Midterm Exam 2

Question 1

The main steps in a PCR reaction are, in order:

Answer 1

Denaturing of the target DNA, addition of primers (single stranded oligonucleotides), annealing of the primers to the DNA, and primer extension

Question 2

Having multiple copies of a gene may translate into a stronger phenotype but this is not always true.

Answer 2

True

Question 3

IF the presence of aroma was always associated with susceptibility to a disease, which of the following is a most likely explanation?

Answer 3

Pleiotropy

Question 4

The reason restriction enzymes are so useful is that they make cuts at specified intervals (e.g. every 20 bp for EcoRI) rather than at specific recognition sequences in the DNA.

Answer 4

False

Question 5

The presence of aroma in rice, according to the assigned paper, is due to

Answer 5

a premature stop codon in the BAD-2 gene

Question 6

The highest temperature (~94oC) in a PCR reaction is used for which of the following

Answer 6

Denaturing template DNA

Question 7

Reverse transcriptase is an enzyme that would most likely be used to create which type of DNA library?

Answer 7

cDNA

Question 8

A molecular marker that is linked to a gene controlling a trait of interest can be useful for marker assisted selection but not as a starting point for eventually cloning the gene of interest.

Answer 8

False

Question 9

A BAC contig is a set of non-overlapping BAC clones created by fragmenting a single sequence of interest into fragments of identical size and then cloning the fragments.

Answer 9

False

Question 10

Where do you think it is most likely to find SSR polymorphisms?

Answer 10

UTRs, introns, or intergenic DNA

Question 11

Molecular marker and morphological marker polymorphism data are both useful for genetic analysis, but molecular markers are more often used because they are more abundant.

Answer 11

True

Question 12

Silent, Missense, and Nonsense mutations can all be generated by SNP polymorphisms

Answer 12

True

Question 13

What is the relationship between restriction enzyme recognition sequence length and the average frequency (in terms of base pairs) at which that enzyme will cut in a sample of genomic DNA?

Answer 13

Shorter length = higher frequency

Question 14

For most sequencing techniques, PCR is a required pre-step, with the goal of amplifying the target DNA that will be sequenced.

Answer 14

True

Question 15

Which of the following suggests that the BAD2 gene is involved in fragrance in rice?

Answer 15

Sequence alignments of the fgr region between a non-fragrant and a fragrant variety revealed polymorphisms in the BAD2 gene.

Question 16

In most plants, a first step in learning how many genes determine a phenotype (either qualitative or quantitative) is to:

Answer 16

clone the genes determining the phenotype

Question 17

If a molecular marker is monomorphic (i.e. no polymorphism) in a particular cross, this indicates that the two parents have the same alleles at the marker locus.

Answer 17

True

Question 18

Mutations are rare and usually deleterious. However, genetic analysis requires abundant DNA polymorphisms. Where would you most likely expect to find abundant polymorphisms that do not lead to major effects on phenotype?

Answer 18

UTRs

Question 19

Non-coding regions of the genome may be more likely to be polymorphic than coding regions and thus more efficient for generating molecular markers. Which of the following is the most likely explanation for this observation?

Answer 19

Mutations are generally random events and since changes in the sequence of non-coding regions are less likely to cause changes in phenotype, these mutations will persist.

Question 20

In the “The gene for fragrance in rice” the authors found that 71% of rice genes were homologs to genes in Arabdopsis. The existence of homologs between two species suggests:

Answer 20

A common ancestor

Question 21

All rice varieties we know of today are cross-pollinated and originated in Asia

Answer 21

False

Question 22

Which of the following suggests that the BAD2 gene is involved in fragrance in rice?

Answer 22

Sequence alignments of the fgr region between a non-fragrant and a fragrant variety revealed polymorphisms in the BAD2 gene.

Question 23

There is not a conversion factor between linkage distance and physical distance. A distance of 1 centimorgan may correspond to 5 kb in one part of the chromosome and 500 kb in another part of the chromosome.

Answer 23

True

Question 24

cDNA libraries are different than gDNA libraries in that they contain only:

Answer 24

DNA sequences that can potentially be expressed

Question 25

In “The gene for fragrance in rice” paper, the authors conclude that the functional basis for the difference between aroma vs. non-aroma alleles is probably

Answer 25

a premature stop codon in the aromatic allele due to deletions which caused a frameshift mutation.

