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Biochemistry 2 > Chapter 4- Flow of genetic information > Flashcards

Chapter 4- Flow of genetic information Flashcards

1
Q

Double helix

A

A helical structure that consists of two complementary nucleic acid strands. This is the form of DNA

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2
Q

Why are nucleic acids suited to their function?

A

They are carriers of genetic information due to their covalent structure

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3
Q

Nucleotide

A

The monomer unit in the nucleic acid polymer. Nucleotides consists of 3 components- a sugar, a phosphate, and a base. It is a nucleoside joined to one or more phosphoryl groups by an ester linkage.

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4
Q

Nucleic acid base sequence

A

The base sequence characterizes the nucleic acid and is a form of linear information

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4
Q

Ribose

A

The sugar found in RNA. An oxygen atom is bound to the 2’ carbon.

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5
Q

Deoxyribose

A

The sugar found in DNA. The 2’ carbon atom of the sugar lacks the oxygen atom

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5
Q

Phosphodiester bridges

A

The hydroxyl (3’-OH) group of the sugar on one nucleotide is esterified to a phosphate group, which is bound to the 5’ hydroxyl group of the next sugar. This forms the backbone of the nucleic acid. Each phosphodiester bridge has a negative charge

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6
Q

Backbone of the nucleic acid

A

The chain of sugars linked by phosphodiester bridges

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7
Q

Purine

A

Adenine (A) and guanine (G) are purine derivatives. Purines are longer than pyrimidines because they have a two ring structure. The nitrogen on the 9th carbon (second ring) forms a bond with the sugar

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8
Q

Pyrimidine

A

Cytosine (C) and Thymine (T) are pyrimidine derivatives. Uracil in RNA is also a pyrimidine. They have a single ring structure. The first nitrogen at the bottom of the ring forms a bond with sugar

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9
Q

How does the covalent structure of RNA differ from that of DNA?

A

RNA contains ribose rather than deoxyribose. RNA also contains uracil instead of thymine

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10
Q

Why are phosphodiester bridges important to the integrity of DNA?

A

Each bridge has a negative charge. The negative charge repeals nucleophilic species like hydroxide ions. These ions can launch a hydrolytic attack on the phosphate backbone. This resistance is crucial for maintaining the integrity of information stored in nucleic acids

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11
Q

Why is DNA likely used as the hereditary material of cells?

A

The absence of the 2’ hydroxyl group in DNA increases its resistance to hydrolysis. Also, DNA is more stable than RNA

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12
Q

Nucleoside

A

A unit consisting of a base bonded to a sugar. There are 4 nucleoside units in RNA and 4 in DNA

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13
Q

4 nucleoside units in RNA

A

Adenosine, guanosine, cytidine, and uridine

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14
Q

4 nucleoside units in DNA

A

Deoxyadenosine, deoxyguanosine, deoxycytidine, and thymidine (thymine nucleosides are rarely found in RNA so by convention the deoxy- prefix is not added)

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15
Q

N-beta-glycosidic linkage

A

The N-9 of a purine or N-1 of a pyrimidine is attached to C-1 of the sugar. It links the base to the sugar in a nucleoside

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16
Q

Nucleoside triphosphates

A

Nucleosides joined to 3 phosphoryl groups- the precursors that form RNA and DNA. There are 4 for DNA and 4 for RNA

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17
Q

4 nucleotide units of DNA

A

The 4 nucleotide units of DNA are deoxyadenylate, deoxyguanylate, deoxycytidylate, and thymidylate

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18
Q

4 nucleotide units of RNA

A

Adenylate, guanylate, cytidylate, and uridylate

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19
Q

Naming system for nucleotides

A

The number of phosphoryl groups and the attachment site are designated (ATP is adenosine 5’-triphosphate)

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20
Q

How long are DNA molecules?

