1
Q
- hardy weinberg doesn’t work with inbreeding
Blending theory
A
traits of parents get mixed like fluid in the offspring, resulting into a new trait that resembles parents
- prediction: if blended like fluid, parents traits are lost in the offspring and cannot be recovered
2
Q
Mendel’s result #1
A
- Mendel was able to control the crosses between different purebred pea plants
- already, blending theory is incorrect because you don’t get a mix of the colours
Prediction: once you blend the offspring, the parental traits get lost
3
Q
Mendel’s result #2
A
Made sure that pollination is self-pollination, so he transferred pollen to the stigma
- what many know as a monohybrid cross
If blending theory is correct, every F2 should be purple. -
argues against blending theory
- majority of them are purple, but significant fraction looked clearly white, identical to the parental
4
Q
Presence of “white flower” in Mendel’s experiment
A
The presence of white flowers indicates that the parental traits are not lost in F1s.
This also suggests that the “element” responsible for the flower color (trait/phenotype) works like particles, is maintained from one generation to the next and can be separate
5
Q
Meiosis (in the germ cell)
A
: is the cellular/molecular base of mendel’s law
During meiosis, a A/a cell produces 4 gametes, 2 X A & 2 X a.
Therefore, meiosis produces the same ratio of A to a, giving
an equal chance of transmitting between A or a
Starts heterozygous, produces 4 gametes, equal # of dominant and recessive
6
Q
Mendel’s conclusions
A
There is a dominant trait that hides the recessive trait (only one shows up in F1)
The elements producing traits are transmitted to the next generation in a predictable pattern
two copies of the “element - associated with a trait/phenotype” that are inherited from the parents but only one of them is visible (dominant/recessive)
- F1 was heterozygous
- only one of the two elements is transmitted to the next generation, and each of the two parental elements had an equal chance of transmitted to the next generation
7
Q
3:1 ratio
A
- in monohybrid cross, because 3 out of the 4 possible combinations contain at least one dominant trait in the offspring, the cross results in the 3:1 ratio of dominant and recessive traits
8
Q
Why did Mendel’s e experiment work?
A
9
Q
Presence of “white flower” in Mendel’s experiment
A
The presence of white flowers indicates that the parental traits are not lost in F1s. This also suggests that the “element” responsible for the flower color (trait/phenotype) works like particles, is maintained from one generation to the next and can be separate.
10
Q
Mendel’s conclusions
A
There is a dominant trait that hides the recessive trait (only one shows up in F1). The elements producing traits are transmitted to the next generation in a predictable pattern: two copies of the “element - associated with a trait/phenotype” that are inherited from the parents but only one of them is visible (dominant/recessive). F1 was heterozygous. Only one of the two elements is transmitted to the next generation, and each of the two parental elements had an equal chance of transmitted to the next generation.
11
Q
3:1 ratio
A
- in monohybrid cross, because 3 out of the 4 possible combinations contain at least one dominant trait in the offspring, the cross results in the 3:1 ratio of dominant and recessive traits.
12
Q
Why did Mendel’s experiment work?
A
- traits affected by only one gene, many traits (people’s height) are affected by multiple genes. Pure genetic background and ability to cross or self pollinate (ability to control cross/mating). Ability to obtain a large # of progeny. - blending theory came up because they looked at complex traits (height) which is impacted by many genes, no clear 3:1 ratio, Mendel was trying to find all purebreds.
13
Q
Element
A
Mendel refers to is the gene: traits determined by a single gene.
14
Q
Meiosis (in the germ cell)
A
: is the cellular/molecular base of mendel’s law. During meiosis, a A/a cell produces 4 gametes, 2 X A & 2 X a. Therefore, meiosis produces the same ratio of A to a, giving an equal chance of transmitting between A or a. Starts heterozygous, produces 4 gametes, equal # of dominant and recessive.
15
Q
Monohybrid cross
A
Monohybrid: heterozygote of a gene - during self pollination only 1 of the 4 gametes transmitted to next gen - randomly chosen, thats why you need a large number to get an idea of what happens during meiosis.
16
Q
Testcross
A
: the cross of an individual to a fully recessive individual. Testcross is useful for determining the genotype a testee.
17
Q
Loss-of-function mutations
A
(molecular term) are often recessive (phenotype) because a single copy of the wild-type (functional gene) gene can often provide necessary and sufficient functions for the organism. Most genes (wild-type) are haplosufficient.
18
Q
Haplosufficient:
A
describes a gene where one functional copy (allele) is enough for a cell or organism to have a normal function or phenotype, even if the other copy is lost or mutated. - even if you have one or two copies of the gene, it provides sufficient function.
19
Q
White flower vs purple
A
White flower plants have loss-of- function mutations of a gene required for pigment production. Because the wild-type gene is haplosufficient, white is recessive to purple.
