Concept 13,14,15,16,17,18 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 - prevents mRNA from getting degraded

Answer 4

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

Answer 5

- mRNA products = referring to what we get AFTER processing pre-mRNA - longer mRNA products means that there must be an issue in PROCESSING - this is because processing = removal of introns + 5' cap + addition of polyA tail - so the protein must have been responsible for mRNA processing

Answer 6

- we know there must be an issue with the processing of pre-mRNA - less stable points that the protein must have been involved in the addition of 5' cap OR polyA tail, as both are involved in stability - BUT since it can't be translated, this proves that the protein must be involved in adding the 5' cap - as this is required for mRNA to bind to the ribosomes

Answer 7

- 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 8

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

Answer 9

- 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 10

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

Answer 11

The genetic code is nearly universal across all life domains.

Answer 12

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

Answer 13

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

Answer 14

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 15

tRNA synthetase

Answer 16

CHARGED when an amino acid is attached to the tRNA

Answer 17

- 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 18

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

Answer 19

Translation → carries the amino acids

Answer 20

A component of ribosomes

Answer 21

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 22

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 23

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 24

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 25

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 26

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

Answer 27

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 28

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 29

- 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 30

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

Answer 31

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 32

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

Answer 33

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 34

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

Answer 35

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 36

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 37

- 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 38

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

Answer 39

KEY POINTS TO CONSIDER 1) tRNA is produced by transcription of the tRNA gene. done by RNA polymerase III 2) tRNA is charged with an amino acid. occurs on the 3' acceptor sequence 3) tRNA binds to the A site (only the 1st time the initiator tRNA binds to the P site) 4) it binds as its anticodon is COMPLEMENTARY to the CODON of the mRNA!!! 5) amino acid is added to the carboxyl end of the growing polypeptide chain via a peptide bond on the 3' end (A site) 6) tRNA moved down to the P site 7) tRNA exits the ribosome, recycled and charged with the same amino acid

Answer 40

EXPLAIN MEANS STATE THE OBVIOUS, IMPACTS, CAUSES ETC. - DETAILED 1) STATE THE OBVIOUS = translation's termination will be affected 2) EXPLAIN = the ribosome will continue translating until it reaches the end of the mRNA 3) IMPACTS = longer polypeptide will be non0functional -> wastes cellular resources

Answer 41

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

Answer 42

- these codons (3 amino acids) show the AMINO ACID that will be produced. - these codons are on mRNA!!!!! - These codons have to be read in the 5' to 3' direction

Answer 43

NO!! - the amino acid binds to tRNA according to its shape, NOT based on its anticodon

Answer 44

UNDERSTANDING: - understand that in the amino acid sequence, Ser should be there, but Tyr is now there - trna holding Tyr is mutated - mutation in anticodon does NOT mean a diff amino acid carried. what it DOES mean is that the anticodon will bind to a different CODON on the mRNA sequence!!! SO... - tRNA charged with Tyr has an anticodon that has mutated. most likely SINGLE POINT MUTATION - this anticodon will now bind to a CODON on the mRNA that is complimentary to the codon CODING for serine, not tyr - this explains why Tyr replaces Ser GETTING THE EXTRA MARKS: - figuring out what the new anticodon for Tyr could be! - mRNA codons that code for serine (get from table) = UCU, UCC, UCA, UCG, AGU, AGC (5' to 3') - so....new anticodon will bind to 5' to 3' UCU. hence mutated anticodon should be 3' to 5' AGA -> so original anticodon must be AUA - new anticodon will bind to 5' to 3' UCC. Hence mutated anticodon will be 3' to 5' AGG -> so original anticodon must be AUG, underwent single point mutation

Answer 45

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

Answer 46

- 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 47

- increased tendency for DNA mutations and other genetic changes (eg. insertions, deletions , or changes in chromosome number)

Answer 48

It happens when these systems are not working properly: - repair DNA - check for errors - control cell division ALL THIS DUE TO DNA DAMAGE --> MEANS DNA CANNOT BE REPAIRED/REPLICATED ACCURATELY

Answer 49

1) cell degenerates/dies 2) cell loses its normal function 3) cell develops cancer --> uncontrolled cel growth

Answer 50

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 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