A

A DNA molecule has to be made up of many nucleotides to carry the genetic information that is necessary for the organism. The E. coli genome is a single DNA molecule which consists of two strands of 4.6 million nucleotides each. The human genome has 3 billion nucleotides in each strand of DNA

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21
Q

DNA abbreviations

A

pApCpG denotes a trinucleotide of DNA that consists of deoxyadenylate, deoxycytidylate, and deoxyguanylate linked by two phosphodiester bridges- the p denotes a phosphoryl group

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22
Q

DNA chain directionality

A

One end of the chain has a free 5’-OH group or a 5’-OH group attached to a phosphoryl group. The other end has a free 3’-OH group. The sequences are typically written from the 5’ to 3’ direction by convention

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23
Q

The central dogma of biology

A

The flow of genetic information is generally from DNA to RNA to proteins

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24
Q

Bases are attached to which atom of the sugar?

A

Carbon atom 1’

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25
Q

Which organism has the largest known chromosomes?

A

The Indian muntjac has some of the largest known
chromosomes, with one over a billion nucleotides in length

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26
Q

X-Ray diffraction photographs of DNA

A

Maurice Wilkins and Rosalind Franklin obtained X-Ray diffraction photographs of fibers of DNA. The diffraction patterns indicated that DNA is formed of two strands that wind in a regular helical structure. This data was used to create the Watson-Crick double helix model

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27
Q

Watson-Crick double helix model of DNA

A

DNA is made of two helical strands, coiled around an axis with a right handed screw sense. The strands are antiparallel. There are sugars and phosphates on the outside of the helix. The bases (purines and pyrimidines) are on the inside and are nearly perpendicular to the axis. There are 10.4 bases per turn, 3.4A base separation between adjacent bases, and the structure repeats at intervals of 34A

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28
Q

End view of the DNA double helix

A

There is a rotation of 36 degrees per base and the bases are stacked on top of one another

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29
Q

What is the diameter of the DNA double helix?

A

20A

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30
Q

Antiparallel

A

The strands of the DNA double helix are antiparallel because they have opposite directionality

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31
Q

Base pairing of DNA

A

Guanine is paired with cytosine and adenine is paired with thymine in order to form base pairs that basically have the same shape. The base pairs are held together with hydrogen bonds. Hydrogen bonds are weak but stabilize the helix because there are so many of them in the DNA helix. Due to the base pairing rules, the ratios of adenine to thymine and of guanine to cytosine are nearly the same in all species

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32
Q

Base stacking interactions of DNA

A

Inside the helix, the bases are stacked on top of one another. This contributes to the stability of the double helix because the formation of the helix is facilitated by the hydrophobic effect. Also, stacked base pairs attract one another through van der Waals forces. Base stacking in DNA is favored by conformations of the somewhat rigid 5 membered rings of the backbone sugars as well

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33
Q

How does the hydrophobic effect impact formation of the DNA double helix?

A

The hydrophobic bases cluster in the interior of the helix away from the surrounding water. Polar surfaces are exposed to water. This is similar to protein folding, where hydrophobic amino acids are found in the interior of the protein

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34
Q

van der Waals forces in DNA molecules

A

The stacked bases in DNA molecules attract one another through van der Waals forces, which stabilize the helix. There is a very small amount of energy associated with a single van der Waals interaction, but it adds up in a double helix since so many atoms are in van der Waals contact

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35
Q

B-DNA helix

A

Also known as the Watson-Crick model. Under physiological conditions, most DNA is in the B form. The B-form is longer and narrower than the A-form helix, and also forms a right handed helix

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36
Q

A-DNA helix

A

A different form of DNA that was found in X-ray diffraction studies. This form is also a right handed double helix made up of anti-parallel strands held together by Watson-Crick base pairing. The A-form helix is wider and shorter than the B-form and its base pairs are tilted rather than perpendicular to the helix axis.

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37
Q

Where is the A form of DNA found?

A

It can be found in dehydrated DNA. Double stranded regions of RNA and some RNA-DNA hybrids have a double helical form that is similar to that of A-DNA

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38
Q

What is the biochemical basis for differences between the two forms of DNA?

