Genetics exam 2 Flashcards


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created 7 weeks ago by sariA04
Ch. 6, 7, 8, 11, 18, and 25
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1

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

Pedigree

2

Why is human genetics hard to follow?

-a small number of offspring

-experimental crosses not ethical

-long generation time (22-23 yrs)

3
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4
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..

5

What are the 4 basic inheritance patterns?

-Autosomal recessive

-Autosomal dominant

- Sex-linked recessive

-Sex-linked dominant

6
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-2 unaffected parents can have affected offspring

-both male and female offspring affected at similar freq.

Autosomal recessive

7

What are examples of autosomal recessive disorders?

-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)

8
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-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

Autosomal dominant

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-unaffected parents can have affected offspring

-Males affected more frequently than females

-Transmitted from mothers to sons

-50% of child will be affected

Sex-linked recessive

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Example of__________

Sex-linked recessive

11

color blindness is___linked and affects _____more than _______

-x

-male

-female

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-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

Sex-linked dominant

13

What are examples of sex-linked dominant disorders?

-Alport syndrome

-Coffin-Lowry syndrome

-Fragile X syndrome

-Incontinentia pigmenti

-Vitamin D resistant rickets

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-only males have it

-pass to all sons, none of daughters

Y-linked trait (dominant)

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-Mothers always pass on

-Fathers never pass on

Mitochondrial mutation

16

What is the chance of their next child being affected?

(a)= affected

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

-gender x condition

17

Look at slide 34 in Ch 6. pt 1

DO IT NOW

18

-non identical twins

-2 zygotic eggs

Dizygotic twins

19

-identical twins

Monozygotic twins

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

concordant traits

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-separate embryotic sacks

dizygotic twinning

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-comes from 1 zygote

-early splitting

Dichorionic diamniotic (monozygotic twinning)

23
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-late split

Monochorionic diamniotic (monozygotic twinning)

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-very late split

-same sack

Monochorionic monoamniotic (monozygotic twinning)

25

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

-overweight

-consistent association

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.

Environmental impact

27

Genes on the same chromosomes tend to

get transmitted together

28

certain combinations of alleles are passed more frequently than others

linkage

29

Complete linkage leads to;

____% non recombinant gametes and progeny

100

30

the crossing over between ______ chromosomes can break linkage

homologous

31

Alleles are passed in different combinations from how they were inherited

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Recombination

32

Recombination leads to:

-recombinant gametes and progeny

33

________ associated with physical change in chromosomes

recombinant phenotypes

34
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recombinant diagram

35
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When a plant is normal with tall leaves and short with dwarf leaves, they are

nonrecombinant progeny

36
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This is an example of

no crossing over

37
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This is an example of:

which results is;

crossing over

-2 nonrecombinant and 2 recombinant gametes

38
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One chromosome contains both wild-type alleles, one chromosomes contains both mutant alleles

coupling (cis configuration)

39
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wild type allele and mutant alleles are found on the same chromosomes

repulsion (trans configuration)

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

determining genetic distance

41

How to calculate recombination frequency

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

42

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

independent assortment

43

Two genes are linked if they are less than____ apart

50cM

44

If the progeny resulted in:

-66

-77

-28

-33

Which are recombinant, and what would the distance be?

-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

45

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

interference

46

How to calculate amount of interference:

coefficient of coincidence= observed DCO/expected DCO =

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

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

47

Distances based on crossover/linkage

genetic map

48

Distances based on actual base-pair differences

physical map

49
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This is an example of

crossovers with three linked loci

50

recombination distance are ______accurate the larger they get

less

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Consider the mutation and cross over

How halotype works

52
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consider how when the top of loci is blue, it shows the blue parents trait

Quantitative trait locus mapping (QTL)

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-controls have little to no genes of the condition

-lower freq. of variant

-variant linked to condition

-cases have the condition

Genome-wide association studies (GWAS)

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Used to assign a gene to a particular human chromosome

-gene present (given)

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

Somatic-cell hybridization

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displays the mutant phenotype

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displays the wild type

57

-prepared from cells halted in metaphase(relaxed)

- numbered based on size

-Sex chromosomes

-based on haploid # (23 chromosomes)

Karyotyping

58

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

-X

-Y

59

-moving, gaining, or losing pieces of a chromosome

-2n=6

-duplication

rearrangement

60

-changes in # of copies of an individual chromosome

-2n+1=7

-trisomy

Aneuploidy

61

-extra whole set of chromosomes

-3n=9

-autotriploid

polyploidy

62

Types of rearramgement:

-duplication

-deletion

-translocation

-inversion

63

Copied region of chromosomes

duplication

64

deleted region of chromosomes

deletion

65

inverting (flipping) region of chromosomes

inversion

66

moving part of chromosome somewhere else

translocation

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-bulge during pairing in meiosis 1

Duplication (segment of chromosome is duplicated)

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-formation of loop during pairing in prophase 1

Deletion

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-turned 180 degrees

inversion

70

Inversion that doesn't include the centromere

paracentric inversion

71

inversion includes the centromere

pericentric inversion

72

If a crossover were to occur in the inverted region,

it would result in a complication

73

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

The ultimate storage problem

74

Is overwound or underwound causing it to twist on itself

Supercoiling DNA

75

-add two turns

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

Positive supercoil

76

-removes 2 turns

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

Negative supercoiling

77

What do these charactericstics describe?

