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Genetics exam 2

front 1

A document to record ancestry, used by genealogists in study of human family lines, and in selective breeding of other animals

back 1

Pedigree

front 2

Why is human genetics hard to follow?

back 2

-a small number of offspring

-experimental crosses not ethical

-long generation time (22-23 yrs)

front 3

back 3

front 4

back 4

..

front 5

What are the 4 basic inheritance patterns?

back 5

-Autosomal recessive

-Autosomal dominant

- Sex-linked recessive

-Sex-linked dominant

front 6

-2 unaffected parents can have affected offspring

-both male and female offspring affected at similar freq.

back 6

Autosomal recessive

front 7

What are examples of autosomal recessive disorders?

back 7

-cystic fibrosis (severe damage to lungs)(causes thickened fluids)

-Sickle cell anemia (affects shape of blood cell)

-Tay-Sachs (toxic build up in brain and spinal cord)

front 8

-Affected individuals have at least 1 affected parent

-Affected parent of either sex can pass on to offspring of either sex

-appears in male and female

- not sex linked

back 8

Autosomal dominant

front 9

-unaffected parents can have affected offspring

-Males affected more frequently than females

-Transmitted from mothers to sons

-50% of child will be affected

back 9

Sex-linked recessive

front 10

Example of__________

back 10

Sex-linked recessive

front 11

color blindness is___linked and affects _____more than _______

back 11

-x

-male

-female

front 12

-affected offspring come from parents

-males pass only to daughters

-females pass to children of either sex

-if dad is affected 100% of daughters are affected

-if mom affected 100% of sons are affected

back 12

Sex-linked dominant

front 13

What are examples of sex-linked dominant disorders?

back 13

-Alport syndrome

-Coffin-Lowry syndrome

-Fragile X syndrome

-Incontinentia pigmenti

-Vitamin D resistant rickets

front 14

-only males have it

-pass to all sons, none of daughters

back 14

Y-linked trait (dominant)

front 15

-Mothers always pass on

-Fathers never pass on

back 15

Mitochondrial mutation

front 16

What is the chance of their next child being affected?

(a)= affected

back 16

(1/2) x (1/2)= (1/4)

-gender x condition

front 17

Look at slide 34 in Ch 6. pt 1

back 17

DO IT NOW

front 18

-non identical twins

-2 zygotic eggs

back 18

Dizygotic twins

front 19

-identical twins

back 19

Monozygotic twins

front 20

-the trait shared by the members of a twin pair

-genetically the same

-environment may be similar

-the increased rate is more likely impacted by genetic factors rather than environmental factors

back 20

concordant traits

front 21

-separate embryotic sacks

back 21

dizygotic twinning

front 22

-comes from 1 zygote

-early splitting

back 22

Dichorionic diamniotic (monozygotic twinning)

front 23

-late split

back 23

Monochorionic diamniotic (monozygotic twinning)

front 24

-very late split

-same sack

back 24

Monochorionic monoamniotic (monozygotic twinning)

front 25

Overweight biological parents tend to have_______children. There is no ___ _____ between the weight of children and that of their adoptive parents.

back 25

-overweight

-consistent association

front 26

This is an example of :
A pregnant woman is concerned about exposure to an environmental substance (drug, chemicals, or virus) that cause birth defects.

back 26

Environmental impact

front 27

Genes on the same chromosomes tend to

back 27

get transmitted together

front 28

certain combinations of alleles are passed more frequently than others

back 28

linkage

front 29

Complete linkage leads to;

____% non recombinant gametes and progeny

back 29

100

front 30

the crossing over between ______ chromosomes can break linkage

back 30

homologous

front 31

Alleles are passed in different combinations from how they were inherited

back 31

Recombination

front 32

Recombination leads to:

back 32

-recombinant gametes and progeny

front 33

________ associated with physical change in chromosomes

back 33

recombinant phenotypes

front 34

back 34

recombinant diagram

front 35

When a plant is normal with tall leaves and short with dwarf leaves, they are

back 35

nonrecombinant progeny

front 36

This is an example of

back 36

no crossing over

front 37

This is an example of:

which results is;

back 37

crossing over

-2 nonrecombinant and 2 recombinant gametes

front 38

One chromosome contains both wild-type alleles, one chromosomes contains both mutant alleles

back 38

coupling (cis configuration)

front 39

wild type allele and mutant alleles are found on the same chromosomes

back 39

repulsion (trans configuration)

front 40

Usually 2 step process:\

1. determine data are consistent with genes assorting independently (chi-squared test)

2. If the data is not consistent with independent assortment, this suggests genes are linked

- determine how closely linked the genes are

back 40

determining genetic distance

front 41

How to calculate recombination frequency

back 41

recombination freq.=(# of recombinant progeny/ total # of progeny)x100%

front 42

No correlation between which alleles are transmitted=50% , recombinant gametes =50cM

back 42

independent assortment

front 43

Two genes are linked if they are less than____ apart

back 43

50cM

front 44

If the progeny resulted in:

-66

-77

-28

-33

Which are recombinant, and what would the distance be?

back 44

-66 and 77 are the highest number= parental progeny

-28 and 33 lowest numbers= recombinant progeny

(28+33)/(66+77+28+33)=0.303 x100= 30.3 cM

front 45

a crossover at one place on the chromosome reduces the likelihood of a second crossover

back 45

interference

front 46

How to calculate amount of interference:

coefficient of coincidence= observed DCO/expected DCO =

back 46

-((2)^3)/(total amount of flies x DCO freq.)