A

Many of the structural differences between the two forms arise from different puckerings of their ribose units. In A-DNA, C-3’ lies out of the plane formed by the other atoms in the ring (a C-3’ endo conformation). B-DNA has C-2’ endo puckering (C-2’ endo conformation), where C-2’ is out of plane. C-3’ endo puckering in A-DNA leads to an 11 degree tilting of the base pairs away from the perpendicular to the helix

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39
Q

Why does RNA tend to take the A-DNA form?

A

Because of steric hindrance from the 2’ hydroxyl group. The 2’ oxygen atom would be too close to the 3 atoms of the adjoining phosphoryl group and to one atom in the next base. In the A form, the 2’ oxygen projects outward, away from other atoms.

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40
Q

Why does dehydration favor the A form of DNA?

A

The phosphoryl and other groups in the A form helix bind fewer water molecules than do those in the B form

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41
Q

Z-DNA

A

A left handed double helix. The phosphoryl groups in the backbone are zigzagged. It is the narrowest of all DNA forms. Its function is unknown, but one Z-DNA binding protein is required for viral pathogenesis of poxviruses (like variola)

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42
Q

Shapes of DNA chromosomes

A

Linear in humans. Bacteria and archaea have circular chromosomes. Chromosomes must be compact to fit inside cells

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43
Q

Superhelix

A

When the axis of the double helix is twisted/supercoiled. This process occurs with the E. coli genome. The relaxed circular DNA and the superhelix form are
topological isomers of each other

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44
Q

Relaxed molecule

A

A circular DNA molecule without any superhelical turns

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45
Q

Why is supercoiling important? (2)

A
  1. A supercoiled DNA is more compact than its relaxed counterpart
  2. Supercoiling may hinder or favor the capacity of the double helix to unwind and affect the interactions between DNA and other molecules
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46
Q

Stem-loop

A

A common structural motif that is formed with two complementary structures in a single strand bind to form a double helix structure. Can be formed from single stranded DNA or RNA. The double helix can be made up of Watson-Crick base pairs. Sometimes, the structures include mismatched base pairs or unmatched bases that bulge out from the helix. Mismatches can destabilize the local structure but introduce deviations from the standard double helical structure that can be important from higher order folding and function

47
Q

Complex structures of RNA

A

Single stranded nucleic acid, most notably RNA, can adopt complex structures when widely separated bases interact. Hydrogen bond donors and acceptors that don’t participate in Watson-Crick base pairing can participate in hydrogen bonds to form nonstandard pairings. Metal ions like magnesium can assist in the stabilization of these elaborate structures. 3 or more bases can also interact to stabilize the structure

48
Q

Why are complex structures important for RNA?

A

The structures allow RNA to perform many functions that double stranded DNA can’t. The complexity of RNA molecules can be similar to that of proteins, allowing them to carry out functions that it was formerly thought only proteins could do.

49
Q

Semiconservative replication

A

Because of the base-pairing rules, the sequence of one
strand determines the sequence of the partner strand. Separation of a double helix into two strands during DNA replication creates two single stranded templates to make new double helices. One strand of each daughter DNA molecule is newly synthesized, while the other is unchanged from the original molecule

50
Q

Meselson and Stahl’s experiment

A

Tested the semiconservative replication hypothesis. Parent DNA was labeled with N15, a heavier isotope of nitrogen, making the molecule denser than normal DNA. The labeled DNA was created by growing in a medium that contained N15 as the only nitrogen source. After the heavy nitrogen was incorporated, the bacteria was transferred to a medium that only contained N14. Density gradient centrifugation established that upon the shift to 14N medium, newly synthesized DNA consisted of DNA with equal parts 15N-DNA and 14N-DNA, a result consistent with semiconservative replication. These results agreed perfectly with the Watson-Crick model for DNA replication

51
Q

How is the double helix disrupted in the laboratory?

A

It can be disrupted by heating a solution of DNA or by adding acid or alkali to ionize its bases.