20
Q
Chromosome Theory of Inheritance
A
Chromosomes, like Mendel’s “elements, come in matched (homologous) pairs in an organism. The members of a homologous pair separate during meiosis, so each sperm or egg receives just one member. Members of different chromosome pairs are sorted into gametes independently of one another in meiosis, just like alleles of different genes in Mendel’s.
21
Q
Thomas Hunt Morgan
A
skeptical about the chromosome theory of inheritance. Believed that we were overlooking other possibilities that may equally give two kinds of germ cells that the Mendelian explanation calls for.
22
Q
Morgan’s experiment #1
A
White is recessive (white flies have mutations in the gene that transports precursors of red pigment). - white has loss of function mutation - that’s why both the offspring have red eyes - red is haplosufficient because it is dominant.
23
Q
Result of Morgan’s experiment #1
A
leads us to know that the trait in drosophilla is X linked.
24
Q
What does morgan’s second experiment show?
A
This might suggest that mendel was wrong, because Mendel’s hypothesis states that the dominant trait should be seen in F1.
25
Why do mendel's experiments still work in terms of Morgan's experiments
Traits affected by only one gene. Many traits (people’s height) are affected by multiple genes. Pure genetic background and ability to cross- or self- pollinate (ability to control cross/mating). Ability to obtain a large number of progeny.
26
What are the findings from this pedigree
This isn’t sex linked; this is a recessive trait. This is autosomal.
27
What does this pedigree show?
This could be dominant or recessive trait - because it is rare, it can’t be recessive - that means it has to be dominant. 1/16 chance for all of the children to be unaffected because III 4 is heterozygous and III 5 is homozygous, so theres a 50/50 chance that the children will be affected, so you multiply 4 x 4.
28
Rare traits?
dominant.
29
common traits?
recessive.
30
what is the probability of the red circle being affecting
On both sides of the pedigree: There is an affected child (aa) born to unaffected parents. Therefore, both parents must be carriers (Aa). So for each affected individual: Their unaffected siblings have probabilities: AA = 1/4, Aa = 1/2, aa = 1/4 (ruled out because they’re unaffected). 👉 Given unaffected: ½ x ¾ = ⅔. Each parent of the “?” individual is: The child of an unaffected sibling of an affected individual. That sibling has a 2/3 chance of being a carrier. If that sibling is a carrier (Aa), they pass the allele with probability 1/2. So for each parent: ⅔ x ½ =⅓. To be affected (aa): Both parents must be carriers. And both must pass the recessive allele ⅓ x ⅓ x ¼ = 1/36.
31
how are genes inherited?
- Genes are inherited from one generation to the next in the form of a chromosome. genes on the same vs different chromosomes - Mendel’s pea plant has 7 pairs of chromosomes.
32
independent assortment
resulting in equal ratio of 4 different genotypes - stating that alleles for different genes segregate (separate) independently during gamete (sperm and egg) formation, leading to new combinations of traits in offspring. This random shuffling happens because homologous chromosome pairs align randomly during meiosis, creating diverse genetic combinations and increasing genetic variation, except for genes that are physically close (linked) on the same chromosome- probability that the paternal chromosome 1 and 2 are transmitted to a single gamete in human = ¼.
33
mitosis
during mitosis, everything lines up, so sister chromatids separate.
34
meiosis
- during meiosis, the homologous chromosomes come together, line up at the metaphase plate individually - meiosis: 2 types of gametes that can be produced, but during anaphase, chromatids can be pulled apart in different ways, which means you can get gametes with different phenotypes - completely random.
35
Probability of the different characteristics
Round, yellow = ¾ x ¾ = 9/16. Round, green = ¾ x ¼ = 3/16. Wrinkled, yellow = 3/16. Wrinkled, green = 1/16.
36
Calculation of probability
Calculation: 2x = # of pairs. X = # of pairs.
37
Dihybrid testcross
produces equal number of parental and recombinant type. The testcross is an easy way to determine the genotypes of gametes that a testee can produce. Due to “independent assortment”, equal number of parental and recombinant types are observed in the progeny of a dihybrid.
38
Recombinants
Recombinants are produced during meiosis - Getting recombinants can be used to map certain genotypes and phenotypes.
39
chi-square test
way of determining if the of observed values significantly deviate from expected. Determines degree of deviation, we want it to be 95%.
40
df: degree of freedom
(number of independent value -1).
41
Chromosome theory of inheritance
Genes are located on chromosomes inside the nucleus, and their behavior during meiosis explains inheritance patterns.
42
Linkage & Mapping
Study of how genes located on the same chromosome are inherited together and how their relative positions can be determined.
43
Sex chromosomes (Drosophila)
Females: XX. Males: XY.
44
pr (purple eyes)
Recessive mutation causing purple eyes instead of red.
45
vg (vestigial wings)
Recessive mutation causing small wings.
