Biochemistry (Molecular ) Flashcards

USMLE STEP 1

1
Q

DNA and cell

A

-There is a lot of DNA in a cell
and since the cell is only a few microns in diameter —-) it needs a way to package DNA into a small space inside +++
they do it by coiling DNA around histones

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2
Q

Histones :

A

-there is a few different types of histones : H2A ,H2B,H3,H4; they aggregate together and DNA wrapps around them to form NUCLEOSOME ,

There is another type of histone called H1 which is not in the nucleosome ,

but why does DNA wrapps around histones ???

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3
Q

the charge of DNA and histones :

A

-DNA is - (negatively ) charged

-Histone is + (positively ) charged

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4
Q

Euchromatin vs heterochromatin :

A

-Euchromatin :acetyle groups , transcriptionaly active(more accessible to the transcription machinery )

-Heterochromatin : more methylation , less acetylation , can’t be transcribed

histone methylation —) makes DNA MUTE ( transcription inactive )

histone Acetyl—–) makes DNA ACTIVE —) transcription active ++++

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5
Q

DNA abreviation :

A

-Deoxyribonucleic acid +++

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6
Q

DNA composition :

A

-DNA is composed of nucleotides

and each nucleotide is composed of :
_Phosphate group
_5-carbon sugar (deoxyribose )
_Nitrogenous base (A, G , C , T )

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7
Q

Mnemonic to remember the name of PYRIMIDINES :

A

'’CUT PYe ‘’
-Cytosin and uracil and Thymin are PYrimidine

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8
Q
A

Nucleotides bind together using phosphate and sugar groups

-5’——–) 3’ , 3’—–) 5’
DNA is antiparallel molecule

-The basis form hydrogen bonds between them
(A–T)
(C—G)

-know that hydrogen bonds are weaker than covalent bonds thus can easily been broken and reformed for transcription

basis form bonds according to the rule of complementary base pairing +++

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9
Q

Facts about DNA :

A

-superorganised molecule

-long molecule (over 2 meters along when fully stretched )
to make this DNA fit into a tiny nucleus inside the cell ,our cells relies on a few packaging tricks +++
we can wrap DNA around histones (because DNA is negatively charged by phosphate , and histone is positively charged , so they attracted and DNA get wrapped around histones

-Nucleosome = DNA around histones (histone + DNA )

nucleosomes get packaged as chromatin fibers

-Chromatin comes in 2 flavours :
_Euchromatin (nucleosomes are loosely packed , they are used to the transcription ++++ )

  _ Heterochromatin( nucleosomes are densely packed , contains genes that cells rarely or never uses to make proteins , so they are not active for the transcription +++)

-the chromatin condenses to give 46 chromosomes ( mazal ma3reftch 3lach had 46 )

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10
Q

Nucleotide composition (DNA )

A

-Base + sugar (desoxyribose) +phosphate

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11
Q

NucleoSide : (DNA )

A

Base + sugar (desoxyribose )

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12
Q

A researcher is investigating the structure of chromatin and finds that DNA binds with several proteins to form a “bead on a string appearance” as detailed in the image below. Which of the following best describes the chemical interaction between DNA and protein that results in the formation of this molecular structure?

A

Response:
Bond between negatively charged phosphate and positively charged arginine and lysine residues

Major takeaway
The nucleosome is held together by ionic bonds between the negatively charged phosphate residues of DNA and positively charged lysine and arginine residues found on histones.

Main explanation
The nucleosome refers to a DNA-protein structure held together by negatively charged phosphate residues on nucleotides and positively charged arginine and lysine residues on histones. The nucleosome serves as the backbone for the tight packaging and coiling of DNA, which gives rise to chromatin.
Chromatin is composed of DNA (approximately 146 base pairs) that loops twice around a histone octamer, thus giving the appearance of a bead on a string. The histone octamer consists of two copies of each of the following histone proteins: H2A, H2B, H3, and H4. By forming tightly packed chromatin, the DNA is able to fit within the confines of the cell nucleus.