52
Q

Density gradient centrifugation can be used

A

Density-gradient centrifugation can be used to distinguish
between DNA that contains 15N and DNA that contains
14N. When second-generation DNA was centrifuged, it produced two bands. One was in the same position as the intermediate band from the first generation, while the second was higher (appeared to be labeled only with N14), confirming the semiconservative replication hypothesis

53
Q

Melting

A

Dissociation of the double helix- it only occurs at a certain temperature.

54
Q

Melting temperature (Tm)

A

The melting temperature (Tm) of DNA is defined as the temperature at which half the helical structure is lost. Inside cells, the double helix is not lost by the addition of heat. Proteins called helicases use chemical energy (from ATP) to disrupt the helix.

55
Q

Hypochromism

A

Stacked bases in nucleic acids absorb less UV light at a wavelength of 260 nm than do unstacked bases. The absorbance of a DNA solution at a wavelength of 260 nm increases when the double helix is melted into single strands

56
Q

Annealing

A

When the temperature is lowered below Tm, complementary strands of nucleic acids spontaneously reassociate

57
Q

DNA polymerase

A

Enzymes that promote the formation of the bonds joining units of the DNA backbone- they catalyze phosphodiester bridge formation

58
Q

Characteristics of DNA synthesis (4)

A
  1. The reaction requires all 4 deoxynucleoside 5’ triphosphates- dATP, dGTP, dCTP, and TTP, as well as a Mg ion
  2. A DNA template strand is used to assemble the new DNA strand
  3. DNA polymerases require a primer to begin synthesis
  4. Many DNA polymerases are able to correct mistakes in DNA by removing mismatched nucleotides. They are able to excise incorrect bases in a separate reaction- this contributes to the low error rate of DNA replication
59
Q

Template strand

A

The preexisting DNA template strand used to make new DNA. DNA polymerases can catalyze the formation of a phosphodiester linkage efficiently only if the base on the incoming nucleoside triphosphate that is complementary to the base on the template strand. The template strand will be complementary to the original strand

60
Q

Primer

A

Used by DNA polymerases to begin DNA synthesis. A primer strand has a free 3’-OH group bound to the template strand.

61
Q

DNA chain elongation reaction

A

Catalyzed by DNA polymerases. The 3’-OH terminus of the growing strand launches a nucleophilic attack on the innermost phosphorus atom of the deoxynucleoside triphosphate. A phosphodiester bridge is formed and pyrophosphate is released. At this point, the reaction is readily reversible. The hydrolysis of pyrophosphate to yield ions of Pi is an irreversible reaction that helps drive the polymerization forward

62
Q

Coupled reactions

A

The second reaction provides the energy to drive the first reaction forward. This occurs with DNA elongation. Strand elongation is driven by Pyrophosphate Hydrolysis

63
Q

What is the direction of DNA elongation?

A

Elongation occurs in the 5’ to 3’ direction

64
Q

Activated precursors

A

dNTPs are considered to be activated precursors because although only one phosphate will be incorporated into the
backbone, the subsequent breakdown of the released pyrophosphate helps to drive the phosphodiester bond
formation

65
Q

Viruses

A

Genetic elements enclosed in protein coats that can move from one cell to another by can’t grow independently. Their genetic material can be DNA or RNA

66
Q

Tobacco mosaic virus

A

An single stranded RNA virus that infects the leaves of tobacco plants. The RNA is surrounded by a protein coat of 2130 identical subunits. An RNA directed RNA polymerase copies the viral RNA. The virus basically instructs the infected cells to commit suicide

67
Q

RNA directed RNA polymerase

A

An RNA polymerase that takes direction from an RNA template. It is responsible for replicating the genome in RNA viruses, and is exclusive to RNA viruses only. They carry out transcription as well

68
Q

Retroviruses

A

Viruses with RNA genomes- their genetic information flows from RNA to DNA rather than from DNA to RNA. Includes HIV and other tumor causing viruses

