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OAT-Biology > Molecular Genetics Workshop > Flashcards

Molecular Genetics Workshop Flashcards

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AP® Biology flashcards Biology 101 flashcards
1
Q

Why is DNA replication semi-conservative?

A

because half of the original DNA molecule is incorporated into each daughter strand.

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2
Q

What are the steps to DNA replication?

A
  1. DNA helix is unwond by helicase where the H bonds are broken.
  2. DNA polymerasesynthesizes new strands complimentary to the original parental strands in the 5’ to 3’ direction beginning at the origin of replication.
  3. replication fork is formed
  4. The leading strand is synthesized continuously using one RNA primer.
  5. The lagging strand is synthesized discontinuously using a RNA primer for every Okazaki fragment.
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3
Q

What is transcription?

A

Theprocess by which information stored in the bases of DNA is copied into RNA.

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4
Q

What are the steps for transcription?

A
  1. RNA polymerase binds to the TATA box in the promoter region of the DNA template strand.
  2. Only one strand of the DNA is used for a given gene.
  3. Nucleotides are added in a 5’ to 3’ direction
  4. Heterogeneous nuclear RNA (hnRNA) is formed
  5. The introns are cleaved out and the exons spliced together to form the mRNA
  6. The 5’ end of the mRNA is capped
  7. The 3’ poly A tail is added
  8. The finished mRNA product leaves the nucleus through nuclear pores.
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5
Q

What is translation?

A

Translation is the synthesis of an amino acid chain using mRNA as a template.

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6
Q

What are the steps of translation?

A
  1. Occurs in the cytoplasm and requires energy (GTP)
  2. mRNA bind to a ribosome, Translation begins when the ribosome encounters a start codon.
  3. tRNA delivers amino acids to the ribosomes.
  4. the tRNA/ amino acid complex temporarily binds the mRNA codon.
  5. An enzyme (peptidyl transferase) fogers a peptide bond between adjacent amino acids.
  6. Pretein synthesis stops when a stop codon is reached.
  7. Post-translational modifications occur to the protein product.
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7
Q

what are the three main classes of small scale mutations?

A
  1. Base substitution
  2. Deletions
  3. Insertions
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8
Q

Describe base substitution

A

one base pair is substituted for another.
Transition- substitution of a pyrimidine (C or T) by another pyrimidine, or of a purine(A or G) by another purine.
Transversion- substitution of a pyrimidine by a purine or of a purine by a pyrimidine.

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9
Q

Describe deletions

A

One or more nucleotides are lost from a sequence

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10
Q

Describe insertions

A

One or more nucleotides are added to a sequence.

This can include transposition- a sequence is inserted at an incorrect location in the DNA.

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11
Q

What is direct repair

A

reverses DNA damage without cutting the deoxyribose phosphate backbone

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12
Q

what is base excision repair (BER)

A

used when incorrect bases are present in the DNA. The damaged base is recognized by a glycosylase and is hydrolytically removed from the deoxyribose phosphate backbone. The leaves an apurinic or apyrimidic site where the purine or pyrimidine was removed. The correct base is then inserted and the break is sealed by DNA ligase.

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13
Q

what is Mismatch repairs

A

method similar to BER to remedy correct pairings of the normal bases.

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14
Q

what is nucleotide excision repair (NER)

A

removes thymine dimer and bulky adducts. The are of DNA surrounding and including the damaged portion is unwond and an endonuclease makes cuts on both the 5’ and 3’ side of the damage. The bases are removed by an exonuclease and DNA is resynthesized, using the sister strand as a template to fill the gap. DNA ligase seals the new section into the backbone.

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15
Q

what is post-replication repair

A

used to repair double strand breaks. This involves a type of recombinational repair in which a single strand of DNA from a homologous chromosomes is used to resynthesize the missing portion. Broken ends can also be rejoined directly and ligated together. However, the original sequence is not always maintained and mutations such as translocations can often occur as a result of this.

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16
Q

what is a nucleosome

A

most basic unit. consst of 8 histone proteins. DNA is wound almost two times around this protein core to produce a “bead-like” structure.

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17
Q

what are the guidelines for cross-over

A
  • it occurs randomly along the entire chromosome.
  • Two genes close together on the chromosome have a low chance of cross-over (the genes are said to be linked).
  • The further apart two genes are on a chromosome, the greater the chance of these genes crossing-over.
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18
Q

what are the events that occur during recombination?

A
  1. Homologous pair line up.
  2. An endonuclease nicks a single strand of DNA on each homolog at the same place.
  3. The homologs exchange strands and are ligated together forming the Holliday structure.
  4. Branching migration occur,incorporating a portion of the opposite strand into each molecule.
  5. Cleavage occurs: if the same strands are cleaved the original chromosomes are reformed. If the opposite strands are cleaved then recombinant chromosomes results.
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19
Q

constitutional chromosomal abnormalities

A

the abnormality is found in all cells of the body.

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20
Q

Somatic chromosomal abnormalities

A

the abnormality if found only in certain cells or tissues.

