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Molecular > Eukaryotic Genomes > Flashcards

Eukaryotic Genomes Flashcards

1
Q

Define genomics

A

The study of genome organisation and the identification of genes and their functions

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2
Q

Give the number of Mbp (million base pairs, the likely number of protein coding genes and how many rRNA, tRNA and small nuclear RNAs they have

A

. 12.1 Mbp (in 16 chromosomes)
. 5775 likely protein-coding genes
. Approx 140 genes for ribosomes RNA, 275 for tRNA, 40 for small nuclear RNAs

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3
Q

How many Mbp and protein-coding genes in the model plant Arabidopsis thaliana?

A

. 125 Mbp

. 25,498 protein-coding genes

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4
Q

How many Mbp and protein-coding genes are in the plant black cottonwood

A

. 485Mbp (genomic size)

. 45,555 protein coding genes

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5
Q

Give the genomic size and the number of protein-coding genes in wheat

A

. 17,000 Mbp (17Gbp)

. 107,891 protein-coding genes

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6
Q

What is the genomic size and number of protein-coding DNA in the invertebrate Nematode?

A

. 97.1 Mbp

. 19,000 protein-coding genes

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7
Q

What is the genome size and number of protein-coding genes in the vertebrate fruit fly?

A

. Approx 180Mbp

. 15,500 protein-coding genes

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8
Q

Compared to other insects what do honeybees have?

A

. A new family of genes coding for major royal jelly protein-component of royal jelly fed to larvae, especially guess larvae
. Fewer genes for taste receptors
. Many more genes for odorant receptors-odours are important in social communication and flower recognition

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9
Q

What is the genome size and number of protein-coding genes in humans?

A

Estimated 3286 Mbp

Approx 21,000 genes

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10
Q

What is the genome size and number of protein-coding genes in pigs?

A

. Approx 2.8 Gbp

. Approx 22,000

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11
Q

What is the genome size or birds and chickens?

A

About 1,000Mbp

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12
Q

What is the genome size or pufferfish (freshwater and saltwater)?

A

About 400Mbp

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13
Q

Where is extra DNA?

A

. Within genes-introns

. Between genes- repetitive DNA

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14
Q

Give a brief overview of what happens in RNA processing in eukaryotes

A
  1. Addition of 7-methyl G cap
  2. Removal of introns
  3. 3’ end cleavage and addition of poly(A) tail
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15
Q

Excision of an intron involves two steps. What are these? (Lariat formation)

A
  1. 5’ cut and formation of a loop with a tail: lariat. Loop is formed by 5’-3’ phosphodiester bond (between the hydroxyl groups)
  2. 3’cut and joining of exons
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16
Q

Intron excision is catalysed by spliceosomes. How does it do this?

A
  1. snRNPs bind to primary transcript
  2. Interactions between snRNPs form spliceosome
  3. Spliceosome cuts 5’ end of intron and forms lariat (by forming a covalent bond)
  4. Spliceosome cuts 3’ end of intron and join exons
    (See last page if lecture 6)

Molecular (16 decks)

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