69
Q

Reverse transcriptase

A

Copies RNA into DNA in retroviruses. This enzyme acts as both a polymerase and an RNase. Retroviruses contain two copies of a single stranded RNA molecule, and reverse transcriptase makes a double helical DNA version of the viral genome. The genome can then be incorporated into the chromosomal DNA of the host and be replicated along with normal cellular DNA. It is an RNA dependent DNA polymerase since is makes DNA instead of RNA

70
Q

2 steps of gene expression

A
  1. An RNA copy is made that encodes directions for protein synthesis (mRNA)
  2. The information in mRNA is translated to synthesize functional proteins
71
Q

Types of RNA (3)

A
  1. Ribosomal RNA (rRNA)
  2. Transfer RNA (rRNA)
  3. Messenger RNA (mRNA)
72
Q

Messenger RNA (mRNA)

A

The template for translation. A distinct mRNA is produced for each gene in eukaryotes. The mRNA of all organisms has structural features, like stem-loop structures, that regulate the efficiency of translation and the lifetime of the mRNA

73
Q

Kilobase

A

A unit of length equal to 1000 base pairs of a double stranded nucleic acid molecule (or 1000 bases of a single stranded molecule). In bacteria, the average length of an mRNA molecule is about 1.2 kilobases

74
Q

Transfer RNA (tRNA)

A

Carries amino acids in an activated form to the ribosome for peptide bond formation. The amino acids are added in a sequence dictated by the mRNA template. There is at least one kind of tRNA for each of the 20 amino acids. tRNA is made up of around 75 nucleotides

75
Q

Ribosomal RNA (rRNA)

A

The major component of ribosomes. Bacteria contain 3 kinds of rRNA- 23S, 16S, and 5S RNA. One molecule of each of these species of rRNA is present in each ribosome. It is the catalyst for protein synthesis and is the most abundant of these 3 types of RNA

76
Q

RNA polymerase

A

Catalyst of transcription- the initiation and elongation of RNA chains. It does not require a primer and can’t correct mistakes as well as DNA polymerase can.

77
Q

Components required by RNA polymerase (3)

A
  1. A template- double stranded DNA is preferred, single stranded DNA can be used, RNA can’t be used
  2. Activated precursors- all 4 ribonucleoside triphosphates
  3. A divalent metal ion, like Mg or Mn
78
Q

Direction of synthesis of RNA

A

5’ to 3’, same as DNA

79
Q

Mechanism of elongation of RNA

A

The 3’-OH group at the terminus of the elongating chain makes a nucleophilic attack on the innermost phosphoryl group of the incoming nucleoside triphosphate.

80
Q

Synthesis of RNA is driven by

A

The hydrolysis of pyrophosphate

81
Q

Evidence that RNA is synthesized from DNA

A

The base composition of newly synthesized RNA is the complement of that of the DNA template strand. There is also evidence from base sequence studies. The nucleotide sequence of a segment of the gene encoding the enzymes required for tryptophan synthesis was determine using DNA sequencing techniques

82
Q

Promoter sites

A

Regions of DNA templates that specifically bind RNA polymerase and determine where transcription begins

83
Q

TATAAT sequence

A

The TATAAT sequence in the Pribnow box- a promoter site in bacteria. Bacteria have two sequences in the 5’ side of the first nucleotide to be transcribed that function as promoter sites

84
Q

TATA box (Hogness box)

A

A promoter site found in eukaryotic genes. Eukaryotic genes also have a CAAT box, and transcription is further stimulated by enhancer sequences

85
Q

Consensus sequence

A

There are often variations in the sequence of a promoter for different genes. The average of such variation is called the consensus sequence. The consensus sequence is the idealized sequence since not all base sequences of promoter sites are identical

86
Q

Prokaryotic promoter sites (2)

A
  1. -35 region- TTGACA
  2. -10 region- TATAAT (Pribnow box)
    RNA starts after this, at the +1 region
87
Q

