Genes, Proteins and Meiosis Flashcards

1
Q

Describe the structure of DNA.

A
  • Double stranded DNA backbone is made of phosphates and deoxyribose sugars.
  • Each phosphate carries a negative charge, so DNA is very highly negatively charged.
  • Each phosphate forms a phosphodiester bond and hydrogen bonds forms between complementary nitrogenous bases.
  • Double strands intertwine to form an antiparallel double helix, as one strand runs 5’ to 3’, and the other runs 3’ to 5’, giving strands a polarity and a direction.
  • Guanine and adenine are the purines. Thymine and cytosine are the pyrimidines.
  • Guanine and cytosine are complementary and form 3 hydrogen bonds. Adenine and thymine are complementary and form 2 hydrogen bonds.
  • Flat edges of the bases are hydrophobic. The sides of the bases and the sugar-phosphate backbone are hydrophilic. This causes the bases to stack together and form a double helix.
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2
Q

Describe the major and minor grooves.

A

Major and minor groves form by the relative distances between the 2 backbones. A larger distance results in a major groove, and a smaller distance results in a minor groove. Grooves are important because proteins and enzymes need to make bonds (mostly hydrogen bonds) with DNA at the grooves. This is most easily done at the major grooves, as they are more open. Amino acids of the sides of the alpha helixes of the proteins can bind to the DNA. Upon binding, proteins can bend the strand and change its conformation, often by neutralising the DNA complex’s charge.

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3
Q

Name the 4 types of histones and what they come together to form.

A

Histones are types of DNA binding proteins. There are 4 types: H2A, H2B, H3 and H4. The four come together in a positively charged octameric complex. Negatively charged DNA wraps round the histone octamer twice. The 2 turns of DNA and octamer is a nucleosome, the building blocks of chromosomes.

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4
Q

Describe DNA replication in prokaryotes and eukaryotes.

A

In bacteria, DNA replication at a unique replication origin and proceeds simultaneously in opposite directions (bidirectionally). In eukaryotes, replication begins at multiple origins, as DNA is so much larger, and proceeds bidirectionally with primers.

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5
Q

Describe the process of DNA replication.

A
  1. Initiator proteins bind to the origin and recruit DNA helicase.
  2. Helicase breaks hydrogen bonds between complementary base pairs and unwinds the double helix to form a single stranded DNA template strand.
  3. This allows primase to make short RNA primers for DNA polymerase.
  4. DNA polymerase makes DNA in the 5’ to 3’ direction and adds deoxynucleotide triphosphates, dNTPs, to the 3’ end.
  5. Energy comes from breaking high energy phosphate bonds in dNTPs, so it is energetically favourable for dNTPs to be added.
  6. DNA polymerase takes over from primase and elongates the RNA primer, forming an RNA-DNA hybrid.
  7. As the DNA unwinds, the process repeats and Okazaki fragments are formed.
  8. Previous Okazaki fragment and old RNA primer on the leading strand.
  9. New RNA primer is synthesised by primase.
  10. DNA polymerase with 5’ to 3’ exonuclease activity adds new primer to new Okazaki fragment.
  11. DNA polymerase finishes DNA fragment and old RNA primer is erased and replaced by DNA.
  12. The two fragments are joined by DNA ligase.
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6
Q

What is the function of telomerase?

A

The last primer is close to the end of the strand and primase has no room to synthesise the end of the strand.

  • Telomerase is a DNA polymerase that uses a short RNA molecule as a template, and not the DNA.
  • It makes repeats on the lagging strand.
  • Telomeric repeats are added to the end and prevents telomeres getting shorter and shorter in each replication cycle.
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7
Q

What are the 4 types of DNA repair?

A
  • Mismatch repair – enzymes cut out incorrect base and resynthesise the DNA that has been cut out.
  • Direct repair – DNA gets damaged by UV light or chemicals and is directly repaired.
  • Excision repair – damaged base can be directly removed by enzymes and re-synthesise the gap.
  • Nonhomologous end-joining – if both strands are damaged, the cell has no template. Randomly joining 2 ends of DNA, which is dangerous and may cause lasting mutations, but it is the only way cells can fix this.
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8
Q

Which RNA polymerases make which RNA molecules?

A
  • RNA polymerase I makes rRNA (ribosomal)
  • RNA polymerase II makes mRNA
  • RNA polymerase III makes tRNA
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9
Q

What must happen before translation can occur?

A
  • Introns are removed by spliceosomes in splicing in the nucleus.
  • A 5’ cap is added to mRNA so that it can be recognised by the ribosome.
  • Polyadenyl is added to the 3’ end to prevent mRNA from being chewed up the endonucleases.
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10
Q

What are the roles of transcription factors?

A
  • Transcription factors are needed to allow RNA polymerase to bind to the promotor region.
  • Transcription factors can activate or suppress the binding to the promotor region, so can allow or prevent RNA polymerase binding.
  • Transcription factors can bind to promotor regions or enhancer regions, which can be anywhere in the sequence.
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11
Q

Define epigenetic regulations.

A

Modifications that amend the gene expression without changing the DNA base sequence.

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12
Q

What is DNA methylation?

A

DNA methylation creates modifications in the promotor regions of genes and is associated with gene silencing.

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13
Q

What are histone modifications?

A

Histone modifications can be methylation, acetylation (associated with gene expression), phosphorylation, etc. This changes packaging and how accessible the gene is for transcription.

