Concept 13, 14 Flashcards

Question

How is RNA processed in eukaryotes? - 5’ end of molecule

Answer 1

- 5’ end of molecule BECOMES…5’ cap - A modified G nucleotide is added to the 5’ end via a triphosphate linkage

Answer 2

3’ end of molecule BECOMES…3’ tail - CLEAVAGE = A polyadenylation signal (e.g., AAUAAA) in the pre-mRNA 3’ end triggers cleavage of around 10–30 nucleotides downstream. - ADDITION = THEN, 150–250 adenine (A) nucleotides added to the 3′ end.

Answer 3

5’ cap helps SMALL SUBUNIT of ribosome attach to the mRNA

Answer 4

- Protects ends of transcript from degradation - 3’ tail signals that mRNA is ready to leave the nucleus

Answer 5

- If 1 nucleotide codes for amino acids = there are 4 bases = 4 possible amino acids - If 2 nucleotides code for amino acids = 16 (4^2) possible amino acids - If 3 nucleotide code for amino acids = 64 (4^3) possible amino acids 3 nucleotides which code for 1 amino acid = CODON

Answer 6

1) Degeneracy / Redundancy 2) Start codon 3) Stop codons 4) Universality

Answer 7

- Codons SHOULD mean 64 amino acids - But we only have 20 amino acids BECAUSE… - Multiple codons can code for THE SAME amino acid Eg. leucine has six codons

Answer 8

- Often occurs in the third position of the codon. Eg. UAA, UAC codes for the same amino acid

Answer 9

The genetic code is nearly universal across all life domains.

Answer 10

- ONE START CODON = AUG Indicates where the open reading frame for translation begins.

Answer 11

THREE STOP CODONS = UAA, UAG, UGA - Signal the end of translation.

Answer 12

1) 5’ to 3’ LEFT TO RIGHT 2) Loop in the middle = anticodon (3 nucleotides) 3) 3’ end = amino acid binding site ALWAYS CCA

Answer 13

tRNA synthetase

Answer 14

CHARGED when an amino acid is attached to the tRNA

Answer 15

- Amino acid binding sites are the same…how about codon/anticodon? - YOU CANNOT infer the amino acid bound to the trna based on its anticodon/codon! No relationship - The amino acid is determined by smth else - So instead the mRNA sequence, which impacts the anticodon, and therefore the type of tRNA and what amino acid it happens to carry!

Answer 16

Transcription → from template DNA strand, used to code for proteins

Answer 17

Translation → carries the amino acids

Answer 18

A component of ribosomes

Answer 19

Eg. UAG and UAA code for the same protein 1) Mutation Buffering - If one base is changed, the amino acid coded for could still be the same! - Mutations likely do not alter the amino acid sequence, reducing harmful effects/loss of function 2) Efficient Translation - Allows multiple tRNAs with different anticodons to bind the same codon, speeding up protein synthesis

Answer 20

Thinking in the context of TRANSLATION - Region of DNA or mRNA that can be TRANSLATED into a protein - this is therefore read by RNA POLYMERASE / ribosome

Answer 21

1) Start codon: Begins with AUG 2) Stop codon: Ends with UAA, UAG, or UGA 3) Length: The sequence between start and stop codons is a multiple of three (triplets = codons). / Length must be divisible by three

Answer 22

RIBOSOMES, during TRANSLATION 1) Bind to the 5′ end of mRNA 2) Scan the mRNA in the 5′→3′ direction for the start codon (AUG). 3) Decode codons (triplets) into amino acids using tRNAs. 4) Stops at the stop codon

Answer 23

Each strand can be read in three possible ways - Rmb AUG is the start codon!! - How it is read depends on whether the coding sequence is read from the first, second, or third nucleotide. 1) Frame 1: Start at 1st nucleotide AUG CCG AUG UAG → Met-Pro-Met-Stop (valid, will produce functional protein). 2) Frame 2: Start at 2nd nucleotide UGU CGA UGU → Cys-Arg-Cys (no stop codon → invalid protein). 3) Frame 3: Start at 3rd nucleotide GCC GAU GUA → Ala-Asp-Val (no stop codon → invalid protein). **only the one starting with ATG will be translated. so...these are reading frames , but the REAL potential reading frames (esp when exam qs ask), is based on where the start codon is found

Answer 24

- Look at mRNA!!!! (not dna) - Look for start codon (AUG)

Answer 25

1) We know rna polymerase works in the 5’ to 3’ direction, hence template must be 3’ to 5’ (bottom) 2) We know mRNA must be 5’ to 3’, hence top is coding/complementary