-((2)^3)/12= 0.67

front 47

Distances based on crossover/linkage

back 47

genetic map

front 48

Distances based on actual base-pair differences

back 48

physical map

front 49

This is an example of

back 49

crossovers with three linked loci

front 50

recombination distance are ______accurate the larger they get

back 50

less

front 51

Consider the mutation and cross over

back 51

How halotype works

front 52

consider how when the top of loci is blue, it shows the blue parents trait

back 52

Quantitative trait locus mapping (QTL)

front 53

-controls have little to no genes of the condition

-lower freq. of variant

-variant linked to condition

-cases have the condition

back 53

Genome-wide association studies (GWAS)

front 54

Used to assign a gene to a particular human chromosome

-gene present (given)

-Human chromosomes present ( find most similar to gene presented)

back 54

Somatic-cell hybridization

front 55

back 55

displays the mutant phenotype

front 56

back 56

displays the wild type

front 57

-prepared from cells halted in metaphase(relaxed)

- numbered based on size

-Sex chromosomes

-based on haploid # (23 chromosomes)

back 57

Karyotyping

front 58

With sex chromosomes, the largest is the ___ chromosome and the short chromosome is the ___ chromosomes

back 58

-X

-Y

front 59

-moving, gaining, or losing pieces of a chromosome

-2n=6

-duplication

back 59

rearrangement

front 60

-changes in # of copies of an individual chromosome

-2n+1=7

-trisomy

back 60

Aneuploidy

front 61

-extra whole set of chromosomes

-3n=9

-autotriploid

back 61

polyploidy

front 62

Types of rearramgement:

back 62

-duplication

-deletion

-translocation

-inversion

front 63

Copied region of chromosomes

back 63

duplication

front 64

deleted region of chromosomes

back 64

deletion

front 65

inverting (flipping) region of chromosomes

back 65

inversion

front 66

moving part of chromosome somewhere else

back 66

translocation

front 67

-bulge during pairing in meiosis 1

back 67

Duplication (segment of chromosome is duplicated)

front 68

-formation of loop during pairing in prophase 1

back 68

Deletion

front 69

-turned 180 degrees

back 69

inversion

front 70

Inversion that doesn't include the centromere

back 70

paracentric inversion

front 71

inversion includes the centromere

back 71

pericentric inversion

front 72

If a crossover were to occur in the inverted region,

back 72

it would result in a complication

front 73

The packaging of tremendous amounts of genetic info. into the small space within a cell has been called

back 73

The ultimate storage problem

front 74

Is overwound or underwound causing it to twist on itself

back 74

Supercoiling DNA

front 75

-add two turns

- occurs when DNA is overrotated; the helix twists on itself

back 75

Positive supercoil

front 76

-removes 2 turns

-occurs when DNA is underwound; the helix twists on itself in the opposite directions

back 76

Negative supercoiling

front 77

What do these charactericstics describe?

-less condensed

-on chromosome arms

-unique sequences

-many genes

-throughout S phase

- often transcription

- crossing over is common

back 77

Euchromatin

front 78

What do these characteristics describe?

-more condensed

-At centromeres, telomeres, and other specific places

- repeated sequence

-Late S phase

-infrequent transcription

-crossing over is uncommon

back 78

Heterochromatin

front 79

Regions of relaxed chromatin where active transcription is taking place

back 79

chromosome puffs

front 80

Chromosome puffs on giant polytene chromosomes isolated from the __ ____ of larval Drosophila

back 80

-salivary glands

front 81

Chromosome fragments that lack ______ are lost in mitosis

back 81

centromeres

front 82

Proposes that mitochondria and chloroplasts in eukaryotic cells arose from bacteria

back 82

Endosymbiotic theory

front 83

back 83

endosymbiotic theory

front 84

Organelles in a ______cell segregate randomly into the progeny cells.

back 84

heteroplasmic

front 85

back 85

understand it

front 86

What is the importance of mutations?

