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LQB182 Exam 2022 QUESTIONS WITH COMPLETE SOLUTIONS

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Storage of DNA info Correct Answer: -in the sequence of bases -5' to 3' -chromosomes= 51 million to 245 million base pairs -information in DNA used for replication of chromosomes, production of proteins and regulation -average gene = 1400 bp The central dogma Correct Answer: replication DNA → RNA → protein (transcription and translation replication only occurs when the cell needs to divide, not when it needs to encode a message Nucleus: replication, transcription Cytosol: translation intron= non-coding sequence stop codon does not create an amino acid, isn't involved in the translation polyA signal = a long chain of adenine nucleotides added to a mRNA molecule to make a 5' to 3' end mRNA, the 3' to 5' DNA template strand must be copied intron sequences removed during splicing so finished transcription product contains only exons. it is then translated, goes through post-translational modification and results in a finished protein Gene transcription Correct Answer: Step 1: making a message eukaryote promoters: upstream of gene, contains TATA box approximately 25 nucleotides upstream of transcriptional start that will be recognised by transcription factors and will attach to it and will bring in the RNA polymerase. (enzyme that makes polymer of RNA) gene transcription: one one strand is copied into a complementary RNA transcript called the template strand. No T in RNA this transcript travels to the cytoplasm where its information into amino acid Step 2: removing introns and protecting ends Spliceosome Ribonucleoprotein: protein and RNA, sn= small nuclear the spliceosome recognises the intron boundary, forms a loops called a lariat and the introns excised and exons joined together. GT and AG are usually between the exon and intron which is how the spliceosome recognizes it. Protecting the mRNA ends 5' methyl cap added to 5' end PolyA tail protects 3' prime end: transcription continues past final end of RNA, polyA signal sequence (AAuAA), identifies cleavage site PolyA polymerase a

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LQB182 Exam 2022 QUESTIONS WITH
COMPLETE SOLUTIONS
Only this content will be on the multiple choice
Storage of DNA info Correct Answer: -in the sequence of bases
-5' to 3'
-chromosomes= 51 million to 245 million base pairs
-information in DNA used for replication of chromosomes, production of proteins and regulation
-average gene = 1400 bp

The central dogma Correct Answer: replication DNA → RNA → protein (transcription and
translation
replication only occurs when the cell needs to divide, not when it needs to encode a message
Nucleus: replication, transcription
Cytosol: translation
intron= non-coding sequence
stop codon does not create an amino acid, isn't involved in the translation
polyA signal = a long chain of adenine nucleotides added to a mRNA molecule

to make a 5' to 3' end mRNA, the 3' to 5' DNA template strand must be copied

intron sequences removed during splicing so finished transcription product contains only exons.
it is then translated, goes through post-translational modification and results in a finished protein

Gene transcription Correct Answer: Step 1: making a message
eukaryote promoters: upstream of gene, contains TATA box approximately 25 nucleotides
upstream of transcriptional start that will be recognised by transcription factors and will attach to
it and will bring in the RNA polymerase. (enzyme that makes polymer of RNA)

gene transcription: one one strand is copied into a complementary RNA transcript called the
template strand.
No T in RNA
this transcript travels to the cytoplasm where its information into amino acid

Step 2: removing introns and protecting ends
Spliceosome
Ribonucleoprotein: protein and RNA, sn= small nuclear
the spliceosome recognises the intron boundary, forms a loops called a lariat and the introns
excised and exons joined together.
GT and AG are usually between the exon and intron which is how the spliceosome recognizes it.
Protecting the mRNA ends
5' methyl cap added to 5' end
PolyA tail protects 3' prime end: transcription continues past final end of RNA, polyA signal
sequence (AAuAA), identifies cleavage site

,PolyA polymerase adds PolyA tail to 3' prime end

From gene to mRNA to protein

Ribosomes are 60% rRNA and 40% protein. large unit (60s) and small unit (40s)
codons are made of 3 bases, 64 codons in the genetic code (64 different options), 3 stop codons
tRNA: transcribed from tandem gene arrays,70-90 nucleotides in length
tRNA structure: at the 3' prime end is the amino acid attachment sites (same in all tRNA),
adaptor molecules (bind to codons in mRNA, donate 1 of 20 amino acids to protein synthesis at
the ribosome during translation)

aminoacyl-tRNA synthetases
-link specific tRNA with correct amino acid
-enzymes that catalyse a two step reaction requiring energy (from ATP)
-20 of them
-one for each amino acid

making tRNA
step1- amino acid and ATP bind in AMinoacyl-tRNA synthetase (2 pyrophosphate is lost, energy
is transferred)
step2- tRNA is recruited into another active site (AMP is displaced, energy used to attach tRNA
to amino acid)

The wobble hypothesis Correct Answer: a single tRNA anticodon may pair with more than one
mRNA
decreases the original number of tRNAs from 61 to 31
3rd position in the codon- called the wobble base

Ribosomes Correct Answer: rRNA= ribosomal RNA, mRNA= messenger RNA, tRNA=transfer
RNA

● Ribosomes are where protein synthesis (translation) occurs
● Each ribosome has a large and small subunit
● Each subunit a hybrid of RNA and proteins
● High molecular weight

31 ribosomal proteins make large subunit in prokaryotes, 21 in small.
49 ribosomal proteins make large subunit in eukaryotes, 33 in small.

