Bis 101 Exam 1- Uc Davis

Front 1

How is genetic information organized?

Back 1

DNA-> Genes-> Chromosome pair-> Cell Nucleus-> cells-> organism

Front 2

What is a gene?

Back 2

A unit of inheritable information (segments along a DNA)

Front 3

Genotype

Back 3

The composition of a gene, or set of genes, for an individual

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Phenotype

Back 4

The physical manifestation of a specific genotype

Front 5

Gene expression

Back 5

Going from genotype to phenotype. A particular gene turns on. Production of an endpoint product driven by information encoded in DNA

Front 6

Central Dogma

Back 6

Transfer of sequence (genetic) info between carrying molecules

replication

DNA------------------>RNA--------------------->PROTEIN

transcription translation

Front 7

Assume that a single gene controls eye color.How many copies of the eye color gene doyou have?

Back 7

2

Front 8

An allele is:

Back 8

A particular variant of a gene

Front 9

In the diploid cells of an organism, there can be_________ different allele(s) of a given single-copynuclear gene.

Back 9

Two. Can have one or two but only up to two

Front 10

Forward Genetics

Back 10

Start with phenotype to inform gene function

Front 11

Reverse Genetics

Back 11

Start out with gene, mutate it, then look at the phenotype

Front 12

Which of the following are previous prevailing models ofinheritance?

A. Traits from both parents blend together in offspring

B. Blended traits are passed on to the next generation

C. Offspring are a combination of both parentsD. Offspring inherit traits from each parent in the form ofdiscrete particles

E. All of the above

Back 12

E

Front 13

Which of the following models are supported by Mendel’sexperiments?

A. Traits from both parents blend together in offspring

B. Blended traits are passed on to the next generation

C. Offspring inherit traits from each parent in the form ofdiscrete particles

D. Both A and B

E. None of the above

Back 13

to be determined

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blank

Back 14

blank

Front 15

True Breeding

Back 15

homozygous (all progenies they create are same)

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Test Cross

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cross to homozygous recessive

Front 17

Mendels 1st Law

Back 17

Each parent has 2 characters (genes) ie. discrete particles.

-Each gamete gets on character

-That organism combines to get 2 traits (gametes combine)

Front 18

Monohybrid

Back 18

heterozygous for one gene trait (3:1)

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Dihybrid

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heterozygous for two genes/traits

(9:3:3:1)-phenotypes

Front 20

Mendels 2nd Law

Back 20

characters (genes) of a trait segregate independently of each other

Front 21

Mendel also did experiments with the flowering plant‘Four-O-Clocks’. In this plant, yellow flowers are dominantto white flowers. When a certain yellow-flowered plantwas self-pollinated, its progeny consisted of 77 yellow- and25 white-flowered plants. What was the genotype of theoriginal yellow-flowered parent?

A. YY

B. Yy

C. yy

D. Either YY or Yy

Back 21

B

Front 22

What proportion of the 77 yellow-flowered progenyfrom the selfing will breed true?

A. 1/4

B. 1/3

C. 1/2

D. 2/3

Back 22

B (?)

Front 23

If the original yellow-flowered parent had beenused in a testcross (instead of selfing), what isthe expected genotypes, and proportions ofthe progeny:

A. 3/4 YY : 1/4 yy

B. 1/2 Yy : 1/2 yy

C. 3/4 Yy : 1/4 yy

D. 1/4 YY : 3/4 Yy

Back 23

B

Front 24

True or False?Mendel’s 2nd law, that characters (genes) of atrait segregate independently of each other, isalways true.

Back 24

False

Front 25

In mice, a recessive mutation in gene T (t) results in tail-lessanimals. A second unlinked recessive mutation in gene F (f)results in fat animals. Indicate the genotypes of the parentsof the following cross:

Phenotype of parents: Tail, Slim x tail-less, fatPhenotype of progeny: 24 Tail, Slim0 Tail, fat25 tail-less, Slim0 tail-less, fat

A. TtFf x Ttff

B. TtFf x ttFf

C. TtFF x ttff

D. Ttff x ttFf

Back 25

C

Front 26

Discontinuous trait

Back 26

traits with only a few possible phenotypesthat fall into discrete classes.Example: all of the traits Mendel tracked

Front 27

Continuous trait:

Back 27

do not fall into discrete classes; a segregatingpopulation will show a continuous distribution of phenotypes.Example: height

Front 28

Law of segregation

Back 28

During meiosis, the alleles for each gene segregate fromeach other so that each gamete carries only one allelefor each gene

Aa-A

-a

Front 29

Law of independent assortment

Back 29

Different gene pairs (aka different chromosome pairs) assort independently into gametesduring meiosis

Front 30

Mitosis

(be able to draw)

Back 30

Interphase->Prophase->Metaphase->Anaphase->Telophase->Daughter Cells

Interphase (2N)->Prophase (4N) = DNA replication

Metaphase (line up)->Anaphase (separate poles)->Telophase (divide into two new cells) = Segregation

Left with two daughter cells (each 2N)

Front 31

Which of the following statements are true?