-less condensed

-on chromosome arms

-unique sequences

-many genes

-throughout S phase

- often transcription

- crossing over is common

Euchromatin

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

Heterochromatin

79

Regions of relaxed chromatin where active transcription is taking place

chromosome puffs

80

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

-salivary glands

81

Chromosome fragments that lack ______ are lost in mitosis

centromeres

82

Proposes that mitochondria and chloroplasts in eukaryotic cells arose from bacteria

Endosymbiotic theory

83
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endosymbiotic theory

84

Organelles in a ______cell segregate randomly into the progeny cells.

heteroplasmic

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understand it

86

What is the importance of mutations?

-provides raw material for evolution

-source of many diseases and disorders

-look at whats wrong when we alter a gene

87

-Do not get passed on to the offspring

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

Somatic mutations

88

-reproductive cell mutations

-passed on

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

germ-line mutation

89

One letter is replaced by another

base substitution

90

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

Transition

91

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

Transversion

92

Frameshift mutations are not_____by 3

divisible

93

in frame insertions and deletions are divisible by

3

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?

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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?

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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?

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97

Repetitive sequence can cause______ of strands during replication

slippage

98

Dna polymerase adds

more repeats

99

unequal cross over misaligns

homologous chromosomes

100
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The new codon encodes a different amino acid; there is a change in amino acid sequence

Missense mutation

101
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The new codon is a stop codon; there is a premature termination of translation

Nonsense mutation

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

Silent mutation

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.

Suppressor mutation

104

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

intragenic

105

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

Intergenic

106

Purines and pyrimidine bases exist in different forms called

tautomers

107

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

EX: T and G binding (wobble)

C and A protonated wobble

Nonstandard base

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____ base pairing leads to a replication error

Wobble

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____ and ____may result from strand slippage

-insertions

-deletions

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____ _____ over produces insertions and deletions

unequal crossing over

111

loss of purine

Depurination

112

loss of an amino group

deamination

113

Used to identify chemical mutagens

Ames test

114

A___ can result in possible wobble mispairing

bulge

115

Many incorrectly inserted nucleotides that escape proofreading are

corrected by mismatch repair

116

Changes nucleotides back into their original structures

Direct repair

117

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

-DNA glycosylase

-AP endonuclease

118

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

-polymerase

-ligase

119
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base excision

120

Removes bulky DNA lesions that distort double helix

Nucleotide-excision

121

Can cause chromosomal rearrangement

transposons

122
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example of transposons causing rearrangement

123

How to calculate the frequencies of observed genotypes?

-# of individuals with genotype divided by total # individuals

f(AA)=

f(Aa)=

f(aa)=

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

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

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

124

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

f(AA)= 482/868=0.555

f(Aa)= 213/868=0.245

f(aa)= 173/868=0.199

125

How to calculate observed allelic frequencies

-# copies of alleles divided by total # alleles

f(A)=

f(a)=

-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)

126

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

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

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

127

What does p and q stand for in this equation?

p+q=1

-p: dominant

-q: recessive

128

equation 1 is used to determine___

p+q=1

equation 2 is used to determine___

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

-frequencies of alleles in population

-frequencies of genotypes in the population

129

What are the Hardy-Weinberg Equilibrium Assumptions ?

-no selection

-no mutation

-no gene flow

-random mating

-infinite population size

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?

-(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

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?

-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

132

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

Genetic drift

133

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

nonrandom mating

134

A tendency of like individuals to mate

-creates more homozygotes then expected

-flies with same eye color mate

Positive assortative mating

135

A tendency of unlike individuals to mate.

-creates more heterozygotes than expected

-fly with red eyes mates with white eyes

Negative assortative mating

136

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

results in more homozygotes

inbreeding

137

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

inbreading coefficient (F)

138
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What is this an example of?

Alleles identical by descent

139
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What is this an example of?

alleles identical by state

140

The more inbreeding the less

heterozygotes

141

Caused by accumulation of harmful recessive alleles

Inbreeding depression

142

What equations are needed when calculating with inbreeding factors?

f(AA)= p^2 + Fpq

f(Aa)=2pq-2Fpq

f(aa)= q^2 +Fpq

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

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

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

non-random mating

145

Individuals from one population entering another population

migration

146

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

the difference in allelic frequency and the extent of migration

147

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

Natural selection

148

One allele is always more fit than the other

-the more "fit" allele increases in frequency

Directional selection

149

The heterozygote is more fit than either homozygotes

-stable equilibrium reached and both alleles remain in population

Overdominance (heterozygote advantage)

150

Heterozygote is less fit than either homozygotes

-unstable equilibrium reached- one allele will eventually win

Under dominance (heterozygote disadvantage)

151
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What will be the allele frequncies at the time reproduction beings ?

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

152

w11<w12>w22 is an example of what?

-overdominance

153

w11<w12<w22 is an example of ?

Directional selection against incompletely dominant allele A1

154

w11>w12>w22 is an example of?

Directional selection against incompletely dominant allele A2

155

w11=w12<w22 is an example of?

directional selection against dominant allele A1

156

w11=w12>w22 is an example of

directional selection against recessive allele A2

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consider this when looking at distributions