46
Two recessive mutations
Used to test whether genes are linked or assort independently.
47
Parental types
Offspring that retain the original allele combinations found in the parents.
48
Recombinant types
Offspring with new allele combinations produced by crossing over. Linked genes produce more parental offspring than recombinant offspring.
49
Testcross
Cross between a heterozygote and a homozygous recessive tester to reveal gamete frequencies.
50
Gene linkage
Genes close together on a chromosome tend to be inherited together.
51
Timing of crossover
Occurs after DNA replication, during prophase I of meiosis. More than one crossover can occur in a single meiosis.
52
Chiasmata
Physical sites where crossing over occurs.
53
Neurospora
A mold that grows primarily as a haploid organism. Products of meiosis remain together, allowing direct observation of segregation patterns.
54
Ascospore
Haploid spores produced by meiosis.
55
Ascus
a Sac containing all products of a single meiotic event. Tetrad: Four haploid products immediately after meiosis. Octad: Eight cells produced after a mitotic division following meiosis. Ordered tetrads: Arrangement reflects the sequence of meiotic divisions.
56
Cross between 2 haploids
Two-chromosome stage Before DNA replication. Four-chromatid stage After replication; crossing over occurs here. Result: Only two of the four chromatids are involved in a single crossover.
57
implications of multiple cross overs
Multiple crossovers can restore parental configurations.
58
Gene distance
The farther apart two genes are, the more likely crossing over occurs between them.
59
Recombination frequency (RF)
Proportional to physical distance between genes. More DNA between genes → higher chance of crossover.
60
Maximum RF
Cannot exceed 50% due to independent assortment. example of calculation: (151 + 154) / 2839 × 100 = 10.7%.
61
v (vermillion)
Mutation causing vermillion eye color.
62
cv (crossveinless)
Mutation causing absence of wing cross veins.
63
ct (cut)
Mutation causing snipped wing edges.
64
RF Between v & cv
Recombinant count (v & cv) 268 recombinants.
RF calculation
268 / 1448 × 100 = 18.5%
Interpretation
Genes v and cv are linked but separated by moderate distance.
65
RF Between v & ct
191 recombinants.
RF calculation
191 / 1448 × 100 = 13.2%
66
RF Between cv & ct
Recombinant count (cv & ct) 93 recombinants.
RF calculation
93 / 1448 × 100 = 6.4%
67
cv, ct, v RF's combined
RF summary
v–cv = 18.5%
v–ct = 13.2%
cv–ct = 6.4%
Smallest distance
Between cv and ct → genes closest together.
Expected vs observed
13.2 + 6.4 = 19.6 ≠ 18.5
Reason
Double crossovers restore parental gene combinations and go undetected.
68
Double recombinants
Produced by two crossovers between v and cv.
Calculation
2(3 + 5) + 268 = 284 recombinants
284 / 1448 × 100 = 19.6%
Key insight
Double crossovers must be added back to estimate true distance.
69
Genetic map
Shows linear order and distances of genes on chromosomes.
Map units
Correspond to recombination frequencies.
Application
Predict inheritance patterns and locate genes.
70
Question
RY = parental = 100%
ry= parental = 10%
Ry = recombinant = 40%
rY = recombinant = 40%
Male Gamete:
RY = parental = 100%
ry= parental = 10%
Ry = recombinant = 40%
rY = recombinant = 40%
Female Gamete:
RY = parental = 100%
ry= parental = 10%
Ry = recombinant = 40%
rY = recombinant = 40%
Probability of round and green and self pollinanted = 0.4 (Ry x Ry) x 0.4 = 0.16
0.4 (Ry x ry) x 0.1 = 0.04
0.4 (ry x Ry) x 0.1 = 0.04
Add all the possibilities: 0.04 + 0.04 + 0.16 = 0.24
71
Molecular Markers
Small DNA sequence differences (polymorphisms) within a species that are present at specific chromosomal locations.
72
Functions of Molecular Markers
* They are present through out the genome.
* Most don’t have biological functions, no phenotype.
* Molecular markers are often seen as bands on a gel.
* They can be used to map a gene (eg. disease causing gene) by determining the linkage between the gene of interest and a molecular marker
- markers are on specific chromosome locations
-things close together will segregate together
73
Simple sequence length polymorphisms (SSLPs)
* Human genomes contain a great deal of repetitive DNA sequences, including multiple repeats of short and simple DNA sequences, called SSLP at specific locations of the genome. (eg, CACACA…. )
* Unrelated people likely have different numbers of these repeats.
- Like anything else on the chromosome, humans have one set of paternal and one set of maternal SSLPs.