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21
Q

Numerical chromosomal abnormalities

A

The gain or loss of complete chromosomes

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22
Q

Structural chromosomal abnormalities

A

The formation of abnormal chromosomes through the misrepair of chromosome breaks or a malfunction during recombination

23
Q

Aneuploidy

A

One or more chromosomes are missing or are present in more than the normal number. Usually results from nondisjunction, which is the failure of paired chromosome to separate in anaphase (usually during meiosis 1)

24
Q

Monosomy

A

The loss of a single chromosome.
Autosomal monosomy is always lethal.
Monosomy of sex chromosome X results in Turner’s syndrome (45,X)

25
Q

Trisomy

A 
The gain of an extra chromosome.
Down syndrome (Trisomy 21)
26
Q

Tetrasomy

A

The gain of an extra pair of homologous chromosomes.

Tetrasomy 9p, Tetrasomy 18p

27
Q

Euploidy

A

An extra, complete set of chromosomes is present or missing.

28
Q

Polyploidy

A

More than two sets of chromosomesare present.

Triploid, tetraploid, pentaploid.

29
Q

Monoploidy

A

A complete chromosome set is missing.

Lethal

30
Q

Mixoploidy Mosaicism

A

Two or more genetically different cell lines within a single individual derived from a single zygote.

31
Q

Mixoploidy Chimerism

A

Two or more genetically different cell lines withing a single individual derived from different zygotes

32
Q

what are the two main ways that structural abnormalities arise:

A

recombination malfunction and mispair of chromosome breaks

33
Q

Inversion malfunction

A

The chromosome segment is rejoined opposite of its normal configuration without loss of genetic information

34
Q

Duplication malfunction

A

segment of the chromosome is repeated

35
Q

Deletion malfunction

A

a segment of the chromosome is lost

36
Q

Translocation malfunction

A

chromosomal materil is exchanged between non-homologous chromosomes.

37
Q

Translocation reciprocal

A

there is no loss of genetic infoormation although gene arrangement is altered.

38
Q

Translocation Robertsonian

A

the short arm of two chromosomes breaks off and the long arms are fused together, this can result in a balanced translocation or a unbalanced translocation in which genetic material is gained or lost.

39
Q

Negative control or repression

A

protein binds to DNA in order to interfere with the binding of the RNA polymerase to the promoter region. This prevents transcription

40
Q

Positive control or activation

A

A protein binds to DNA in order to facilitate the binding of the RNA polymerase to the promoter region. This initiates transcription.

41
Q

What is a Polymerase Chain Reaction (PCR)?

A

used to selectively amplify target DNA

42
Q

What are the steps for PCR?

A
  • denaturing (94C) the original DNA strand.; this produces two complimentary single strands of DNA.
  • Annealing (54C) primers to the single stranded DNA; the specific singl stranded primers that are used attached to the complimentary portion of single stranded DNA,.
  • DNA polymerase attaches to the prime/DNA complex and moves down the DNA chin reading the template strand and adding the appropiate complimentary base. This externsion (72C) produces a new double stranded DNA molecule from each strand of the original DNA causing an exponenial increase in the number of copies of a particular gene.
43
Q

DNA sequencing

A
  • same as PCR.
  • run in the presence of small amt of ddNTPs which lack hydroxyl group at the 3’ position as well as dNTPs. The addition of one ddNTPs to the growing chain will terminate chain elongation since no further nucleotides can be added without the 3’ hydroxyl group.
44
Q

Mitochondrial DNA (mtDNA)

A
  • maternally inherited.
  • mitochondria from sperm are destroyed by the egg after fertilization leaving only maternal mtDNA originally found in the unfertilized egg.
45
Q

Cystic Firbrosis (CF)

A

Caused by loss of function mutations in the CFTR gene and it is the most common lethal genetic disease in the US. The normal gene codes for a membrane protein that transport chloride ions in and out of cells. This has the greatest effect on the epithelial cells lining organs and it is particular evident in the lining of the lungs.

46
Q

Huntington disease (HD)

A

results from a gain of function mutations that produces an autosomal dominant disorder characterized by neurodegeneration. Caused by expansion of a triplet repeat, whcih is a set of three nucleotides (CAG) that code for glutamine.

47
Q

what are proto-oncogenes

A

non-mutant versions of genes that control cell proliferation.

48
Q

what are oncogenes activated through

A

Amplification
Point mutations
Chromosomal translocations
Trasnposition

49
Q

Amplification

A

Hundreds of extra copies of a gene may be present.

50
Q

Point mutations

A

can lead to excessive cellular response

51
Q

chromosomal translocations

A

a novel gene is created.

52
Q

transposition

A

a gene is moved from a relatively inactive area of chromatin to an active area where it is consequently transcribed in abnormal amounts.

53
Q

How can loss of function in a tumor suppressor gene can occut

A

they can occur through:

  • Deletion of a portion of chromosome containing the gene.
  • A point mutation within the DNA sequence of the gene
  • Methylation of DNA the prevents the tumor suppressor gene from being transcribed.
54
Q

what are the four approaches to gene therapy:

A
  • Inserting a normal gene into the genome to replace a nonfunctional gene.
  • Replacing a nonfunctional gene through homologus recombination.
  • Repairing the mutant gene by reversing the original mutation.
  • Altering the regulation of a gene.

OAT-Biology (13 decks)

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