Eukaryotic promoter sites (2)

A
  1. -75 region- CAAT box (sometimes present)
  2. -25 region- TATA box (Hogness box)
    RNA starts after this, at the +1 region
88
Q

Hairpin structure of the mRNA transcript

A

RNA polymerase proceeds along the DNA template, continuing transcription, until it synthesizes a terminator sequence. A base paired hairpin on the newly synthesized RNA molecule encodes a termination signal to stop transcription. The hairpin is formed by base pairing of self complementary sequences that are rich in G and C. New RNA spontaneously dissociates from RNA polymerase when this hairpin is followed by a string of U residues, and the DNA helix reforms. In some cases, the protein rho is used for termination instead

89
Q

Where are the start and stop signals for transcription encoded?

A

In the DNA template. The simplest stop signal is the transcribed product of a segment of palindromic DNA

90
Q

mRNA modifications

A

In eukaryotes, mRNA is modified after transcription. A cap structure is added to the 5’ end- it is a guanosine nucleotide attached to the mRNA with an unusual 5’ triphosphate linkage. A poly A tail (sequence of adenylates) is added to the 3’ end

91
Q

tRNA structure

A

tRNAs contain an amino acid attachment site and a template recognition site (an anticodon). The amino acid attachment site is a CCA arm- an adenylate preceded by two cytidylates. The anticodon is on the opposite site of the tRNA from the CCA arm. The tRNA molecule has a cloverleaf structure in general, with many hydrogen bonds between bases

92
Q

tRNA function in translation

A

A tRNA molecule carries a specific amino acid in an activated form to the ribosome. The carboxyl group of this amino acid is esterified to the 3’ or the 2’ hydroxyl group of the ribose unit of an adenylate at the 3’ end of the tRNA molecule

93
Q

Aminoacyl-tRNA

A

Catalyzes the joining of an amino acid to a tRNA molecule to form an aminoacyl-tRNA. This is an esterification reaction driven by ATP cleavage

94
Q

Anticodon

A

A sequence of 3 bases that acts as the template recognition site on tRNA. The anticodon recognizes the complementary codon on mRNA

95
Q

Codon

A

A 3 base sequence on mRNA that is complementary to the anticodon, used for translation. One codon corresponds to one amino acid in the protein sequence

96
Q

Genetic code

A

The relation between the sequences of bases in DNA and the sequence of amino acids in proteins

97
Q

Features of the genetic code (5)

A
  1. 3 nucleotides (a codon) encode an amino acid
  2. The code is nonoverlapping- 3 nucleotides correspond to one amino acid, the next 3 are totally separate and encode a different amino acid
  3. The code has no punctuation. The sequence of bases is used sequentially from a fixed starting point
  4. The code has directionality (5’ to 3’)
  5. The genetic code is degenerate- most amino acids are encoded by more than one codon. There are 64 possible base triplets but only 20 amino acids
98
Q

Which direction is mRNA read during translation?

A

5’ to 3’

99
Q

Synonyms

A

Codons that specify the same amino acids. CAU and CAC are synonyms for histidine

100
Q

What is the biological significance of the extensive degeneracy of the genetic code?

A

If the code were not degenerate, 20 codons would designate amino acids and 44 would lead to chain termination. The probability of mutating to chain termination would therefore be much higher with a nondegenerate code. Chain termination mutations usually lead to inactive proteins. However, substitutions of one amino acid for another are typically harmless. The code is constructed such that a change in any single nucleotide base of a codon results in a synonym or an amino acid with similar chemical properties

101
Q

Ribosomes

A

Large molecular complexes assembled from proteins and ribosomal RNA. This is where mRNA is translated.