* The lengths of these regions can be detected by PCR amplification and resolving on a DNA gel
- have a very specific location on the chromosome
- number of repeats is where variation occurs
74
Fingerprints
The patterns produced on the gel by SSLPs at the multiple location of the genome can be function as DNA fingerprints
Short tandem repeats that are employed by
the U.S. database for violent unsolved crime
Distant cousins of the criminal likely share some of these markers
75
Single nucleotide polymorphisms (SNPs)
* The genomic sequences of two unrelated people are about 99.9% identical. Most of the 0.1% differences are single-nucleotide variants (SNVs).
* To be classified as a SNPs, a variant must be found in at least 1 percent of any population, common among people. If not, a variant may be a mutation of an individual.
* Some SNPs are more common in a specific population/ethnicity
- mutation can’t be used as a marker
* SNPs could be in the intergenic (anywhere in between) region, within a gene or in a regulatory region near a gene.
* Most SNPs have no effect on health or development.
* Some SNPs, especially the ones found within a gene or in a regulatory region near a gene, may play a role in susceptibility of a disease or sensitivity to an external factor
76
Mapping of disease gene with SNPs
- Two closely located genes or markers (eg. SNPs) are inherited together, eg. A disease-causing gene and SNPs near the gene. Scientists look for SNPs that are linked to the disease, inherited with the disease phenotype. Because the disease-causing gene must be located near those SNPs
77
Who is affected based on this diagram?
Pedigree
- dominant trait
P = gene responsible for disease phenutype
SSLP marker
- analyze the gene gives western blot
- 1,2,56 affected: they must carry marker from female parent common to those 4 children
78
Mapping CF by looking at molecular markers
-Years of linkage analysis of molecular markers in many CF families.
-Comparing normal vs CF individual on different chromosomal locations
-Molecular cloning of the gene
-Identification of three nucleotide deletion, loss of Phe50
- this family can compare DNA markers that are common to the affected person
- can identify candidate gene
The DNA marker associated with the CF chromosomes carrying F508:
The “1s” and “2s” inside the colored bars represent schematic alleles of DNA markers around the CFTR gene; the numbers on the right denote the counts of CF and normal (N) chromosomes with the corresponding DNA marker haplotypes shown on the left.
79
Tibetan-Chinese SNP detection
When Spanish colonists established towns high up in the Andes mountains of South America, Spanish parents have hard time having child.
Some SNPs in a gene called EPAS1 occur at very different frequencies in Tibetans (87 percent) and Han Chinese (9 percent) EPAS1 regulates the number of red blood cells that our bodies produce in response to the level of oxygen in our tissues.
80
Mutations in Halposufficient genes
are recessive:
- Haplosufficient genes: Single copy of wild type is sufficient enough to carry out the gene
Thats why mutations don’t matter
81
Dominant Mutations:
Having a single copy of the mutation produces a phenotype
(disease) despite having a wild-type copy of the gene.
- loss of function mutations could be dominant wild type
82
- Haploinsufficient genes
: occurs when one functional copy of a gene (allele) isn't enough to produce the necessary amount of gene product (like a protein) for normal cell or body function, leading to a disease or developmental issue
- whether you have two copy or one copy, whether there is a phenotype is associated or not
83
Mutations in genes coding ribosomal subunits in Drosophila
are often dominant, haploinsufficiency
- there are many genes that code for ribosomes
- turns out, if you have a loss of function mutation, a minute phenotype occurs
-if you have a loss of function mutation over one of the ribosomes you end up with bristles that are a bit shorter.
This is heterozygous for the loss-of-function mutant of those ribosomal subunit meaning because you need a lot of protein synthesis, its viable but theres a phenotype associated with it
- in drosophila it is haploinssuficent when it comes to the ribosomes
So loss of function could be dominant in this case because the gene wild type gene is haplo insufficient.
84
Dominant Negative
functions of mutations that happens to proteins that homodimerize
describes mutations where the faulty protein product interferes with the function of the normal protein from the other gene copy, creating a dominant trait despite the presence of one working allele
85
P53
Many p53 mutant alleles in cancer function as a dominant negative
P53 is a transcription factor that binds DNA as a homotetramer
has a missense mutation that makes DNA binding domain inactive
- loss of function mutation because DNA binding activity is lost
- if this occurs on one of the two alleles, the mutant p53 will be expressed, and can form a tetramer complex with the proteins
because for P53 to function properly and bind DNA with high affinity, you need to form a tetramer with functional DNA biding domain. Then this is this mutant form that that can no
longer bind DNA will interfere because it can bind to wild type form from your wild type interfere with the wild type to bind DNA because it will physically interact
86
Incomplete or Partial Dominance:
doesn’t behave completely like dominant
if you have one copy, only one functional copy, it doesn't produce the pigment as much as though as
it as the amount of of the pigment when there's a 2 copy of the functional gene and the phenotype is visible
87
Compare wild type to heterozygeous = you will see partial dominance
is a genetic pattern where a heterozygote (carrying two different alleles for a trait) shows an intermediate or blended phenotype between the two homozygous parents, like red and white flowers producing pink offspring, unlike complete dominance where one allele fully masks the other
88
Codominance
(both alleles are expressed/detected): Three alleles determine
the blood type: i, IA & IB
- an inheritance pattern where two alleles are expressed equally, and neither allele is dominant or recessive
- The gene responsible for the blood type encodes a glycosyltransferas
Blood: A and B are dominant over O, but codominant with each other in conventional assay
89
IV Recessive lethal:
Homozygous mutations causing lethality in the animal, either recessive or dominant mutations
90
Conditional alleles
engineered gene variants that remain functional until activated or inactivated by a specific trigger, allowing scientists to study gene function in precise tissues or at specific times, bypassing issues like early lethality, often by using site-specific recombinases
Phenotype associated with mutations depend on temp; molecular reason: is because of the different protein stability
91
Auxotrophs
organisms that lost the ability to synthesize certain substances required for their growth
92
Arginine in conditional alleles
Initially in minimal media where only sugars are provided, certain organisms cannot grow.