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13
Q

A researcher is studying the structure of the nucleus under an electron microscope. He detects an area of heterochromatin, at the periphery of the nucleus that he identifies as the inactivated X-chromosome (the Barr body). Which of the following cellular mechanisms best explains the appearance of this structure on electron microscopy?

A

Histone methylation

Major takeaway
Histone methylation refers to the process of adding a methyl group to histone subunits of the nucleosome. Methylation is largely responsible for the appearance of heterochromatin on electron microscopy.

Main explanation
Histone methylation is the process of adding methyl groups to histone subunits of the nucleosome (DNA, histone complex). Methylation is largely responsible for the appearance of heterochromatin (tightly packed DNA) on electron microscopy and contributes to the formation of the Barr body (inactivated X chromosome) in females.
In addition, the Barr body undergoes DNA methylation–a form of irreversible gene silencing without actually changing the sequence of DNA. Similar to histone methylation, DNA methylation causes tight coiling of DNA around histones. Both DNA methylation and histone methylation lead to the condensed and sterically inaccessible form of DNA called heterochromatin, which is located typically at the periphery of the nucleus.
In contrast, histone acetylation removes positive charge from histones; thereby, decreasing its binding affinity for negatively charged DNA and increasing the transcriptional activity of uncoiled DNA. Histone deacetylation has a similar effect to that of histone methylation. Deacetylation increases the positive charge of histones, thus enabling histones to bind more avidly to negatively charged DNA and decreasing transcription.

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14
Q

A researcher is studying the thermal properties of small sequences of deoxyribonucleic acid found in Gram-positive bacteria. Which of the following sequences of DNA will have the highest melting point?

A

Major takeaway
DNA sequences with a higher G-C content are more chemically stable and have a higher melting point, due to additional hydrogen bonds when compared to A-T base pairs.

Main explanation
Deoxyribonucleic acid (DNA) is made up of nucleotide base pairs that bond together via hydrogen bonding. This chemical interaction underlies the base-pairing rules such that only adenine may hydrogen bond with thymine and only cytosine may hydrogen bond with guanine.

In general, DNA sequences with a higher guanine to cytosine base pair content are considered more chemically stable. This is due to the fact that guanine to cytosine base pairs contain three hydrogen bonds, whereas adenine to thymine base pairs contain two hydrogen bonds. Thus, DNA sequences with a higher G-C ratio will have a higher melting point. In the question stem above, choice A has the highest G-C content.

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15
Q

Note DNA and genes

A

each DNA is coding for a lot of genes (thousands of genes )
but there should be a process to control this process of genes

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16
Q

note on DNA and genes

A

DNA is wrapping around each histone (octamer ) twice = NUCLEOSOME ++++

nucleosome wrapped around very tightly giving CHROMATIN (either euchromatin or heterochromatin )

our DNA contains a lot of genes , and the cell function is related to the types of genes they code for = exactly to the protein they made +++ —) a cell can do its function by activating some genes and closing or silencing other genes

if this process is not made correctly , we can have a cell that does a lot of functions not related to it
so the final appearence of a cell depends on which genes are activated ++++(PHENOTYPE )

all of this happens through EPIGENETICS ++++

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17
Q

Definition of epigenetic :

A

Mechanism through wich we can selectively activate or silence genes without modifying the nucleotide sequence++++

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18
Q

in epigenetic , we can modify either Histone or DNA in a way that we activate the genes we want to express and we deactivate genes we want to delete ++

A

-we made this in :
1- HISTONES : histones can either release their DNA or lock down their DNA , in a result of easier or not easier transcription

so : _Acetylation : when we add a group of acetyl to the nucleosome —–) less attraction between DNA and histone —–) DNA loosen up from histones —-) easy transcription +++++

deacytelation is the contrary , we remove acetyl group , therefore increase attraction between DNA and histone , hard transcription