back 86

-provides raw material for evolution

-source of many diseases and disorders

-look at whats wrong when we alter a gene

front 87

-Do not get passed on to the offspring

-passed to new cells through mitosis creating a clone of cells having mutant gene

back 87

Somatic mutations

front 88

-reproductive cell mutations

-passed on

-meiosis and sexual reproduction allow mutation to be passed down to about half the members of next generation

back 88

germ-line mutation

front 89

One letter is replaced by another

back 89

base substitution

front 90

A to G & C to T (purines switch with purines) (pyrimidine switches with pyrimidine)

back 90

Transition

front 91

A or G to C or T (purines switch with pyrimidines) (pyrimidine switch with purines)

back 91

Transversion

front 92

Frameshift mutations are not_____by 3

back 92

divisible

front 93

in frame insertions and deletions are divisible by

back 93

3

front 94

Original sequence:

GGG AGT GTA GAT CGT

What would the sequence look like if a base substitution were to occur on the following nucleotide?

back 94

front 95

Original sequence:

GGG AGT GCA GAT CGT

What would the sequence look like if a base insertion were to occur on the following nucleotide?

back 95

front 96

Original sequence:

GGG AGT GTT AGA TCG T

What would the sequence look like if a base deletion were to occur on the following nucleotide?

back 96

front 97

Repetitive sequence can cause______ of strands during replication

back 97

slippage

front 98

Dna polymerase adds

back 98

more repeats

front 99

unequal cross over misaligns

back 99

homologous chromosomes

front 100

The new codon encodes a different amino acid; there is a change in amino acid sequence

back 100

Missense mutation

front 101

The new codon is a stop codon; there is a premature termination of translation

back 101

Nonsense mutation

front 102

The new codon encodes the same amino acid; there is no change in amino acid sequence

back 102

Silent mutation

front 103

a mutation that hides or suppresses the effect of another mutation

Ex: wild type has red eyes (A+B+), and a mutation causes white eyes (A-B+). This allows for a fly with (A-B-) to have red eyes even if it has two mutant phenotypes.

back 103

Suppressor mutation

front 104

Second mutation in the same gene "fixes" problem caused by first mutation

back 104

intragenic

front 105

second mutation in a different gene. "fixes" the problem caused by the first mutation

back 105

Intergenic

front 106

Purines and pyrimidine bases exist in different forms called

back 106

tautomers

front 107

____ _____ pairings can occur as a result of the flexibility in DNA structures

EX: T and G binding (wobble)

C and A protonated wobble

back 107

Nonstandard base

front 108

____ base pairing leads to a replication error

back 108

Wobble

front 109

____ and ____may result from strand slippage

back 109

-insertions

-deletions

front 110

____ _____ over produces insertions and deletions

back 110

unequal crossing over

front 111

loss of purine

back 111

Depurination

front 112

loss of an amino group

back 112

deamination

front 113

Used to identify chemical mutagens

back 113

Ames test

front 114

A___ can result in possible wobble mispairing

back 114

bulge

front 115

Many incorrectly inserted nucleotides that escape proofreading are

back 115

corrected by mismatch repair

front 116

Changes nucleotides back into their original structures

back 116

Direct repair

front 117

A damaged base can be removed by______, and ______ cleaves the deoxyribose sugar.

back 117

-DNA glycosylase

-AP endonuclease

front 118

DNA _______ adds a new nucleotide and is then sealed by DNA _____ to restore the original sequence.

back 118

-polymerase

-ligase

front 119

back 119

base excision

front 120

Removes bulky DNA lesions that distort double helix

back 120

Nucleotide-excision

front 121

Can cause chromosomal rearrangement

back 121

transposons

front 122

back 122

example of transposons causing rearrangement

front 123

How to calculate the frequencies of observed genotypes?

-# of individuals with genotype divided by total # individuals

f(AA)=

f(Aa)=

f(aa)=

back 123

-f(AA)= # of AA individuals/#of total individuals

-f(Aa)= # of Aa individuals /# of total individuals

-f(aa)=# of aa individuals /# of total individuals

front 124

A population of 868 individuals has 482 AA individuals, 213 Aa individuals, and 173 aa individuals. What are the frequencies of each genotype?

back 124

f(AA)= 482/868=0.555

f(Aa)= 213/868=0.245

f(aa)= 173/868=0.199

front 125

How to calculate observed allelic frequencies

-# copies of alleles divided by total # alleles

f(A)=

f(a)=

back 125

-f(A)=(2 x #AA individuals + #Aa individuals)/(2 x total # of individuals)

-f(a)=(2 x #aa individuals + #Aa individuals)/(2 x total # of individuals)

front 126

A population of 868 individuals has 482 AA people, 213 Aa people, and 173 aa people. What are the frequencies of each allele?

back 126

-f(A)= (2 x 482 +213)/(2 x 868)=0.678

-f(a)= (2 x 173 + 213)/(2 x 868)=0.322

front 127

What does p and q stand for in this equation?

p+q=1

back 127

-p: dominant

-q: recessive

front 128

equation 1 is used to determine___

p+q=1

equation 2 is used to determine___

p^2+2pq+q^2=1

back 128

-frequencies of alleles in population

-frequencies of genotypes in the population

front 129

What are the Hardy-Weinberg Equilibrium Assumptions ?

back 129

-no selection

-no mutation

-no gene flow

-random mating

-infinite population size

front 130

0.000625 Caucasians suffer from cystic fibrosis an autosomal recessive disease.

(1)What is the frequency of the disease-causing allele in the population?

