliced from pre-mRNA so the resulting mRNA has fewer nucleotides
24
Q

Name the cell organelle involved in translation

A

Ribosome

25
Q

A codon found on a section of mRNA has the sequence AUC. List the sequence of bases found on:
A - the tRNA anticodon that attaches to this codon
B - the template strand of DNA that formed the mRNA codon

A

A - UAG on tRNA
B - TAG on DNA

26
Q

Describe the role of tRNA in the process of translation

A
  • tRNA molecule attaches an amino acid at one end and has a sequence of 3 bases (called anticodon) at the other end
  • the tRNA molecule is transferred to a ribosome on an mRNA molecule
  • the anticodon on tRNA pairs with the complementary codon sequence on the mRNA
  • further tRNA molecules, with amino acids attached, line up along the mRNA in the sequence determined by the mRNA bases
  • the amino acids are joined by peptide bonds
  • therefore the tRNA helps to ensure the correct sequence of amino acids in the polypeptide
27
Q

A strand of mRNA has 64 codons but the protein produced from it has only 63 amino acids. Suggest a reason for this difference

A
  • one of the codons is a stop codon that indicates the end of polypeptide synthesis
  • stop codons do not code for any amino acid so there is one less amino acid then there are codons
28
Q

Name the structures X and Y

A

X = ribosome
Y = mRNA

29
Q

Recall the chemical group shown on the end of the polypeptide chain

A

Amino group

30
Q

Determine the anticodon sequence on tRNA molecule 4

A

AUG

31
Q

Deduce the sequence of the first 5 amino acids in the polypeptide

A

Val-Thr-Arg-Asp-Ser

32
Q

Determine the sequence of bases on that portion of DNA from which codons 1-3 are transcribed

A

CAATGGGCT

33
Q

A DNA mutation results in the base cytosine being replaced by uracil in codon 8. Explain the significance of this change

A
  • the mutation changes CAG to UAG
  • UAG is a stop codon that signifies the end of an amino acid sequence at which point the polypeptide is complete and is ‘cast off’
  • the polypeptide chain is therefore shorter than it should be and may not function as normal
34
Q

Another mutant form of a gene causes the inversion (reversal) of the code for the amino acid glutamine (Glu)
A - consider all possible outcomes from this change and explain the effect on the polypeptide in each case
B - If the polypeptide formed from this mutant gene forms part of an enzyme, suggest 2 reasons why it might fail to function. Explain your answer

A

A - Glutamine has 2 codes (GAG and GAA). The reversal of GAG produces the same codon and so still translates as glutamine and hence the polypeptide that is formed = unchanged. Whereas reversal of GAA —> translates to a different amino acid and so produces a different polypeptide (different primary structure - which may effect bonding and therefore the tertiary structure)
B -
1. enzyme function depends on substrate becoming loosely attached to an enzyme within the active site, if mutation changes amino acid it would affect the bonding between the amino acids and the polypeptide won’t’ be formed in the same way, leading to an altered active site shape, meaning the substrate can’t fit.
2. Glutamine may have been one of the amino acids in the active site to which the substrate attaches. If its replaced by another it may mean that, although the shape of the active site is unchanged, the substrate cannot attach normally

A Level Biology: DNA, genes and protein synthesis (6 decks)

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