Answer 26

SITUATION 1 - No stop codon → ribosomes translate the entire mRNA (GOES TO THE VERY END) → doesn’t stop until the very end of mRNA → long, useless protein. SITUATION 2 - Introduce a new stop codon → Translation stops early → creates a truncated prote

Answer 27

- ORF = what can be TRANSLATED into a protein - Protein is made using mRNA - Rmb mRNA is read by ribosomes in the 5’ to 3’ direction!!! (translation) BUT… - There are TWO strands of DNA - Rmb EITHER one of them can be used for TRANSLATION / TEMPLATE - It is only ONCE you decide which strand you use, then you will set the strands as template or coding Just make sure to read it in the 5’ to 3’ direction!!! AND… - Reading frames always start from a START CODON (in this case it is ATG) SO… - Read top strand from 5’ to 3’ (LEFT TO RIGHT) - Top = template - Bottom = coding - Read bottom strand from 5’ to 3’ (RIGHT TO LEFT)

Answer 28

- A large subunit - A small subunit Both subunits are made of: - rRNA (catalyzes the peptide bond formation between amino acids.) - Ribosomal proteins

Answer 29

FROM LEFT TO RIGHT E site: Exit site P site: Peptidyl site A site: Aminoacyl site Each site is a binding site for tRNA

Answer 30

YES!!! - mRNA is read by the ribosome in the 5'→3' direction - absolutely matters because this ensures proper protein synthesis.

Answer 31

SMALL subunit of the ribosome has to bind to something in the mRNA - Euk = 5’ cap - Prok = Shine-Dalgarno sequence - Once binded, it will move along the mRNA to find AUG (start codon) - Once it finds AUG, the LARGE SUBUNIT will bind to the small subunit. A complete ribosome will be formed!

Answer 32

1) Initiation - Small subunit 2) Elongation - Large subunit - Moving 3) Termination

Answer 33

SMALL SUBUNIT 1) Small subunit binds to 5’ cap 2) Small subunit will move along mRNA until it finds AUG (start codon) 3) Once on AUG, the large subunit will bind to the small subunit

Answer 34

LARGE SUBUNIT 1) The first tRNA molecule will bind to P site (tRNA anticodon will be complementary to the codon AUG. Carry Met amino acid) 2) Next tRNA molecule binds to the A site 3) Bond forms between amino acids on P and A site 4) Amino acid chain will sit on the tRNA on the A site MOVING 5) Ribosome moves from left to right 6) So first tRNA molecule sits on E site, second tRNA molecule sits on P site 7) tRNA will leave. Above there is 2 beads in the amino acid chain 8) New tRNA molecule will enter the A site

Answer 35

- Ribosomes reaches a STOP CODON - BUT…there is no tRNA that recognizes/binds to a stop codon. - Protein called RELEASE FACTOR enters the A site - No peptide bond is formed - So when the previous tRNA moves from the P site to the E site, the entire peptide is released

Answer 36

- Initiator tRNA will bind to AUG codon - This tRNA will be charged with methionine

Answer 37

Large subunit (60S ribosome) Small subunit (40S ribosome):

Answer 38

NO - This is REGULATED - Only expressed when protein is needed - Time of development, environmental conditions, cell type

Answer 39

- Transcription activated only at certain times - mRNA processing can be impacted at different times - mRNA produced can be stable, or degraded over time - Translation efficiency may be changed, based on whether the protein is needed or not

Answer 40

1) Beneficial: Provides advantages for selection. 2) Neutral: No detectable change in phenotype. 3) Deleterious: alters cell function (of the protein) Most mutations don't show obvious phenotypes.

Answer 41

1) Spontaneous (de novo) = Occur with no outside influence. 2) Induced mutations

Answer 42

alterations in DNA structure caused by internal cellular processes rather than external environmental factors.

Answer 43

1) Deamination 2) Transposons 3) Depurination (hydrolysis)

Answer 44

- Removal of an amino group (NH2) from a nucleotide. - Cytosine becomes uracil. - overtime will become a C-->T mutation

Answer 45

- Hydrolysis of nucleotide - Specifically cutting the bond between PENTYL SUGAR + DEOXYRIBOSE

Answer 46

- GENES that code for enzymes needed for movement. - A transposon will COPY itself - Then it will PASTE itself into a new POSITION of the genome.

Answer 47

Protein-coding sequence could be interrupted If it lands within a gene, it can cause functional problems.