-methylation : if we add one group of methyl we loosen DNA ,
but if we add 2 or more methyl groups , we lock down DNA and therefore harder transcription ++++

_Epigenetic can include also direct modifications of the DNA +++ (we either activate a portion of dna genes or suppress them ) without changing genes sequence or modifying genes sequence +++
we can modify in the CpG sites by adding a methyl groupp to Cytosine (without modifying DNA sequence )
-DNA methylation —–) can silent the gene expression

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19
Q

Epigenetic changes are (reversible or not )

A

REVERSIBLE
(they can be changed throughout the life of the individual , and also related to some risk factors (diet , physical activity and others )))

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20
Q

Nucleotide synthesis :

A

-either from scratch (de nove synthesis )

-salvage pathway (recycles nucleotides that are already semi degraded ++)

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21
Q

An investigator is studying nucleic acids. Which of the following best differentiates a nucleotide from a nucleoside?

A

Presence of a phosphate group

Major takeaway
Nucleosides are made of a nitrogenous base and a five-carbon carbohydrate (deoxy)ribose. A nucleotide is simply a nucleoside with an additional phosphate group(s).

Main explanation
Nucleosides are composed of a nitrogenous base, either a pyrimidine (cytosine, thymine for DNA, uracil for RNA) or purine (adenine, guanine), and a five-carbon carbohydrate (deoxy)ribose. A nucleotide is simply a nucleoside with additional phosphate group or groups.

Polynucleotides containing the carbohydrate ribose are known as ribonucleotides or RNA. If the 2′ hydroxyl group (OH) is removed, the polynucleotide deoxyribonucleic acid (DNA) results.

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22
Q

A 27-year-old woman gravida 1 para 0 at 20 weeks gestation comes to her physician for follow-up. The physician ordered genetic testing for Lesch-Nyhan syndrome due to a positive family history and results positive for the mutation associated with this disease. Which of the following biochemical steps becomes impaired as a result of this mutation?

A

Hypoxanthine to inosine monophosphate

Major takeaway
Lesch-Nyhan syndrome is a genetic disorder due to an X-linked recessive mutation in the HPRT gene with a deficiency in hypoxanthine-guanine phosphoribosyltransferase (HGPRT), an enzyme that mediates the purine salvage pathway, where purines are recycled and used again in DNA and RNA synthesis. Without HGPRT, all purines are broken down into uric acid, resulting in hyperuricemia very early in life.

Main explanation
This patient has a family history of Lesch-Nyhan syndrome and is currently pregnant with a child who is also affected.

Lesch-Nyhan syndrome is an X-linked recessive disorder due to mutation in the hypoxanthine-guanine phosphoribosyltransferase-1 gene (HPRT1) on the X chromosome. This results in the absence of hypoxanthine-guanine phosphoribosyltransferase (HGPRT), an enzyme that is involved in the purine salvage pathway where purines are recycled and used again to make DNA and RNA.
Synthesis of purines to make DNA and RNA can be from scratch (de novo synthesis) or from recycling. The purine recycling pathway is also known as the purine salvage pathway. Here guanine is converted to GMP, and hypoxanthine is converted to IMP and then AMP. Then both AMP and GMP can be used again to make DNA. This step is mediated by the HGPRT enzyme.

In Lesch-Nyhan syndrome, HGPRT is absent. Leading to the inability of cells to recycle purines. This leads to the accumulation of hypoxanthine and guanine due to two mechanisms:
1. Inability to recycle guanine and hypoxanthine
2. Increased de novo synthesis of purines, which are then converted to hypoxanthine and guanine

Hypoxanthine and guanine will eventually be converted to uric acid; therefore, patients with Lesch-Nyhan syndrome develop hyperuricemia, usually very early in life.

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23
Q

The difference between :
-DNA replication
-DNA transcription
-Translation

A

-Replication duplicates the entire DNA molecule for cell division.