But if you provide arginine right, meaning although they can synthesize alginine themselves, they'll survive.
But if you withdraw arginine from the from the media and keep them, then that will become non function and often case lethal.
Conditional alleles temperature controlled phenotype: Tyrosine kinase that is active at a lower temperature
93
Penetrance
the percentage of individuals with a given allele who exhibit the phenotype of that allele
e.g. BRCA2 mutations predispose to breast, ovarian and pancreatic cancers
94
Expressivity
the degree to which a given allele is expressed at the phenotypic level, the intensity of the phenotype.
95
What does “genes affecting the same characteristic but working independently” mean?
Two genes influence the same trait but act independently, producing a classic Mendelian 9:3:3:1 ratio
96
Which two genes control corn snake coloration?
Orange gene (O/o): controls orange pigment
Black gene (B/b): controls black pigment
97
What phenotype results from genotype O/– ; B/–?
Camouflage
98
What phenotype results from genotype o/o ; B/–?
Black
99
What phenotype results from genotype O/– ; b/b?
Orange
100
What phenotype results from genotype o/o ; b/b
Albino
101
What is the phenotypic ratio for independently acting O and B genes?
9 Camouflage : 3 Black : 3 Orange : 1 Albino
102
Are genes O and B in the same pathway?
No. They are involved in the same biological process but not the same pathway
103
What is the “one-gene–one-enzyme” hypothesis?
Each gene encodes a single enzyme that catalyzes one step in a biochemical pathway.
104
What pathway is used to illustrate one-gene–one-enzyme?
Arginine biosynthesis: precursor → ornithine → citrulline → arginine
105
What is the purpose of a forward genetic screen?
To identify genes involved in a biochemical pathway by isolating mutants.
106
How are mutants generated in a forward genetic screen?
By mutagenesis (e.g., X-rays) followed by screening for growth defects.
107
Why is minimal medium used in auxotrophy screens?
It reveals mutants unable to synthesize essential compounds.
108
What does failure to grow on minimal medium indicate?
A nutritional (auxotrophic) mutant
109
Why are mutants tested on supplemented media?
To identify which compound restores growth, revealing the defective step.
110
Which supplement restores growth in arginine auxotrophs?
Arginine
111
What does “+” mean in supplementation tables?
Growth occurs
112
What does “–” mean in supplementation tables?
No growth
113
What supplements rescue arg-1 mutants?
Ornithine, citrulline, and arginine
114
What supplements rescue arg-2 mutants?
Citrulline and arginine
115
What supplements rescue arg-3 mutants?
Arginine only
116
What is a complementation test used for?
To determine whether mutations are in the same gene or different genes.
117
What result indicates complementation?
Wild-type phenotype in the F1
118
What result indicates failure to complement?
Mutant phenotype in the F1.
119
What does failure to complement imply?
Mutations affect the same gene.
120
What does a 9:7 ratio indicate?
two genes act in the same biochemical pathway. occurs because Both gene products are required for the final phenotype.
121
What happens if the regulatory gene is mutated?
Target gene is not properly expressed.
122
What phenotypic ratio results from regulator–target interaction?
9 normal : 7 mutant
123
What is epistasis?
When one gene masks the phenotypic effect of another gene.
124
What is recessive epistasis?
The recessive genotype of one gene masks another gene’s phenotype
125
What phenotypic ratio is produced by recessive epistasis?
9:3:4
126
What is a suppressor mutation?
A mutation that reverses the effect of another mutation.
127
Does a suppressor restore the original genotype?
No, it restores phenotype, not genotype.
128
What is a modifier mutation?
A mutation that changes the degree of expression of another mutation.
129
Where do modifier mutations often occur?
Regulatory sequences
130
What is synthetic lethality?
Two mutations that are viable alone but lethal together.
131
What biological structures often show synthetic lethality?