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14
Q

What are aminoacyl-tRNA synthetase?

A

At the 3’ end of tRNA, there is a specific amino acid. Aminoacyl-tRNA synthetases are very specifically shaped enzymes that join the correct specific tRNA and amino acids together.

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15
Q

Name and describe the 3 active sites of the ribosome.

A
  • A site – aminoacyl-tRNA synthetase joins
  • P site – peptidyl transferase joins. This is where a peptide bond forms.
  • E site – empty tRNA
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16
Q

Describe the process of translation.

A
  1. Initiator tRNA has a small ribosomal subunit that scans the mRNA to find the start codon, complementary to the anticodon.
  2. Large subunit joins to begin translation.
  3. Second tRNA that has an amino acid joins.
  4. A peptide bond forms between the 2 amino acids.
  5. Peptidyl transferase activity transfers the amino acid from the initiator RNA to the next amino acid, joining them.
  6. tRNA moves to its exit site and is rejected so that another tRNA can join.
  7. This continues until a stop codon, as it has no complementary anticodon.
17
Q

Define polyribosomes.

A

Multiple ribosomes can attach and multiple copies are formed. This can amplify protein signal and can make many proteins from 1 mRNA molecule.

18
Q

Name 5 antibiotics that inhibit translation. How do they do this?

A
  • Tetracycline – blocks the binding of aminoacyl-tRNA synthetases at site A.
  • Streptomycin – blocks transition to a chain elongation complex.
  • Chloramphenicol – inhibits peptidyl transferase.
  • Cycloheximide – blocks translocation.
  • Erythromycin – blocks translocation.
19
Q

What are miRNA and lncRNA, and what do they do?

A

miRNA (micro RNA) control translation and lncRNA (long non-coding RNA) control gene expression

20
Q

How do miRNA control translation?

A
  1. miRNA bind to mRNA and turn off transcripts, by either targeting them for degradation or pausing translocation.
  2. Precursor miRNA is made by RNA polymerase. These are processed and exported to the cytoplasm to make single stranded RNAs.
  3. These bind to risc proteins.
  4. RNA-protein complex finds complementary RNAs and chops them up.
  5. This pauses translocation or rapidly degrades mRNA.
21
Q

What are 2 other uses of miRNA?

A

miRNAs can be used in drugs to target specific disease causing proteins and are also good biomarkers.

22
Q

How does meiosis create genetic diversity?

A

The combination of 2 haploid germ cells produces a single celled diploid zygote, which then divides by mitosis to form a multicellular animal. Meiosis produces 4 haploid cells from 1 diploid cell, with the aim of producing new combinations of genes.

23
Q

Describe meiosis in males.

A
  1. DNA replication – maternal and paternal homologs are replicated to form pairs of identical sister chromatids, held together at the centromere.
  2. The duplicated homologs align and crossover may occur.
  3. In the first cell division, 1 pair of sister chromatids is distributed to each daughter cell.
  4. In the second cell division, 1 sister chromatid is distributed to each daughter cell.
  5. Each haploid daughter produced has a unique genome/each has a different combination of alleles present in the original diploid cell.
24
Q

Describe meiosis in females.

A
  1. The first stage, where the DNA is replicated and cross over might have taken place, has already been completed either before or very shortly after birth. The primary oocytes produced are held at this stage.
  2. Later in life, once a female reaches maturity, individual primary Oocytes pass through the later stages of this process.
  3. The duplicated homologues get separated in anaphase 1, but the cell division is asymmetric.
  4. So only one large cell gets produced containing one of the duplicated homologues and most cell contents, and the other cell gets the other duplicated homologues but practically nothing else.
  5. This is called a polar body, and they either apoptose or they get destroyed within 24- 48 hours.
  6. Then that cell can divide again, and that second meiotic cell division is again asymmetric resulting in one ovum and one polar body.
  7. Theoretically, if this first polar body hasn’t been destroyed yet it will undergo this second meiotic cell division too, but that will result in 2 polar bodies again.
25
Q

Describe the process of crossing over in creating genetic diversity.

A
  1. During meiosis, the duplicated maternal and paternal homologs pair up and are held close together.
  2. Proteins called Spo11 and Mre11 act together to make breaks in the DNA.
  3. The broken ends are converted to single stranded DNA, which invades the equivalent region of DNA on the other homolog.
  4. Further processing produces Holliday/4-way DNA junctions.
  5. Cutting the Holliday junctions can produce cross over products with maternal DNA at one end and paternal DNA at the other.
  6. Also produces non-crossover products, where only small regions of DNA are exchanged.
26
Q

Define aneuploidy.

A

Errors in meiosis can lead to loss or gain or chromosomes.

27
Q

Describe the effects of autosomal aneuploidy.

A

Autosomal aneuploidy can cause developmental problems, such as trisomy 21 in humans causes Downs syndrome and trisomy 18 in cattle causes brachygnathia.

These syndromes are causes partly by an imbalance in gene expression caused by the presence of an extra copy of genes on the triplicated chromosome.

28
Q

Describe the effects of aneuploidy in sex chromosomes.

A

Can give rise to ‘intersex’ individuals:

  • Klinefelter’s syndrome is a 65, XXY stallion, an infertile male phenotype
  • Turner’s syndrome is a 63, X0 mare, a small infertile female phenotype