- 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 57

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

Answer 58

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

Answer 59

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 60

1) WHICH CELLS DOES IT OCCUR? - SOMATIC = Occurs in cells that undergo mitosis - GERMLINE = Occurs in specialised reproductive cells that give rise to gametes 2) IMPACT? - SOMATIC = LOW IMPACT = Mutation only passed to daughter cells produced via mitosis -> mutation only present in the organism in which they occur - GERMLINE = WIDESPREAD IMPACT = Mutation passed to gametes, which can be passed to the next generation

Answer 61

- mutations can have little to no effect on the organism - mutations can have positive or negative effects

Answer 62

SMALL - mutations affect either one or a few nucleotides LARGE - mutations impact chromosome regions or entire chromosome structures

Answer 63

- substitutions - insertions deletions

Answer 64

- depends on if mutation occurs in important regulatory genes (eg. promoters) - depends if they change the polypeptide sequence of protein coding genes

Answer 65

- more likely to have a negative effect on the organism - affects more genes due to size involved + result in chromosomal imbalance

Answer 66

genome instability, cell death, or cancers.

Answer 67

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 68

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

Answer 69

- 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 70

= 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 71

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 72

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 73

- 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 74

- 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 75

- 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 76

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

Answer 77

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

Answer 78

- 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 79

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

Answer 80

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 81

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 82

- 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 83

This means we would have never evolved!

Answer 84

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

Answer 85

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

Answer 86

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

Answer 87

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 88

FIRST QUESTION - 1 amino acid = 3 nucleotides - so..192 amino acids --> 192 x 3 = 576 nucleotides + STOP codon = 579 nucleotides/bases ON THE MRNA - now finding this on the DNA - but since they ask base pairs no need to multiply by 2

Answer 89

GENE = BEFORE TRANSCRIPTION + TRANSLATION, MAKE SURE TO TRACK THE WHOLE PROGRESS FROM DNA -> MRNA 1) TRANSCRIPTION - some part of the mRNA will be removed as they are introns 2) TRANSLATION - some parts of the mRNA will not be translated (as it comes before/after the START/STOP codon)

Answer 90

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

Answer 91

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

Answer 92

- A SNP is when ONE base (A,T,C,G) are different between individuals in a population OR within cells in a single organism - 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 / cells in an organism -> Eg. out of 100 people, everyone mostly have a C, but more than 1 person will have a T

Answer 93

- 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 94

Ie. 2 alleles - Carry 2 copies of every gene

Answer 95

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

Answer 96

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 97

- 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 98

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

Answer 99

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 100

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

Answer 101

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 102

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 - ie. because NEITHER allele is dominant

Answer 103

- 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 104

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

Answer 105

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 106

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

Answer 107

Mature form = original form of the organism HUMANS/PLANTS = mature form is diploid. Next generation is haploid FUNGI = mature form is haploid. Diploid is a brief state

Answer 108

Cannot be sister chromatids They are basically just replicated versions of each other!

Answer 109

2N Diploid Means there are 2 alleles (Eg. Aa) This means there are 2 SETS of CHROMOSOMES in total in a CELL =2? Means there is a total of 2 CHROMOSOMES in a cell (ie. 1 homologous pair) For example, humans would be 2n = 46

Answer 110

Mitosis = somatic cells (all non-reproductive cells) Meiosis = germline cells (reproductive cells)

Answer 111

1) Cell growth (of the cell itself!) 2) Tissue repair 3) Cell proliferation = increase in no. of cells

Answer 112

- Producing gametes Used in sexual reproduction

Answer 113

BLENDING MODEL THEORY - Offspring are a blend of the parents phenotype Eg. red flower + white flower = pink flower offspring - REAL LIFE - Eg. continuous variables, height over time in families

Answer 114

PROVED TO BE FALSE!!! - Disproved by Mendel pea experiment - only found 2 phenotypes Eg. bred yellow + green plants together BASICALLY DID A MONOHYBRID CROSS, THE 2 STEPS

Answer 115

MONOHYBRID CROSS OBSERVATIONS: 1) Noticed one phenotype disappeared in the F1 generation 2) 3:1 green to yellow in F2 generation

Answer 116

BASIC QUALITIES OF ALLELES 1) Each individual has 2 alleles for each gene 2) Each allele comes from each parent 3) These alleles separate so each gamete has 1 allele - So that when gametes fuse, we have 2 alleles again! RANDOM SEGREGATION 4) As random segregation of alleles, parents are EQUALLY LIKELY to pass on a dominant/recessive allele