Multiprotein complexes
132
Why is synthetic lethal not the same as recessive lethal?
Each mutation alone is non-lethal.
133
What genotype is missing in synthetic lethal crosses?
Double homozygous mutant (0 class)
134
Epistasis
normal dihybrid ratio altered from 9:3:3:1 to 9:3:4
135
Emergent population level properties and processes
* Genotype and allele frequencies.
* Change in genotype and allele frequencies over time (evolution).
* Population to population variation (e.g., variation between populations in genotype and allele frequencies
136
Process of evolutionary change:
* Change in population level variation due to MUTATION
* Change in population level variation due to MIGRATION.
* Change in population level variation due to NATURAL SELECTION
* Change in population level variation due to chance (GENETIC DRIFT)
137
Genetics at the population level:
1. Concentrates on collections of individuals and their genetic properties, especially the frequencies of alleles, SNPs, genotypes, haplotypes in time and space (geography, habitats)
2. Studies the origin, maintenance, and change of allelic and genotypic variation in populations.
3. Makes use of models to study the processes that influence population genetic composition and make predictions about how it will change (e.g., in response to selection, migration, etc
138
Notation for genotype frequencies:
f(AA), f(Aa), f(aa)
sum : fAA + fAa + faa = 1.0
139
For a gene with two alleles A and a
* The frequency of one allele (e.g., allele A) is denoted p.
* The frequency of the other allele (e.g., allele a) is denoted q.
* Note that since p + q = 1, we can solve for q = 1-p
140
Hardy Weinberg Principle:
the constancy of allele frequencies when there are no acting evolutionary forces
141
Hardy Weinberg genotypes:
Freq(AA) = p^2
Freq (Aa) = 2pq
Freq(aa) = q^2
*And from then on, these frequencies will remain unchanged (provided there is continued random mating and absence of evolutionary forces)
142
Calculate allele frequencies of A, B, O: according to Hardy Weinberg principle
f(A) = 0.1 + 0.½ + 0.3/2 = 0.30
f(B) = 0.2 + 0.2/2 + 0.3/2 = 0.45
f(O) = 0.1 /2 + 0.2/2 + 0.1 = 0.25
143
Cystic Fibrosis (CF) occurs in 4 out of 10,000 persons. Assume that the genotypes at the locus underlying CF are present in Hardy-Weinberg proportions in the human population. What is the frequency of carriers (heterozygotes) in the population?
If the freq(aa) = 0.0004, the frequency of the a allele must therefore be q = (0.0004)/2 That is, q = 0.02
So, 2pq = 2(0.98)(0.02) = 0.0392
144
A suspect in a crime has the following genotype at three separate and unlinked biallelic loci, A, B and C: A1/A2; B2/B2; C2/C2. Blood found at the scene of the crime is seen to match this genotype. The allele frequencies of alleles A1, B1, C1 in the population are 0.9, 0.8, and 0.7, Respectively. What is the probability that this match occurs in the general population? (Assume Hardy-Weinberg proportions for each locus)
Answer:
If A1 = 0.9, A2 = 0.1
If B1 = 0.8, B2 = 0.2
If C1 = 0.7, C2 = 0.3
Freq (A1/A2) = 2pq = 2 (0.9)(0.1) = 0.18
Freq (B2/B2) = p2 = 0.22 = 0.04
Freq (C2/C2) = q2 = 0.33 = 0.09
Probability (A1/A2; B2/B2; C2/C2) = 0.18 x 0.04 x 0.09 = 0.000648
145
Inbreeding:
-a form of non-random mating
hardy weinberg doesn’t work with inbreeding
- Inbreeding, identity by descent, and increased homozygosity (e.g., half brother-half sister mating, also called half-sib mating)
-Inbreeding increases the probability that two alleles at a locus will be copies of an allele present in an ancestor.
- Such copies are said to be “identical by descent” (IBD)
146
Inbreeding Coefficient
F = the overall probability that the two alleles inherited by a given individual will be identical by descent
Half sib mating (calculation of the inbreeding coefficient: F quantifies the overall probability
that the two alleles inherited by a given individual will be identical by descent.
147
Probability that individual I inherits two copies of grandma’s pink allele
= ½ + ½ + ½ + ½ = 1/16
- has to go from A → B, A → C, B→ I, C→ I
148
Sister brother mating. What's joffrey's inbreeding coefficient
For joffrey:
Probability that he inherits from grandparent A = ½ + ½ + ½ = ⅛
B = ½ + ½ + ½ = ⅛
⅛ + ⅛ = ¼
149
What is the F of child of 1st cousins:
1/16
150
population that switches over to inbreeding from random mating
Inbreeding changes genotype frequencies (fewer heterozygotes than expected under HW conditions)
- In a population that switches over to inbreeding from random mating you would observe in the following generation:
That is, there would be: MORE HOMOZYGOTES, FEWER HETEROZYGOTES
compared with a random mating population
151
New Hardy Weinberg equations when a population moves from inbreeding to random mating
f(A1A1) = p2 + Fpq
f(A1A2) = 2pq (1-F)
f(A2A2) = q2 + Fpq
152
Inbreeding Depression
(reduction in the viability of inbred individuals) occurs because most deleterious conditions are recessive and require two copies of the mutation to be expressed
* Cystic fibrosis
* Sckle cell anemia
* Tay-Sachs disease
* Gaucher disease
* Alcaptonuria
* Albinisms
* etc.