Answer 48

Caused by external agents (MUTAGENS ). 1) Nitrous acid: Deamination 2) Cigarette smoke (benzopyrene) 3) Radiation (X-rays, gamma rays): 4) Ultraviolet radiation (sun)

Answer 49

1) NITROUS ACID = deamination 2) BENZOPYRENE = add chemical group to guanine 3) X-RAY RADIATION = break DNA’s sugar phosphate backbone 4) UV = impact covalent bonds between adjacent thymine bases

Answer 50

GERMLINE - Occurs in germline cells (ie. cells that give rise to gametes) - NOT the gamete itself - Passed to OFFSPRING. - Every single cell in offspring has the mutation SOMATIC - Occurs in somatic cells - Passed to daughter cells in MITOSIS only - NOT present in all cells of an organism

Answer 51

genome instability, cell death, or cancers.

Answer 52

1) Base Excision Repair - replacing wrong bases w/ the right ones 2) Nucleotide Excision Repair - cut out big chunks of wrong dna, replace it with correct section 3) Mismatch Repair (mismatch caused by DNA polymerase)

Answer 53

DNA is double stranded, and the complementary strand can be used as a template for repair.

Answer 54

- Has 2 strands - Complementary strand + template strand - So contains 2 copies of genetic info - So if one of the strands are damaged, complementary strand can be used as a guide for restoration - INCREASES the ACCURACY of dna repair mechanisms

Answer 55

= a genetic alteration that disrupts gene function, leading to harmful effects on an organism's fitness, survival, or health. These mutations can: Disrupt Protein Function: - Cause loss of protein production (e.g., null mutations). - Generate non-functional proteins (e.g., misfolded enzymes). - Create proteins that interfere with normal cellular processes (e.g., dominant-negative mutations).Germline vs. Somatic Mutations

Answer 56

NO!!!! READING FRAME - All possible ways to SPLIT DNA INTO CODONS - The "reading frame" is how you split it into 3-letter codons (e.g., ATG-CCG-TAA vs. TGC-CGT-AAT). OPEN READING FRAME - Section of DNA that CAN code for a protein/be translated

Answer 57

1) Go in 5' to 3' direction (mRNA is in this direction as this is how rna polymerase works) 2) Look for START CODON (ATG in dna, AUG in mRNA)

Answer 58

- SILENT - NO CHANGE in the AMINO ACID of the product - Amino acid sequence is still the same Eg. AUG and ACG will code for the same amino acid

Answer 59

- Change in the AMINO ACID of the product - BUT this shouldn’t change the entire POLYPEPTIDE - Hence nonsense mutations (truncated proteins) is NOT considered a non-synonymous mutation, as it changes the ENTIRE protein, stuffs it up, not just a few amino acids

Answer 60

- Change in ONE SINGLE NUCLEOTIDE (not amino acid like non-synonymous) - MUST be a SUBSTITUTION - Amino acid sequence changes by ONE amino acid Eg. Gln, Pro Ala → Gln, Arg, Ala

Answer 61

1) Protein GAINS a new function 2) Protein has a LOSS of function

Answer 62

- Changes a functional codon to a STOP codon IMPACT - truncates the protein (shorter) → non-functional protein

Answer 63

- Insert/delete nucleotides that are NOT divisible by 3 - This should cause the amino acid sequence to COMPLETELY change - Because…insertion/deletion changes your READING FRAME (ie. how you SPLIT your sequence into codons/3 bases)

Answer 64

YES!! The reading frame (ie. HOW you read the triplets) would TOTALLY change

Answer 65

YES!! - It just totally depends on HOW the nonsense mutation came about (ie. how was the stop codon introduced) - SUBSTITUTION - then the reading frame wouldn’t have changed OR - INSERTION - although at the insertion point a stop codon was introduced and the protein is truncated, if you look AFTER the mutation point, you will find that the reading frame completely shifts!

Answer 66

ANS = INSERTION OF 6 NUCLEOTIDES - Genetic code read in multiples of 3s (ie. codons) - Eg. insert 6 nucleotides. You will have 2 more amino acids. - BUT the rest of the amino acid sequence will remain the same - ie. READING FRAME will stay the same - So things like removing 9, adding 6 etc. will less likely be a frameshift mutation - If you are adding/removing something not of a multiple of 3, then you would COMPLETELY change the entire order of amino acid sequence - The other 3 - after the mutation, the reading frame would totally change (IE. THE CODONS)

Answer 67

- Talking about the ENTIRE amino acid sequence - No more ATG (which creates the START CODON) - NOT a mutation that results in a stop codon in the middle of the sequence, as there is still one/few amino acids before it

Answer 68

This means we would have never evolved!