Answer 117

When both parents are homozygous + are told all offspring are black - Therefore all offspring must be Bb

Answer 118

Remember each gamete will have ONE ALLELE So 4 gametes all will have a // Remember each gamete will have ONE ALLELE 2 gamete cells will have A (2xA) 2 gamete cells will have a (2xa)

Answer 119

1) ENVIRONMENT 2) DIPLOID OR HAPLOID 3) XY SYSTEM 4) RATIO OF X CHROMOSOMES TO AUTOSOMES 5) ZW SYSTEM

Answer 120

1) ENVIRONMENT Eg. higher temps = male 2) DIPLOID OR HAPLOID Diploid = female, haploid = male 3) XY SYSTEM XX in humans = female XY in humans = male 4) RATIO OF X CHROMOSOMES TO AUTOSOMES Eg. for every 2 X chromosomes = 1 autosome 5) ZW SYSTEM ZZ = male ZW = female

Answer 121

Some genes related to gender BUT NOTE: Can also carry genes for physical traits, inherited diseases etc.

Answer 122

Egg cell = 1 X chromosome Sperm cell = either ONE X or Y chromosome (meiosis, parent cell = XY) So…fertilisation determines gender

Answer 123

WORKS IN THE SAME WAY AS HUMANS - GAMETES WOULD HAVE 1 OF THE 2 ALLELES // BETWEEN 4 DAUGHTER CELLS So… 2 gametes = X, 2 gametes = O

Answer 124

Pure breeding Cross 2 individuals which are homozygous (F1) F2 = all heterozygous individuals

Answer 125

CROSS OFFSPRING IN THE F1 GENERATION (IE. BOTH PARENTS ARE HETEROZYGOUS) Complete dominance = 3:1 ratio in offspring Incomplete dominance = 1:2:1 ratio in offspring

Answer 126

Used to determine if an individual of dominant phenotype is heterozygous or homozygous (Bb or BB) This individual will cross with one of their parents (homozygous) BB or bb Recessive homozygous is good This is because recessive homozygous parent will always give a b allele. So this will DIRECTLY reveal what the parent is giving, B or b, which tells us if it is homozygous or heterozygous If dominant homozygous, we can’t directly tell If complete dominance, 1:1 ratio of offspring If incomplete dominance, 1:2:1 ratio of offspring

Answer 127

Allele that causes death of the organism Contains a gene that is essential for survival

Answer 128

- Homozygous and heterozygous both die - Therefore never identified in populations as the offspring die immediately

Answer 129

Huntington’s disease - Dominant lethal allele always present since birth, BUT DO NOT DIE IN THE WOMB - BUT only express/show disease past the age of 40 - So will pass the lethal allele to their offspring + be transmitted

Answer 130

Homozygous = result in death Heterozygous = does not result in death

Answer 131

Agouti - Gene in mice - Will determine coat colour - BUT certain mutations of alleles will be LETHAL A = grey, AY = yellow AY = dominant allele for fur colour AY = recessive lethal allele

Answer 132

- Do a test cross - Get a grey mice (known recessive) - And then get your yellow mice - Use the yellow mice that produces offspring that are half yellow, half grey (1:1)

Answer 133

- NOT NECESSARILY Eg. you calculated 0.14% for a child to have the disease. BUT maybe all kids in the family will have the disease, which is 100% // - Small numbers (eg. one family) tends to give very different results - The more families you look at, the average result will get CLOSER AND CLOSER to the theoretical probability

Answer 134

OPPOSITE OF POLYGENIC TRAIT - Single gene affects 2 or more phenotypic traits that seem unrelated

Answer 135

SOX10 gene causes WS4 syndrome WS4 syndrome includes: - White forelock - Different coloured eyes - Areas of unpigmented skin - Lack of nerve components in the large intestine impacted bowel function - Hearing issues

Answer 136

- SOX10 gene expresses proteins used in EARLY DEVELOPMENT Eg. products needed in pigment cells, certain nerve tissues, future ear

Answer 137

DISTANCE!! - The further the distance between 2 gene loci ON THE SAME CHROMOSOME, the higher the likelihood of crossing over + recombination

Answer 138

Distance is calculated in % However each 1% = 1 map unit = 1 centimorgan (cM)

Answer 139

- diagram that shows the relative positions of genes on a chromosome - Created based on how frequently they are inherited together

Concept 13,14,15,16,17,18 Flashcards

(165 cards)