102
Q

fMet-tRNA

A

fMET is a modified amino acid found at the beginning of polypeptide chains in bacteria. The initiator tRNA carries fMet, and this fMet-tRNA recognizes the start codon AUG. The Shine-Dalgarno sequence identifies the actual start codon, since AUG also codes for methionine

103
Q

Shine-Dalgarno sequence

A

A purine rich sequence in bacteria that precedes the initiating AUG codon. It base pairs with a complementary sequence in an rRNA molecule

104
Q

Start codon in eukaryotes

A

The AUG closest to the 5’ end of an mRNA molecule acts as the start signal for protein synthesis in eukaryotes. It establishes the reading frame for translation

105
Q

Stop codons (3)

A

UAA, UAG, and UGA designate chain termination. These codons are read by specific proteins called release factors instead of by tRNA molecules. Binding of a release factor to the ribosome releases the newly synthesized protein

106
Q

Universality of the genetic code

A

Most organisms use the same genetic code. This universality accounts for the fact that human proteins, like insulin, can be synthesized in the bacterium E. coli and harvested for the treatment of diabetes. However, genome sequencing studies have shown that not all genomes are translated by the same code. Ciliated protozoa differ from most organisms in that UAA and UAG are read as codons for amino acids rather than stop signals. UGA is the only termination signal. Mitochondria also use variations of the genetic
code, so the genetic code isn’t completely universal

107
Q

Why has the genetic code remained nearly invariant throughout evolution?

A

A mutation that altered the reading of mRNA would change the amino acid sequence of most proteins synthesized by that particular organism. Many of these changes would probably be harmful, so there would be selection against this type of mutation

108
Q

Introns

A

Noncoding regions of a gene. Introns were initially detected by electron microscopy studies. The average human gene has 8 introns, while some have more than 100. Intron sizes range from 50-10,000 nucleotides

109
Q

Exons

A

Coding regions of a gene that are expressed. Because eukaryotic genes are interrupted by noncoding sequences, they can be described as discontinuous

110
Q

Detection of introns by electron microscopy

A

An mRNA molecule was hybridized to genomic DNA containing the corresponding gene in these studies. A single loop of single stranded DNA is seen if the gene is continuous. Two loops of single stranded DNA and a loop of double stranded DNA are seen if the gene contains an intron. There are more loops if more than one intron is present

111
Q

Spliceosomes

A

Assemblies of proteins and small nuclear RNA molecules. RNA plays a catalytic role. Spliceosomes recognize signals in the nascent RNA that specify splice sites. Introns nearly always begin with GU and end with an AG that is preceded by a pyrimidine-rich tract. This is a consensus sequence that acts as the signal for splicing

112
Q

Splicing

A

The removal of introns from the primary transcript of newly synthesized RNA. Splicing creates mature RNA

113
Q

Alternative splicing

A

When cells splice their primary transcript in an alternative manner to form a set of proteins that are variations of a basic motif without requiring a gene for each protein. It generates mRNAs that are templates for different forms of a protein. Because of alternative splicing, the proteome is more diverse than the genome in eukaryotes

114
Q

Exon shuffling

A

The idea that new proteins arose in evolution by the rearrangement of exons encoding discrete structural elements, binding sites, and catalytic sites. It is a rapid and efficient way of generating novel genes because it preserves functional units but allows them to interact in new ways.

115
Q

Generation of the tissue plasminogen activator (TPA) gene

A

Generated by exon shuffling. The gene for TPA encodes an enzyme that functions in hemostasis. This gene consists of 4 exons. One is derived from the fibronectin gene that encodes an extracellular matrix protein, one from the epidermal growth factor gene, and two from the plasminogen gene- the substrate of the TPA protein

116
Q

Multiple myeloma

A

A cancer that forms in a plasma cell (a type of WBC that makes antibodies), causing overproduction of the IgG antibody. Cancerous plasma cells then build up in the bone marrow. RBCs become stacked like poker chips, and the patient is at high risk for infection. Symptoms- hypercalcemia, anemia, bone lytic lesions (punched out holes on X-Ray)

117
Q

Helicase

A

DNA helicases are essential during DNA replication because they separate double-stranded DNA into single strands, allowing each strand to be copied

Biochemistry 2 (12 decks)

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