-Transcription converts a specific gene’s DNA sequence into mRNA.

-Translation uses the mRNA sequence to synthesize a protein.

24
Q

The DNA in the cell is the genetic portion of the cell

A

It is what it makes the cell what it is

25
Q

Cell cycle

A

-Can be divided into 2 steps
_Interphase (G1,S,G2,)
during S phase , DNA replicated

_Mitosis (cell division ) : prophase , metaphase , anaphase , telophase

26
Q

Anti-metabolite chemotherapies have an effect on cancer during the …. phase of the cell cycle.

A

Anti-metabolite chemotherapies have an effect on cancer during the S phase of the cell cycle.

27
Q

The pathway in which a cell grows, copies its genetic material, and splits into two daughter cells is the …….

A

The pathway in which a cell grows, copies its genetic material, and splits into two daughter cells is the cell cycle.

27
Q

Hydroxyurea affects cancerous cells during the ….. phase of the cell cycle.

A

Hydroxyurea affects cancerous cells during the S phase of the cell cycle.**

Note : Hydroxyurea

28
Q

The two processes of the ….. phase of the cell cycle are mitosis and cytokinesis.

A

The two processes of the mitotic phase of the cell cycle are mitosis and cytokinesis. (the second phase )

29
Q

Antimetabolite antineoplastics are medications which act in the …….-phase of the cell cycle.

A

Antimetabolite antineoplastics are medications which act in the S-phase of the cell cycle.

30
Q

cell cycle is composed of 2 phases :

A

-Interphase

-Mitosis

31
Q

DNA replication happens from :

A

5’—————-) 3’

32
Q

The steps of DNA replication :

A

-Initiation: the particular area of dna where we begin the replication is the spots that are rich in A-T (because A and T had 2 hydrogen bonds and C-G had three hydrogen bonds so it’s easier for helicase to cut in A-T rather than in C-G ,) that’s why the point of initiation is where we have a lot of A-T
-there is multiple origin of replication (rich in A , T )

-Elongation :

-Termination

33
Q

Telomeres

A
  • do not code for RNA (they can’t make DNA transcription or translation )

-with each replication cycle , the telomeres will shorten but it’s okay because they are not coding for any genes or any genetic material , they help to prevent gene loss

(telomeres shorten with every DNA replication +++ , we can prevent that with telomerase enzyme which performed reverse transcription
so highly divided cells (like cancer cells or stem cells ) have a lot of telomerase activity , therefore they replicate without ending the telomeres and arriving at the haflick limit where they can enter with the genes that are active and modify them ++++=

34
Q

what it is called when we go from RNA to DNA ???

A

Reverse transcription

35
Q

Gene expression

A

is the process of decoding the information stored in the DNA (make proteins )
_Transcription
_Translation

36
Q

DNA repair happens at :

A

G0

37
Q

The DNA damage :

A

-Single-stranded damage

-Double-stranded damage

(the cell has mechanisms to adress both situations , to repair the damage DNA and minimize the consequences)

38
Q

Single strand damage can happen :

A

-Endogenous : errors in DNA replication

-Exogenous : harmful chemical or physical agents

the cell can repair these single stranded DNA by 3 mechanisms :
_Mismatch repair
_Nucleotide excision repair
_Base excision repair

39
Q

DNA mutation link video :

A

https://www.youtube.com/watch?v=vl6Vlf2thvI

40
Q

exon and introns :

A

exons : coding sequences

Introns : Non coding sequences

RNA after transcription get spliced and remove Introns from the RNA , making Messenger RNA (mRNA ) —) much more efficient RNA

41
Q

An investigator teaches the cell cycle to his student. Which of the following processes occurs during the longest phase of interphase in the cell cycle?