A. Chromosome segregation patterns in mitosis are the basisfor Mendel’s law of segregation.

B. Chromosome segregation patterns in mitosis are the basisfor Mendel’s law of independent assortment.

C. Mitosis generates two daughter cells that are geneticallyidentical to the parent cell.

D. In mitosis, homologous chromosomes segregate away fromone another

E. All of the above

Back 31

C

D not true because sister chromosomes not homologous chromosomes segregate away from one another; sister is AA or Aa, homologs are Aa

Front 32

Synapsis

Back 32

pairing of homologous chromosomes (in meiosis)

Front 33

Crossover

Back 33

physical linkage between homologs

Front 34

Meiosis

Back 34

Meiosis I- Interphase->Prophase 1-> Metaphase 1->Anaphase 1->Telophase 1

Meiosis II- Telophase 1->Prophase 2->Metaphase 2->Anaphase 2->Telophase 2->Products of meiosis

Meiosis 1:Interphase= replication; Metaphase 1= pairing; Anaphase 1 and Telophase 1= segregation

Meiosis 2: Metaphase 2->Anaphase 2->Telophase 2= segregation

Front 35

Which of the following statements is true?

A. Homologous chromosomes associate only in meiosis.

B. DNA replication occurs only in mitosis.

C. Sister chromatids separate only in meiosis.

D. Chromosomes line up on the cell equator only inmeiosis.

E. DNA replication occurs twice in meiosis

Back 35

A

Front 36

A diploid cell contains three pairs of homologouschromosomes. Each pair is heterozygous for a pairof alleles: Aa, Bb and Cc respectively. After meiosis,how many different combinations of these allelescould be produced in the haploid gametes?

A. 2

B. 4

C. 8

D. 16

E. 64

Back 36

C-8

B/C 2 x 2 x 2=8

___ x ___ x____ =

each one has two possible combinations

Front 37

You’ve isolated a mutant defective in meiosis that displays unpairedhomologous chromosomes in meiosis I. For this mutant, what doyou predict the wild type gene encodes?

A. A protein required for kinteochore formationB. A protein that is part of sister chromatid cohesion

C. A protein that is part of the synaptonemal complex (SC)

D. A protein required for DNA replication.

E. All of the above

Back 37

C- because synaptonemal complex is meiosis

SCC is mitosis

Front 38

Non-Disjunction

Back 38

the failure of homologouschromosomes to segregate properly in meiosis

Front 39

True or False: All traits will segregate the same regardless of whetherthey are inherited from the maternal or paternal parent.

Back 39

False b/c some traits are linked to Y chromosome

Front 40

Turners syndrome

Back 40

XO

Front 41

Klinefelter’s Syndrome

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XXY

Front 42

Wild type

Back 42

most common variant

Front 43

You are a fly geneticist and notice two different fly variants in yourbottles: one fly is uncoordinated (u) and the other fly has orange eyes(o). You are interested in determining the nature of these characters.You propagate the flies to obtain true breeding lines. You then crossan orange-eyed female to an uncoordinated male. You find thatall the F1 flies have brick-red eyes and are coordinated.Indicate all possible genotypes for the parents (P) and offspring (F1):

A. P: o/o; u/u (female) x o+/o+; u+/u+ (male) F1: o+/o; u+/u

B. P: o+/o+; u+/u+ (female) x o/o; u/u (male) F1: o+/o; u+/u

C. P: o/o; u+/u+ (female) x o+/o+; u/u (male) F1: o+/o; u+/u

D. P: o+/o+; u/u (female) x o/o; u+/u+ (male) F1: o+/o; u+/u

E. C and D

Back 43

C

Front 44

As a good geneticist, you perform the reciprocal cross (uncoordinatedfemales x orange-eyed males) and obtain red-eyed coordinatedfemales and red-eyed uncoordinated males. Based on thisinformation, indicate the genotype of parents (P) and offspring (F1):

A. P: o+/o+; Xu/Xu x o/o; Xu+/Y F1: o+/o; Xu+/Xu and o+/o; Xu/Y

B. P: o+/o+; u/u x o/o; u+/u+ F1: o+/o; u+/u

C. P: o/o; u+/u+ x o+/o+; u/u F1: o+/o; u+/u

D. P: u/u; Xo+/Xo+ x u+/u+; Xo/Y F1: u+/u; Xo+/Xo and u+/u; Xo/Y

E. C and D

Back 44

A

Front 45

Diseases due to a single mutation/gene(monogenic traits)

Back 45

Cystic Fibrosis

Sickle-cell anemia

Huntington’s disease

Duchenne Muscular Dystrophy

Front 46

Diseases due to multiple mutations/genes:

Back 46

Cancer

Diabetes

Mental Disorders

Autism

Front 47

What type of inheritanceis this?1

A) Autosomal recessiveB) Autosomal dominantC) X-linked recessive

D) X-linked dominant

E) None of these

Back 47

A

Front 48

Autosomal Recessive

Back 48

Affected children often from unaffected parents.• On average 1/4 of children are affected in mating of 2 heterozygotes.

