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genes1 Flashcards

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AP® Biology flashcards Biology 101 flashcards
1
Q

Define gene.

A

= Specific base sequence of DNA on a chromosome, which codes for the amino acid sequence of a polypeptide and for a functional RNA (ribosomal RNA + tRNAs etc.).

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2
Q

Define locus.

A

= Fixed position of a specific gene on a chromosome.

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3
Q

How do genes determine nature and development of organisms?

A
  1. Genes determine proteins of organisms, including enzymes.
  2. They therefore control chemical reactions.

=> therefore responsible for chemical activities and development of organisms.

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4
Q

Define triplet/codon.

A

= Sequence of 3 bases coding for a specific amino acid.

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5
Q

Define degenerate.

A

= Most amino acids coded for by more than one triplet.

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6
Q

Define non-overlapping.

A

= Each base in the sequence only read once and is only involved in the codon for one amino acid.

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7
Q

Define universal.

A

= Each triplet codes for the same amino acids in all organisms —> indirect evidence for evolution.

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8
Q

Does all eukaryotic DNA code for proteins?

A

No, a lot of it doesn’t.

  • There are non-coding multiple repeats of base sequences between genes (VNTRs).
  • Even within a gene, only some sequences - exons, actually code for amino acid sequences - separated by one or more non-coding sequences called introns.
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9
Q

Comment on nature of initial DNA bases/amino acids in a sequence.

A
  • Start of DNA coding sequence for a polypeptide is always the same triplet —> codes for methionine amino acid.
  • If methionine does not form part of the final polypeptide, it is removed later.
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10
Q

Define stop codons.

A
  • Triplets/base sequences that do not code for any amino acids and mark the end of the polypeptide chain.
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11
Q

Describe prokaryotic DNA.

A
  • DNA molecules are short, circular and not associated with proteins.
  • NO CHROMOSOMES.
  • No true nucleus.
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12
Q

Describe eukaryotic DNA.

A
  • In a nucleus.
  • Long, linear, highly coiled and folded.
  • Linear DNA associated with histone proteins to form chromosomes.
  • Mitochondria and Chloroplasts of eukaryotic cells also contain DNA which is similar to prokaryotic DNA - short, circular and not associated with proteins.
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13
Q

Describe chromosome structure. How does no. of chromosomes vary?

A
  • Centromere connected to threads called chromatids.
  • Chromosomes are visible as distinct structures when a cell is dividing —> rest of the time, they are widely dispersed throughout the nucleus.
  • DNA double helix is wound around histones to fix it in position —> DNA-histone complex is then coiled —> coil is then looped and further coiled before being packed into a chromosome —> lots of DNA condensed into a single chromosome.
  • Chromosomes contain many genes —> each gene occupies a specific locus on the chromosome.
  • Number of chromosomes always same in all cells of same species (except gametes - 1/2) but varies between species.
    => 46 in humans.
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14
Q

Define homologous pairs (of chromosomes).

A

= Pair of chromosomes of which one is maternal (from egg) and one is paternal (from sperm).

  • Carry the same genes at same loci, but could be different alleles of the same genes.
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15
Q

Define allele.

A

= One of a number of alternative forms of a gene.

=> different allele - different DNA base sequence —> different amino acid sequence so produces a different polypeptide.

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16
Q

Define mutation.

A

= Random, spontaneous change in the base sequence/quantity of DNA.

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17
Q

Explain how a mutation could lead to malfunctioning proteins being produced.

A
  1. Different allele.
  2. Different amino acid sequence.
  3. Different polypeptide.
  4. Polypeptide may not function properly/at all.
  5. If enzyme - different tertiary structure (as different primary structure).
  6. Active site shape changed —> substrate may no longer fit —> E-S complex can no longer be formed.
  7. Enzyme may no longer function —> can have serious consequences for the organism.
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18
Q

Describe structure and function of RNA.

A
  • Ribose sugar molecule (pentose).
  • Phosphate group.
  • Nitrogenous base - uracil instead of thymine.
  • transfers the coded information on DNA in the nucleus to ribosomes where it is translated into proteins.
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19
Q

Define genome.

A

= Complete set of genes in a cell (including those in mitochondria and/or in chloroplasts).

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20
Q

Define proteome.

A

= Full range of proteins that a cell is able to produce.

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21
Q

Describe DNA structure.

A
  1. Double polynucleotide strand.
  2. Large molecule.
  3. Double-helix molecule.
  4. Pentose sugar = deoxyribose.
  5. A–T and C—G.
  6. Found mostly in nucleus.
  7. Quantity is constant for all cells of a species
    (except gametes).
  8. Chemically very stable.
22
Q

Describe mRNA structure.

A
  1. Single polynucleotide chain.
  2. Smaller than DNA but > tRNA.
  3. Single helix molecule (except in some viruses).
  4. Pentose sugar = ribose.
  5. A–U and C—G.
  6. Manufactured in nucleus but found throughout cell.
  7. Quantity varies from cell to cell and with level of metabolic activity.
  8. Less stable than DNA and tRNA —> molecules usually broken down in cells within a few days.
23
Q

Describe tRNA structure.