A

Synthesis of RNA, proteins, and cell organelles

Major takeaway
Interphase is the longest part of the cell cycle and is a state of preparation, during which the cell carries out its cell functions, grows and replicates its DNA to prepare for mitosis. It can further be broken down into three subphases: G1, S, and G2.
Main explanation
The cell cycle refers to the events that somatic cells go through from the moment they are formed until they divide into two identical daughter cells. This cycle varies in length depending on the type of cell. The cell cycle has two phases: interphase and mitosis. Interphase is the longest part of the cell cycle and refers to a state of preparation, during which the cell carries out its normal cellular functions, grows, and replicates its DNA to prepare for mitosis. It can further be broken down into three subphases: G1, S, and G2.

42
Q

Eucaryotic and prokaryotic cells :

A

-Prokaryotic : bacteria cells

-Eucaryotic : human cells

43
Q

Transcription in the prokaryotic cells :

A

-uses RNA polymerase holoenzyme to make all the subunits of RNA : tRNA , mRNA, rRNA

(The process of RNA transcription in eucaryotic cells is different )

44
Q

There are 3 types of RNA :

A

-rRNA

-mRNA

-tRNA

those 3 RNA are involved in the translation process (either by making ribosomes , or the portion to be translated or the amino acid necessary for translation )

45
Q

A very important note in the transcription process :

A
  • RNA polymerase 2 read the template strand of DNA from 3’—) 5’ , and then he forms mRNA from 5’—–)3’

(the opposite ) very important ++++

46
Q

2 clinical vignettes or infos about RNA polymerase inhibition :
-what’s the name of the drug that inhibit RNA polymerase ?
-what’s the name of the poison that inhibit RNA polymerase ?

A

-Ammanitin (poison in mashrooms ) —-) it inhibit RNA polymerase in eukaryotic cells

  • Rifampicin (Antibiotic , antituberculeux ) inhibit RNA polymerase in prokaryotic cells (cells in the bacteria , therefore bacteria cannot synthesize proteins , is not able to do its function anymore
47
Q

A 12-year-old black boy of African descent presents to the emergency department with severe abdominal and joint pain. He has had these episodes twice before and has family members who have suffered similar symptoms. The patient’s hematocrit is 28% with evidence of sickle cells on a peripheral blood smear. Following acute management, he is sent for genetic testing and sequence data analysis reveals a mutation where glutamine is substituted with valine. Which of the following best describes the underlying mechanism of this patient’s disease?

A

Missense mutation

Major takeaway
Missense mutation leads to nucleotide substitution that results in changed amino acids (called conservative if the new amino acid has a similar chemical structure). Examples include sickle cell disease (substitution of glutamic acid with valine).

Main explanation
This patient’s presentation is most consistent with a diagnosis of sickle cell anemia, which is caused by a single nucleotide substitution of glutamic acid with valine in codon six. This represents a missense mutation because the encoded amino acid is changed by the mutation.

Other single nucleotide substitutions include:

Silent mutation: A point mutation that forms a triplet that codes for the same amino acid. Often involves a base change at the tRNA wobble position (3rd position of the codon).
Nonsense mutation: Nonsense mutations are caused by a mutation of a residue into a stop codon (UAA, UAG, and UGA). This causes premature termination of the affected protein and therefore a defect in its production. One example of a disease caused by nonsense mutations is ß-thalassemia.

Other genetic mutations include:

Other genetic mutations include:

Frame-shift mutation: Deletion or insertion of a number of nucleotides not divisible by three. This leads to the misreading of all nucleotides downstream. Proteins may be shorter or longer, and their function may be disrupted or altered. Examples include Duchenne muscular dystrophy or Tay-Sachs disease.
Splice site mutation: An alteration (especially point mutations) in the nucleotide sequence required for splicing (e.g., the exon-intron border or at the junction). This results in defective mRNA (e.g., due to a retained intron) leading to shortened proteins that are either defective or exert an altered function. Examples include Gaucher disease, Marfan syndrome, dementia, and epilepsy

48
Q

A researcher is studying the nitrogenous base production pathway. He adds an anti-hepatitis C medication to a tissue sample and observes that the synthesis of guanosine monophosphate is inhibited. Which of the following enzymes is normally responsible for the synthesis of guanosine monophosphate?