Answer 69

- Nonsynonymous is the change in AMINO ACIDS - not a whole polypeptide (which is what frameshift mutation does)

Answer 70

- Any SINGLE nucleotide change - Could be substitutions, insertions, deletion (indel)

Answer 71

1) SMALL MUTATIONS - point mutation - insertion - deletion 2) LARGE MUTATIONS

Answer 72

1) GENE DUPLICATION - an entire gene is copy + pasted (so you have 2x of the same gene) 2) INVERSION - change the order of a whole gene so its backwards 3) GENOME DUPLICATION - double the no. of chromosomes in an individual

Answer 73

- This is found in each CELL - Each cell (except red blood cells and gametes) contains a full copy of this genome

Answer 74

1) SNP - Difference in a SINGLE base 2) CNV - Difference in whole sections of DNA

Answer 75

- A SNP is when ONE base (A,T,C,G) are different between individuals in a population - Caused by SINGLE BASE SUBSTITUTIONS -> Eg. C in dna sequence may be T in somebody else - BUT To be classified as a SNP, the variant must occur in ≥1% of the population -> Eg. out of 100 people, everyone mostly have a C, but more than 1 person will have a T

Answer 76

- Your DNA has extra or is missing certain genes. - Within the population, you can increase/decrease the “copy number” of a gene -> Eg. Some individuals may have multiple copies of a gene or none at all Why It Matters: - Extra Copies: Might make you better at something (e.g., digesting starch). - Missing Copies: Could cause problems (e.g., disease). Example:- Some people have extra copies of a "starch-digesting gene" because their ancestors ate lots of potatoes/rice, while others don’t

Answer 77

Ie. 2 alleles - Carry 2 copies of every gene

Answer 78

- Alleles: Different versions of the SAME gene - They are basically MUTATIONS of each other

Answer 79

1) Homozygous: When the two alleles are the same (BB or bb). 2) Heterozygous: When the two alleles are different (Bb). 3) Hemizygous: Organism has only one allele

Answer 80

- Sex chromosomes vs autosomes -> Eg. female is XX, carry diff versions of the SAME X gene, alleles could be homozygous OR heterozygous -> Eg. male is XY, the genes are totally different, therefore only have one allele for the X chromosome gene

Answer 81

Homozygous for the trait of interest (coat color).Homozygous for the trait of interest (coat color)

Answer 82

Through GENETIC CROSSES EXAMPLE = Crossing a pure-breeding black-haired dog (BB) with a pure-breeding brown-haired dog (bb). - F1 Generation: All pups are black-haired (Bb). This indicates that the black coat trait is dominant to the brown coat trait, which is recessive.

Answer 83

1) DOMINANT + RECESSIVE 🐶 = black, brown dogs 2) INCOMPLETE DOMINANCE 🌸 = red, white, pink flowers 3) CODOMINANCE 🐔 = black + white feathers

Answer 84

DOG EXAMPLE… - Dominant allele → codes for a enzyme that results in black pigment - Recessive allele → MUTATED dominant gene. Changes function of enzyme, so black pigment can no longer be produced PHENOTYPE - Dominant + recessive allele → black pigment overrides the brown pigment → because black pigment is produced in LARGER QUANTITIES

Answer 85

FLOWER EXAMPLE - Red allele (R) → red flower - White allele (W) → white flower PHENOTYPE - RR → red flower → high level of pigment → 2 red allele doses - WW → white flower → no level of pigment → no red allele dose - RW → pink flower → intermediate pigment → 1 red allele dose - We call this a DOSAGE EFFECT

Answer 86

- When the heterozygote displays an INTERMEDIATE PHENOTYPE between the two homozygous phenotypes. - The phenotype will be a blend of both parents. - Neither allele is dominant or recessive.

Answer 87

Wjhen both traits are expressed in the heterozygote. The heterozygote's phenotype shows both traits evident in homozygous individuals.

Answer 88

CHICKEN EXAMPLE - Black allele → black feathers - White allele → white feathers PHENOTYPE - BW → black AND white feathers (not grey) - BECAUSE…Black and white pigments are DISTINCT, SEPARATE, produced in different cells, and also not in equal quantities

Answer 89

ABO blood groups in humans. You will be AB blood type You won’t be a mixture of A+B

Concept 13, 14 Flashcards

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