Pro Vs Euk, + more ribosomes Correct Answer: Pro: three smaller rRNA and fewer proteins
-small subunit called 30S
-large subunit called 50S come together to make 70S complex
-transcription and translation are coupled, no nuclear membrane, it occurs in open cytoplasm
area

Euk: Four larger rRNAs and more proteins

,-Small subunits=40S
-Large subunits=60S =80S
-nuclear envelope separates transcription and translation

S=Svedberg unit = rate of sedimentation of a particle in a centrifuge, an indicator of size, how
heavy is it.

Ribosome synthesising: Nucleolus is actually a dense region of the nucleus where massive
transcription of ribosomal RNA genes is occurring.

Nucleolar Organiser
Nucleolus is actually a dense region of the nucleus where massive transcription of ribosomal
RNA genes is occurring.

45S rRNA gene arrays gather at the nucleolus. 45S rRNA genes are encoded by 10 chromosomes
- genes are arrayed at the ends, region called nucleolar organisers. All the nucleolar organisers
together form the nucleolus.

Chromosomes with rRNA genes
5 clusters of ribosomal RNA genes
13p12, 14p12, 15p12, 21p12 and 22p12

How ribosomes are made Correct Answer: The 45S rRNA transcript + ribosomal proteins from
cytoplasm = ribonucleoprotein particle
45S rRNA transcript and particle are processed into immature large and small ribosomal
subunits.
Final functional form only as each is individually transported through the nuclear pores into the
cytoplasm.

-Continuous transcription of multiple copies of 45S rRNA genes occurs in the nucleolus
-The 5.8S, 18S & 28S rRNA are processed and packaged with the ribosomal proteins in the
nucleolus to form a large ribonucleoprotein particle
-5S rRNA transcribed elsewhere in the nucleus now binds
-Inactive large & small ribosomal subunits separate (18S rRNA in small, rest in large) &
subunits transfer to cytoplasm through the pores in the nuclear membrane
-Small & large subunits only begin to translate mRNA after they reach the cytoplasm
(functional!)

Translation Correct Answer: The molecular mechanism by which the primary amino acid
sequences of proteins are synthesised from mRNA
Begins at the ribosome in the cytoplasm
Needs:
- accessory proteins for initiation (IF), elongation (EF) & termination (RF) of nascent
polypeptide
-mRNA (codons from gene)

, -a pool of activated tRNAs (adaptor between mRNA nucleic acid sequence & amino acid
sequence)
-energy (lots of ATP & GTP)

3 main steps in translation: initiation, elongation, termination
The main steps are similar but not identical for prokaryotes and eukaryotes

INITIATION
Formation of initiation complex
-Assembly of everything needed to begin translation
mRNA, initiator tRNA (fmet - prokaryotes, Met - eukaryotes), and the ribosome (initiation
factors, and GTP (energy) aids in the formation of the initiation complex)

-AUG - initiation codon, where translation starts, Methionine is first amino acid in polymer
Downstream of sequence that functions like a "promoter" equivalent for translation - binds the
small ribosomal subunit
Prokaryotes = Shine-Dalgarno sequence (Aussies!) eg. AGGAGG The prokaryote ribosomes
finds the start sequence by finding the Shine-Dalgarno sequence.
Eukaryotes - Ribosomal binding site (5' cap or 6-7 base upstream of start codon) then Kozak
sequence eg (gcc)gccRccAUGG (R=purine)
The Prokaryote shine-dalgarno sequence
The 30S Ribosomal subunit recognises the Shine-Dalgarno sequence in the 5'UTR and then
locates the nearest AUG codon downstream to it.

Initiator complex
-Small subunit +
initiation factors +
energy (GTP)

-Binds mRNA at Shine-Dalgarno
(or Kozak)
and Initiator tRNA
(Met)

-Large subunit binds
Initiator tRNA into P site
Energy used


The tRNA anticodon is complementary to the mRNA codon

Only the initiating Met-tRNA can bypass the A site and dock directly on the P site All
subsequent charged tRNA dock at the A site

Elongation
-Next tRNA with complementary anticodon enters ribosome and sits in A site

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