A

Inosine monophosphate dehydrogenase

Major takeaway
De novo synthesis for both purine and pyrimidine bases begins with ribose-5-phosphate. Phosphoribosyl pyrophosphate (PRPP) synthetase catalyzes the conversion of ribose-5-phosphate to PRPP.

49
Q

A study is conducted on the effects of media on the production of B-galactosidase in E. Coli. When the bacteria are inoculated in medium A, the production of B-galactosidase is minimal. However, when the bacteria are inoculated in medium Y, many B-galactosidase is produced. Which of the following is likely present in medium Y?

A

Response : -low glucose , -High lactose concentration

Major takeaway
Gene expression of the structural genes in the lac operon is controlled positively and negatively by two distinct mechanisms. The presence of glucose negatively influences gene expression, and the presence of lactose positively influences gene expression.

Main explanation
The lac operon represents a model for the regulation of gene transcription in E. Coli and functions based on the concentration on the availability of both glucose and lactose to the cells. The lac operon consists of the promoter, operator, and three structural genes, lacZ, lacY, and lacA. These genes produce proteins that allow E.Coli to utilize lactose for energy production. The lacZ gene leads to the production of B-galactosidase, an enzyme that breaks down lactose into glucose and galactose. The lac Y gene produces B-galactosidase permease, allowing lactose to enter the cell. Upstream of these genes lies the promoter region (where the RNA polymerase binds and initiates transcription), followed by the operator region.
The lac repressor protein and catabolite activator protein (CAP) are two leading players that regulate these genes. The repressor protein binds to the operator region and, when bound, prevents RNA polymerase from transcribing the genes of the lac operon. This protein remains attached to the operator region unless lactose is present in abundance. If lactose is present, it binds to the repressor protein detaching it from the operator region, allowing the RNA polymerase to proceed and initiate transcription of the lac operon genes.
The catabolite activator protein (CAP) binds to DNA at a slightly higher location than the lac operon promoter region. It positively regulates the lac operon, increasing the gene expression of the lac operon structural genes. The CAP protein only binds to the DNA when the high cyclic adenosine monophosphate (cAMP) concentration. Glucose inhibits cAMP production and thus, by surrogate effect, diminishes transcription of the lac operon. On the other hand, if glucose concentration is low, then cAMP production increases, allowing the CAP to bind to its region in the DNA and increasing the gene expression of the lac operon structural genes.

Several scenarios exist for understanding the transcription pattern of these genes:

  1. Glucose high, lactose high: The repressor protein will not bind to the operator region because lactose is present, allowing the RNA polymerase to initiate gene transcription. However, because glucose is present, cAMP is low, and the CAP cannot bind to the DNA and positively increase gene transcription. Therefore, the structural genes are expressed at a low rate, producing the proteins in small amounts.
  2. Glucose high, lactose low: In this case, minimal to no gene expression and protein production will occur because the repressor is bound to the operator region (since lactose is low), and therefore, RNA polymerase cannot initiate gene transcription.
  3. Glucose low, lactose low: This case will amount to minimal gene expression since the lactose is low, and the repressor protein is bound to the operator preventing the RNA Polymerase from gene transcription.
  4. Glucose low, lactose high: here, the function of the lac operon is at its highest. Because of the presence of lactose, the repressor protein will be detached from the operator region, allowing RNA polymerase to function in gene transcription. Also, because of low glucose, cAMP levels in the cell increase, allowing CAP to bind and positively impact the gene transcription in the lac operon, increasing gene expression.
50
Q

Zak’s mnemonic to remember the stop codons for the translation to stop is :

A

-UAA : (you are away )
-UGA : (you gone away )
-UAG : (You are gone )

50
Q

The Translation process :

A

-Initiation : -Ribosome grabs mRNA and start the translation

it starts from AUG which codes for methionine +++ (so most proteins actually starts with methionine )

translation continues until the ribosome runs to a stop codon : 3 stop codons (UAA, UGA ,UAG)

51
Q

(Exons/Introns) ……..? are the sequences of genes that code for specific amino acid sequences.