(2) find value of p

(3)What proportion of individuals are carriers?

back 130

-(1) q^2 =0.000625, q=√0.000625 =0.025

-(2) p+q=1 -> p=1-0.025= 0.975

-(3) 2pq= 2 x 0.975 x 0.025 = 0.049 , about 5% carry the cystic fibrosis mutation

front 131

in cats, all-white is dominant over colors other than all white. In a population of 100 cats, 19 are all white. Assuming that the population is in HArdy-Weinburg equilibrium, what is the frequency of the all-white allele in this population?

How many white cats would be expected to be genotype AA? Aa?

back 131

-white =19 out of 100, so 81 of the cats are not white(recessive)

-81/100 = q=0.9

-p=1-0.9 = 0.1 are white and dominant

front 132

Can cause severe changes in allelic frequencies and are more severe in smaller populations

back 132

Genetic drift

front 133

Will alter the genotype frequencies, but not the allele frequencies of a population

back 133

nonrandom mating

front 134

A tendency of like individuals to mate

-creates more homozygotes then expected

-flies with same eye color mate

back 134

Positive assortative mating

front 135

A tendency of unlike individuals to mate.

-creates more heterozygotes than expected

-fly with red eyes mates with white eyes

back 135

Negative assortative mating

front 136

-increases the likelihood of getting two copies of the same allele from the same ancestor

results in more homozygotes

back 136

inbreeding

front 137

a measure of how likely it is that two alleles are identified by descent

back 137

inbreading coefficient (F)

front 138

What is this an example of?

back 138

Alleles identical by descent

front 139

What is this an example of?

back 139

alleles identical by state

front 140

The more inbreeding the less

back 140

heterozygotes

front 141

Caused by accumulation of harmful recessive alleles

back 141

Inbreeding depression

front 142

What equations are needed when calculating with inbreeding factors?

back 142

f(AA)= p^2 + Fpq

f(Aa)=2pq-2Fpq

f(aa)= q^2 +Fpq

front 143

What will the genotype frequencies be after 1 generation of random mating, followed by 1 generation of inbreeding between siblings?(F=0.25)

p=0.6 q=0.4

back 143

Fpq= 0.25 x 0.6 x 0.4=0.06

F(AA)= p^2 =0.6^2=0.36 +0.06= 0.42

F(Aa)= 2pq= 2 x 0.6 x 0.4=0.48 -(2 x 0.06) =0.36

F(aa)=q^2=0.4^2=0.16 + 0.06 = 0.22

front 144

-Doesn't change allele frequencies by itself but changes distribution of who has them

-more homozygous

-recessive alleles that are not good will more likely show

back 144

non-random mating

front 145

Individuals from one population entering another population

back 145

migration

front 146

The amount of change in allelic frequency due to migration between populations depends on

back 146

the difference in allelic frequency and the extent of migration

front 147

changes in allele frequency due to different levels of fitness (ability to survive and reproduce)

back 147

Natural selection

front 148

One allele is always more fit than the other

-the more "fit" allele increases in frequency

back 148

Directional selection

front 149

The heterozygote is more fit than either homozygotes

-stable equilibrium reached and both alleles remain in population

back 149

Overdominance (heterozygote advantage)

front 150

Heterozygote is less fit than either homozygotes

-unstable equilibrium reached- one allele will eventually win

back 150

Under dominance (heterozygote disadvantage)

front 151

What will be the allele frequncies at the time reproduction beings ?

back 151

1. divide by the most fit number in this case 180

AA=180/180=1

Aa=162/180= 0.9

aa= 36/180= 0.2

2. A=(180+180+heterozygote)/(2x total number of individuals)

(180 + 180 +162) / (162+180+36) x 2= 0.69

a= (36 +36+ 162) / (756)= 0.31

front 152

w11<w12>w22 is an example of what?

back 152

-overdominance

front 153

w11<w12<w22 is an example of ?

back 153

Directional selection against incompletely dominant allele A1

front 154

w11>w12>w22 is an example of?

back 154

Directional selection against incompletely dominant allele A2

front 155

w11=w12<w22 is an example of?

back 155

directional selection against dominant allele A1

front 156

w11=w12>w22 is an example of

back 156

directional selection against recessive allele A2

front 157

back 157

consider this when looking at distributions