153
Factors influencing allele frequencies:
Mutation, Migration, Genetic Drift, Natural Selection
154
Typical rate of mutation per base pair of DNA per generation
10^-9 mutations/base pair/generation
Allele Frequency Change from Mutation:
A1 → A2
mutation: rate u = 10-5 per generation
p(t+1) = pt(1-u) = pt - upt
p2= p1 (1-u)
p3 = p2(1-u)
155
Migration
bidirectional (back and forth between 2 isolated populations), unidirectional, higher than mutation rates,
- migration patterns can help explain spatial variation in the frequency of the B allele
- can be opposed by selection
- bidirectional migration isn’t really isolated because there’s back and forth between both
-migration can be bad too
Migration can be frequent or infrequent. When frequent and non- directional it tends to homogenize the allele and genotype frequencies of separate populations. Migration can be maladaptive.
156
Genetic Drift:
a mechanism of evolution characterized by random, chance-based fluctuations in allele frequencies (gene variants) within a population from generation to generation
Random genetic drift is strong in small populations or by migrants on an island
* Genetic drift moves allele frequencies up or down with equal probability. It is strongest in small populations.
157
Mutation = greatest contributor to genetic variation
- Mutation is the ultimate source of genetic variation, but occurs rarely at a single locus, and thus does not alter allele or genotype frequencies rapidly.
158
Random Genetic Drift
Random genetic drift is strong in small populations
Random genetic drift is weaker in large populations
Each colour represents the
frequency of one of the alleles
at a biallelic locus
159
Genetic Drift
high frequency of retinitis pigmentosa in the inhabitants of tristan da cunha
160
Evolutionary processes influence genotype and allele frequencies in populations
[Page 152]
161
Natural Selection
- can lead to adaptation
- Natural selection “fine tunes” phenotypes to their environment
162
Population genetic models of selection
* Help us to see how adaptation comes about
* Predict the outcome of selection (i.e., the equilibrium allele frequency at a locus)
* Predict rate of evolutionary change (i.e., the rate of change in allele frequency at a locus)
* Examine the interaction of selection with other population processes.
163
Natural selection in mid-19th
Britain favoured the dark form of the Peppered Moth
- During the era of heavy coal burning in Britain, it took about a hundred years for the white form of the Peppered Moth to be nearly replaced by the dark form
164
Natural Selection through relative fitness
Natural selection typically works by favouring the form with the highest relative fitness (relative to the survival probability of the form that has the highest survival)
165
Using a model of selection to calculate genotype frequencies after selection
Answer
166
w(average)
selection can shift allele frequencies
167
How will natural selection act to alter frequency of allele a (“light”) in a coal-polluted forest environment over many generations?
increasing f of dark form and decreasing of light form
168
Directional selection
(e.g., selection of one favoured allele) is common
Directional selection will eventually drive the less fit form from the population leaving the more fit form to predominate.......adaptation in progress.
continual change in one direction from selection
169
Balancing selection
“Balancing selection” is different, as it can favour two or more alleles.
Examples:
* Negative –frequency dependent selection
* Heterozygote advantage
170
negative frequency dependent selection
(mediated by learned predator-avoidance behaviour)
type of balancing selection
171
negative frequency dependent selection example #2
(mediated by self-incompatibility and rare mate type advantage)
172
Heterozygote advantage
Balancing selection
173
Model for heterozygote advantage
AA phenotype 1 w(AA) = 1-t
Aa phenotype 2 w(Aa) = 1
aa phenotype 3 w(aa) = 1-s
174
Deleterious mutation-selection interaction (why bad things happen to good populations)
Answer
175
The B locus has two alleles B and b with frequencies of p = 0.95 and q = 0.05, respectively. The genotypic relative fitnesses at this locus are wB/B = 0.8, wB/b = 1.0, and wb/b = 0.7. What will the frequency of the b allele be in the next generation?