A
  1. Single polynucleotide chain.
  2. Smaller than mRNA and DNA.
  3. Clover-shaped molecule.
  4. Pentose sugar = ribose.
  5. A–U, C—G.
  6. Manufactured in nucleus but found throughout cell.
  7. Quantity varies from cell to cell and with level of metabolic activity.
  8. Stability:

DNA > tRNA > mRNA.

24
Q

Outline basic process of protein synthesis.

A
  1. DNA provides instructions in long sequence of bases.
  2. Complementary section of part of this sequence made in a pre-mRNA molecule => transcription.
  3. pre-mRNA is spliced —> mRNA.
  4. mRNA used as a template to which complementary tRNA molecules attach.
  5. Amino acids carried by tRNA molecules linked together to form polypeptide => translation.
25
Q

Outline process of Transcription.

A

= process of making pre-mRNA using part of the DNA as a template.

  1. DNA-helicase enzyme acts on specific region of the DNA molecule, causing H-bonds to break —> 2 strands separate and nucleotide bases in that region exposed.
  2. Nucleotide bases on template strand (one of 2 DNA strands) pair with complementary free RNA nucleotides in nucleus.
  3. RNA-polymerase then moves along strand, forming phosphodiester bonds between nucleotides, forming phosphate-sugar backbone.
    => forms pre-mRNA molecule.
  4. DNA strands H-bonds reform after RNA-polymerase has built the relevant part of the strand of pre-mRNA —> double helix reforms.
  5. When the RNA-polymerase reaches a particular stop codon/triplet code, it detaches —> completing production of pre-mRNA.
26
Q

Outline process of Splicing.

A
  • Eukaryotes only - most prokaryotes don’t have introns in their DNA.
  • Exons code for proteins, whereas introns do not.
  • If introns were not removed, it would prevent polypeptide synthesis.
    1. Intron base sequences removed and the functional enzymes are joined together.
    2. mRNA molecules are too large to diffuse out of the nucleus —> after splicing, they leave via nuclear pores —> outside nucleus, mRNA is attracted to ribosomes which it attaches to for translation.
27
Q

Outline process of Translation.

A
  • Approx 60 different tRNAs, each with a specific anticodon —> attaches to a specific amino acid —> each amino acid therefore has one or more tRNA molecules.
    1. Ribosome attaches to start codon at one end of the mRNA molecule.
    2. tRNA molecule with complementary anticodon sequence moves to ribosome and pairs with the codon on mRNA - this tRNA carries a specific amino acid.
    3. tRNA molecule with complementary anticodon pairs with the next codon on the mRNA —> tRNA molecule carries another (next in sequence) amino acid.
    4. Ribosome moves along mRNA, bringing 2 tRNAs together, forming a peptide bond.

(done using an enzyme - hydrolysis of ATP provides required energy).

  1. Ribosome moves to the 3rd codon, linking 2nd and 3rd amino acids. tRNA molecules released where they go to cytoplasm and join with complementary amino acids.
  2. Process continues with up to 15 amino acids added per second, until the desired polypeptide chain produced.
  3. Up to 50 ribosomes can pass immediately behind 1st —> identical polypeptides can be made simultaneously.
  4. Polypeptide synthesis continues until ribosome reaches a stop codon —> ribosome, mRNA and final tRNA molecule all separate and polypeptide chain is now complete.
28
Q

How are some proteins assembled?

A
  • Sometimes a single polypeptide chain is a functional protein.
  • Multiple polypeptide chains often linked together to give a quaternary structure functional protein.
    1. Polypeptide coiled/folded to produce secondary structure.
    2. Secondary structure coiled/folded further to produce tertiary structure.
    3. Different polypeptide chains along with any prosthetic (non-protein) groups are linked to form the quaternary structure.
29
Q

Define gene mutation.

A

= Change to one or more nucleotide bases/change to sequence of the bases in DNA (in a gene).

  • Can arise spontaneously during DNA replication and include base substitution/deletion.
30
Q

Outline substitution mutations. Consequences?

A
  1. Nucleotide in a DNA molecule replaced by another nucleotide that has a different base.
  2. Can result in change of one amino acid in resulting polypeptide produced as one triplet codon is changed,

—> if the particular amino acid is important in forming bonds that determine the tertiary structure, protein may have a different shape and may not function properly.

31
Q

Why do some substitution mutations not lead to different polypeptides being produced?

A
  • Genetic code is degenerate - not all base substitutions cause a change in the sequence of encoded amino acids —>
  • If new base triplet codes for same amino acid as the initial triplet => silent mutation.
  • Mutation will have no effect on final polypeptide produced.
32
Q

Outline deletion mutations. Consequences?

A
  1. Nucleotide (and base) lost from normal DNA sequence.

=> frameshift mutation —> when n bases deleted where n is not wholly divisible by 3.