• Equal numbers of affected males and females.• Depending how rare the affected allele is, matings of affected individuals tounrelated, unaffected individuals usually produce normal offspring.Conversely, consanguineous marriages have a much higher chance to produceaffected offspring.

Front 49

Autosomal Dominant

Back 49

• Affected children have at least one affected parent.

• Usually 1/2 children are affected. Most dominant disease traits are rare so mostaffected individuals are heterozygotes.

• Equal numbers of affected males and females.• Except for the very rare case of a new mutation, unaffected parents do not produceaffected children (or the disease is not fully penetrant).

Front 50

What type of inheritanceis this?

A) Autosomal recessive

B) Autosomal dominant

C) X-linked recessive

D) X-linked dominant

E) None of these

Back 50

C

Front 51

X linked recessive

Back 51

• Many more males than females affected.

• None of the offspring of an affected male are affected.

• None of the sons of an affected male are affected or will pass on the disease allele.

Front 52

X linked dominant

Back 52

• Affected males have daughters that are all affected.

• Affected heterozygous females pass the disease to 1/2 sons and1/2 daughters. Most dominant disease traits are rare so thatmost affected individuals are heterozygotes.

Front 53

What type of inheritance is this?

A) Autosomal recessiveB) Autosomal dominantC) X-linked recessive

D) X-linked dominant

E) None of these

Back 53

E

Front 54

Creighton and McClintock:

Back 54

Evidence that recombination of geneslinked on a chromosome occurs through physical exchange ofchromosome segments with its homologous partner

Front 55

Linkage and Recombination

Back 55

1. Parental phenotype is more common than recombinant inmost crosses.

2. During meiosis, some alleles assort together if they areadjacent to each other on the chromosome.

3. The closer two genes physically are on a chromosome, themore likely they segregate together in meiosis.

4. Recombinant progeny are produced by crossing over inmeiosis.

Front 56

Crossing Over

Back 56

When homologs pair at Meiosis 1, they exchange corresponding chromosome parts by breakage and reunion

Front 57

Parental:

Back 57

chromosome allele configuration is identical to what parent had- 2 or more genes

Front 58

Recombinant

Back 58

different combination of linked alleles than parents

Front 59

True (A) or False (B)Crossing over is more likely to occur betweengenes that are farther apart on a chromosome.

Back 59

True

Front 60

You perform the cross:AB/ab x ab/aband obtain 1000 progeny in the following amounts:442 AB/ab 458 ab/ab46 Ab/ab 54 aB/ab. What is the map distance between these two genes?A. 10 cMB. 20 cMC. 40 cMD. 50 cM

Back 60

A. 10 cM

46 + 54= 100

100/1000=.1

.1x100= 10

Front 61

Gene Linkage

Back 61

genes reside on the same chromosome ~in close proximity to each other

Front 62

chiasmata

Back 62

physical exchange in crossing over

Front 63

Cis

Back 63

WT (or mutant, as long as same 2 alleles) alleles are on the same homolog RT/rt

Front 64

Trans

Back 64

WT (or mutant) alleles are on opposite homologs Rt/rT

Front 65

Recombination Frequency

Back 65

The % for which you had recombination between the two loci

Front 66

If genes are unlinked

Back 66

1:1:1:1 parental to recombinant

Recombination Frequency >50%= unlinked

RF<50%= linked- 2 genes on same chromosome

distance is directly proportional to frequency (greater one distance is greater other)

Front 67

RF from a three points cross allows us to determine

Back 67

1. If genes are linked

2. if so, The physical distance between loci

3. The phsyical order of genes on a chromosome (or physical order of loci)

Front 68

Solving three point cross problems

Back 68

1. Define parental class

2. Define the double cross-over class - least frequent

3. Define order of loci in parents (informed by DCO) (gene that switched linkage= middle locus)

4. Determine linkage by add recombinants/ total # progeny

Front 69

Back 69

B

Front 70

Back 70

A

Front 71

Back 71

A

Front 72

Interference

Back 72

A crossover/region of the genome (reduces) affects the likelihood of another CO nearby

I=1-COC (coefficient of confidence)

COC=# observed DCO/expected DCO (based data)

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Back 73

D

Front 74

Back 74

D

Front 75

Simple Trait

Pleiotropy

Complex Trait

Back 75

Simple Trait- one gene-> one phenotype

Pleiotropy- one gene-> many phenotypes

Complex Trait- many genes-> one phenotype

Front 76

Back 76

B

each would have wild type copies of the other- called complementation

Front 77

Complementation Test

Back 77

allows us to sort mutants in groups (genes).

alleles of the same gene will NOT compliment

Mutations in different genes compliment