A

(Exons/Introns) Exons are the sequences of genes that code for specific amino acid sequences.

52
Q

(Exons/Introns) ………,are repetitive, interspersed segments of DNA that do not code for amino acid sequences.

A

(Exons/Introns) Introns are repetitive, interspersed segments of DNA that do not code for amino acid sequences.

53
Q

The initiator tRNA can begin mRNA translation when it encounters the start codon triplet ……..

A

The initiator tRNA can begin mRNA translation when it encounters the start codon triplet AUG.

mnemonic to remember the start codon :
Students begin school in AUGust.

54
Q

During the process of translation, the following codon is found in the mRNA in the A site of the ribosome: UGA. Which of the following steps is expected to occur?

A

Response : Release factors recognize this as the stop signal for protein synthesis

Major takeaway
Stop codons in the A site are recognized by release factors, which then stops the process of translation.

Main explanation
Stop codons are signals on mRNA that signal the termination of mRNA translation and protein synthesis. There are three known stop codons, which are the UAG, UGA, and UAA. When the stop codon is in the A site of the ribosome, no complementary tRNA with an anticodon sequence matches these stop codons. Instead, proteins found in the cell called release factors recognize these stop codons and stop the process of translation; catalyze the breakdown of the growing polypeptide from the tRNA in the P site, and release mRNA from the ribosome, thereby terminating translation.

55
Q

A group of researchers are studying the process of protein synthesis from mRNA in cells. They find that after mRNA leaves the nucleus, it binds with a ribosomal-tRNA complex to carry out the translation. Which of the following is true regarding this process?

A

Response :
Translation begins at a specific codon sequence that encodes for methionine

Major takeaway
In both prokaryotes and eukaryotes, the start codon in the mRNA signals where translation will start. It is made of AUG and encodes for methionine.

Main explanation
Translation refers to the process of protein synthesis and occurs in the cytoplasm of all cells. Following transcription, mRNA or messenger RNA is produced. The four nucleotides found in mRNA are adenine (A), guanine (G), uracil (U), and cytosine (C). Each three-nucleotide sequence in mRNA comprises a codon, and each codon encodes for a specific amino acid, which serves as the building block for proteins.
Protein synthesis requires mRNA to serve as the coding sequence for amino acids in accordance with its codon sequence; tRNA or transfer RNA acts as a carrier of amino acids and brings them to the ribosomes for translation.

Ribosomes serve as the site of protein synthesis in both prokaryotes and eukaryotes. In prokaryotes, the ribosomes are designated as 70S, while in eukaryotes it is larger and designated as 80S. In both prokaryotes and eukaryotes, the ribosomes have two distinct subunits. In prokaryotes, the smaller subunit is designated as 30S and the larger subunit is designated as 50S. While in eukaryotes, the smaller subunit is designated as 40S and the larger subunit is designated as 60S.

The process of translation begins once mRNA leaves the nucleus and attaches to a ribosome. The first step is called initiation, where ribosomes attach to mRNA. In order to determine where to start translation, a specific sequence in the 5’ untranslated region (5’ UTR) is identified downstream from the start codon. The start codon specifically implies where the translation will start and is made of AUG, which encodes for methionine.

The process of initiation in both prokaryotes and eukaryotes requires ribosomes, initiation factors (IF), mRNA, and methionyl tRNA. First, IF binds to the small ribosomal subunit. Then this complex allows for the binding of mRNA with the aid of the 5’ UTR (e.g. 5’ cap in eukaryotes) and the initiator tRNA. After this step, the larger ribosomal subunit joins the complex and forms a complete ribosome that is ready for translation with mRNA and initiator tRNA attached to it.