=> amino acid sequence of polypeptide is completely different —> produced protein is unlikely to function properly.

=> all triplets in sequence read differently as each has been shifted to the left by one base.

33
Q

What can increase mutation rate? Give examples.

A

Mutagenic Agents:

  1. UV light.
  2. Ionising radiation.
  3. Chemicals.
  4. Some viruses.
34
Q

Define chromosome mutations.

A

= Changes in the number/structure of whole chromosomes.

35
Q

Outline polyploidy chromosome mutation.

A

= Change in whole sets of chromosomes.

  • Occurs when organisms have three or more sets of homologous pairs of chromosomes rather than the usual 2.
36
Q

Outline non-disjunction chromosome mutation.

A

= Change in the number of individual chromosomes.

= Gamete has +/- 1 of a particular chromosome.

  • Occurs when individual homologous pairs of chromosomes fail to separate during meiosis.
37
Q

How is meiosis different from mitosis?

A

Meiosis leads to 4 genetically different haploid daughter cells.

Mitosis leads to 2 genetically identical diploid daughter cells.

38
Q

Outline Meiosis I.

A
  1. Homologous chromosomes pair up and their chromatids wrap around each other.
  2. Equivalent portions of chromatids can be exchanged in crossing-over.
  3. Homologous chromosomes separate, with one chromosome from each pair going to each of two daughter cells
39
Q

Outline Meiosis II.

A
  1. Chromatids are separated by a further division into 4 cells, each with a single chromatid.
40
Q

Why do we half the chromosome number in meiosis?

A
  1. To produce haploid gametes —> so zygote and offspring formed have the correct diploid number of chromosomes as haploid gametes from mother and father fuse together, restoring diploid number of chromosomes.
  2. Ensures one allele from each of mother and father for each gene.

=> maintains a constant number of chromosomes in adult species.

41
Q

State factors of meiosis (+ later fertilisation) which lead to genetic variation.

A
  1. Independent segregation of homologous chromosomes.
  2. Genetic combination by crossing over.
  3. haploid gametes fuse randomly at fertilisation —> produces more genetic variation in offspring.
  4. gametes usually come from different parents —> 2 different genetic makeups combined so even more variety as a result.
42
Q

Outline how independent segregation leads to more variation.

A

In meiosis I:

  1. Homologous chromosomes line up alongside partner.
  2. 23 homologous pairs side by side - chromosomes randomly arranged, with one of each pair passing to each daughter cell.

—> whether dominant/recessive allele is passed on to each daughter cell depends on how chromosomes line up.

  1. => independent segregation - combination of maternal and paternal (of parent cells) chromosomes that go in to daughter cells at meiosis I is random/matter of chance.

=> produces new genetic combinations in daughter cells.

43
Q

Outline how crossing over leads to more variation.

A
  1. Each chromosome lines up alongside homologous partner in meiosis I.
  2. Chromatids of each pair twist around each other.
  3. Tensions created (due to twisting) + portions of chromatids break off.
  4. Broken portions might then rejoin with the chromatids of its homologous partner => recombination.
  5. Usually equivalent portions (same genes) of homologous chromosomes that are exchanged.
  6. New genetic combinations of maternal and paternal alleles produced.
44
Q

Why is crossing over important?

A

Without crossing over and recombination, possible that only 2 genetically different types of daughter cell would be produced => crossing over increases genetic variation even further.

45
Q

Comment on possible chromosome combinations following meiosis.

A
  • Number of possible combinations of chromosomes for each daughter cell = 2^n where n = pairs of homologous chromosomes.
  • Variety increased further through random pairing of male and female gametes —> number of possible combinations of chromosomes in zygote (2^n) ^2 where n = haploid number of organisms.
46
Q

Define genetic diversity. Comment on its impact on natural selection.

A

= Total number of different alleles in a population.

=> Genetic diversity is a factor that enables natural selection to occur.

47
Q

Why is higher genetic diversity good?

A
  1. Wider variety of alleles in a population.
  2. More likely that some individuals in a population will survive an environmental change.
  3. Because, wider range of alleles and characteristics in a population —> increased probability that an individual will have a characteristic that suits it to the new environmental conditions.
48
Q

Define population.

A

= Group of organisms of the same species in a particular habitat at one time.

49
Q

Comment on reproductive success and its impact on allele frequency.

A
  1. Within any population of a species, there will be a gene pool containing a wide variety of alleles.
  2. Random mutation of alleles within gene pool
    - –> new allele - usually harmful - but in certain environments new allele > old allele therefore giving the organism an advantage over other individuals in the population.
  3. These ^ individuals will be better adapted —> more likely to survive and reproduce = greater reproductive success.
  4. Over many generations, numbers of individuals with new advantageous allele will increase at the expense of the individuals with the less advantageous alleles.
  5. => allele frequency of new advantageous allele increases, while that of the old allele decreases.

NB => what is advantageous depends on environmental conditions at any one time.

50
Q

Define selection.

A

= Process by which organisms better adapted to their environment are more likely to survive and reproduce.

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