Answer
176
Selective sweeps
can reduce diversity in genomic regions, especially those with low rates of recombination—they leave behind a trace of past events that have influenced the genome
Note that this one chromosome becomes common as a result of the sweep in the Population (Population A) where the nucleotide mutation is strongly selected
(In Population B the nucleotide mutation did not occur
177
Complex (quantitative) trait genetics
The nature of quantitative variation:
Distribution of phenotypic states is often “bell-shaped”
178
Simple Trait Vs Complex
population variation controlled by one/two genes vs many genes
179
Quantitative genetic variation
: Important for understanding the evolution of common diseases, crop improvement, and the rate of evolutionary change
180
types of complex genetic traits
continuous and threshold
181
continuous
body size, running speed
182
threshold
cancers, heart diseases
183
Variance
a measurement of how far each number in a data set is from the mean average and thus from every other number in the set
184
Phenotypic variance of a trait
How much is due to differences in the environment that individuals experience and how much is due to genetic differences among individuals experience and how much is due to genetic differences among individuals
185
Multifactorial model for complex trait variation
phenotypics variance can be decomposed in variance that is attrituble to environmental factors and genetic
186
bell shape distributions of complex traits
allelic segregation in F2, 0 up to 4 long alleles
187
bell shape continued
Answer
188
central limit theorum
statistical principle stating that under certain conditions, the distribution of the sum of large number of independent and identically distributed random variables tends to approach a normal distribution
189
experiment to estimate genetic and environmental variance
- additive
-variants are additive as both genetic and environmental variants create an effect
190
Tobacco corolla length experiment
P1 and P2 are different inbred lines
( genetically homozygous... at some loci they will be homozygous for different alleles)
- F1 hybrid = heterozygous
F2 progeny segregates for alleles that differ in P1 and P2, larger variance - spread across means
191
sources of variance
parents: environmental differences between individuals
F1: environmental differences between individuals
F2: environmental and genetic differences between individuals
192
Broad sense Heritability formula
H^2 = Vg/Vx
measures total phenotypic variants that are due to genetics
- bounded by 0 and 1
- dimensionless
- standardized parameter
Vg = genetic variance
Vx = phenotypic variance
193
You have a population of genetically identical mice of strain A. You also have a second population of genetically identical mice of strain B. The mean and variance of body weight of the strain A population are 30 g and 400 squared grams, respectively, while the mean and variance of body weight of strain B are 10 g and 400 squared grams, respectively. You cross the strains to obtain an F1, which you allow to interbreed, and then raise the F2 progeny. The mean and variance of body weight of the F2 population is 20 g and 1200 squared grams. What is the broad-sense heritability of body weight in F2 generation?
Ve (environmental variants) = 400 squared grams (from parents who are inbred)
Vx = 1200 squared (from F2 population)
Vx = Ve + Vg, so that means Ve = 800g
H^2 = 800/1200 = 0.67
194
MZ and DZ twins
identical = monozygotic twins from same embryo splitting
fraternal same sex and opposite sex = dizygotic
195
broad sense heritability in twins
if twins tend to share environmental features, increase correlation and broad heritability
H^2 = r
196
- most similar heritability traits in twins
fingerprint count, autism, height, schizophrenia
197
Trait correlations for MZ twins raised apart should be smaller than those of DZ twins raised apart (T or F)
False
198
Trait correlations for MZ twins raised together should be smaller than those of MZ twins raised apart (T/F)
F
- shared environmental variation so the correlation is higher
199
Heritability estimates obtained based on one set of MZ twins raised apart are in general applicable to the human population as a whole (T/F)
F
- twins don't represent whole population
200
QTL mapping
...resembles classical mapping approaches (controlled crosses used)...co-segregation of a marker with a trait’s phenotypic value helps you to locate the region of the genome where a gene with influence on the phenotype resides....there may be many such genes (called quantitative trait loci or QTL’s)
201
Association mapping or genome-wide association studies (GWAS)
...relies on population surveys that look for correlation between markers and phenotypic trait values in samples taken from large populations.
202
principles of QTL mapping
Co-segregation (co-inheritance) of phenotypes and marker alleles (usually SNPs)
203
Where is the gene that influences “hairiness”?
at the top of the chromosome
- crossing over occuring individually at each meiosis
- co-segregation informing regardless affecting phenotype
- all the tops of the chroosmes that are hairy have the top as bronze from parent 1
204
QTL mapping: crosses of two inbred lines
inbred lines that differ in the trait of interest and their molecular marker genotypes
- key idea of QTL mapping: if a marker is close to a gene that effects somsething then there will be different averages of traits if a marker is far away, there's no difference in traits
- more markers in genomic region means it is more likely to hit a responsible gene
205
Where is the gene that influences tomato size?
since M3 has the highest difference between mean of B/B and B/S through QTL mapping, we know thats where the gene is that influences the size
206
*Two populations of sheep have the same mean and variance for body weight, but population A has a larger heritability for body weight than population B. In each population, a breeder creates a new population by selecting the top 10% heaviest animals to mate. The population derived from population A will have a larger mean body weight than the population derived from population B (T/F)
T
Because both populations experience the same selection differential, the population with higher heritability will show a greater response to selection, resulting in a higher mean body weight in the next generation
BIOL 202
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