Question 26

A key consideration in choosing a vector (for example, a plasmid or a bacterial artificial chromosome) for propagating and maintaining a population of DNA fragments is

Answer 26

the size of the DNA insert

Question 27

The rice BAD (betaine aldehyde dehydrogenase) gene is part of ____ - a group of genes with similar sequence and function, which commonly arise as a result of gene or genome duplication events.

A gene cluster

Answer 27

A gene family

Question 28

If you wanted to study the inheritance of rice, you could assess aroma in the F2 generation and use a chi square test to determine the number of genes involved. However, you would not be able to determine the DNA basis of the aroma vs. non-aroma, using this method.

Answer 28

True

Question 29

Which of the following is FALSE about the Sanger sequencing technique?

Answer 29

The reading frame of the gel starts at the longest fragment and ends at the shortest fragment

Question 30

What makes TAQ polymerase so special and useful for PCR?

Answer 30

Works at high temperatures

Question 31

Restriction enzymes:

Answer 31

Are isolated from bacteria, where they are part of a defense system against pathogens

Question 32

Molecular marker polymorphisms are always within the coding regions of genes.

Answer 32

False

Question 33

Rice is a model plant for the grasses because of its:

Answer 33

Small genome size, short generation time, amenability to genetic transformation, and economic importance

Question 34

In the “The gene for fragrance in rice” the authors found that 71% of rice genes were homologs to genes in Arabdopsis. The existence of homologs between two species suggests:

Answer 34

A common ancestor

Question 35

Genetic maps (also called linkage maps) are in units of kilo base pairs, while physical maps are in units of centimorgans.

Answer 35

False

Question 36

A Simple Sequence Repeat (SSR) marker is based on

Answer 36

conserved sequences flanking a repeat motif sequence

Question 37

The main reason why the corn genome is much bigger than the rice genome is that the corn genome contains more:

Answer 37

Transposable elements

Question 38

According to the authors of “The gene for fragrance in rice” paper, there is no evidence that the BAD gene has pleiotropic effects on fragrance and stress tolerance. In other words, fragrance and stress tolerance seem to:

Answer 38

Not be determined by the same gene

Question 39

Changes in DNA sequence that lead to new molecular marker alleles at a locus may not cause changes in phenotype.

Answer 39

True

Question 40

In the PCR process which of the following determines what DNA sequence will be amplified by the polymerase?

Answer 40

primers

Question 41

What is the role of vectors (e.g. plasmid or BAC) in DNA library preparation?

Answer 41

propagate and maintain DNA fragments generated by the restriction digestion

Question 42

The Illumina HiSeq is based on the principle of ionic current modulation.

Answer 42

False

Question 43

Which sequencing technique produces the longest reads?

Answer 43

Nanopore

Question 44

In the Ilumina HiSeq, each DNA fragment results in:

Answer 44

2 short reads, one from the 3’end and another from the 5’ end

Question 45

Illumina sequencing is a better option for determining the correct sequence of recessive alleles due to loss-of-function mutations than Nanopore sequencing, because Nanopore sequencing requires simultaneous integration of phenotype and genotype.

Answer 45

False

Question 46

In the Illumina Hiseq technique, it is essential that the DNA fragments are abundantly replicated in clusters because the fluorescence signal from one DNA molecule alone is not strong enough to be visualized. Which of the following plays an essential role in ensuring that the fragments are replicated in clusters?

Answer 46

Adaptors

Question 47

The principle of sequencing by ionic current modulation is mainly based on the disruption of the ionic current being independent of the shape of the molecule (A, T, C or G) translocating through the pore.

Answer 47

False

Question 48

Which of the following is FALSE about the Illumina HiSeq technique?

Answer 48

Fluorescent tags are added on the terminal phosphate of each nucleotide

Question 49

If I wanted to look for polymorphisms between two protein sequences, which would be the best platform to do this?

Answer 49

BLASTp

Question 50

If I wanted to find the DNA sequence of the Vrs1 gene in barley, which is the best platform to conduct this search?

Answer 50

NCBI Nucelotide

Question 51

In a NCBI nucleotide entry, if you click on cds you will get:

Answer 51

The sequence of a gene, including introns (if the gene has introns and the sequence is from genomic DNA).

Question 52

The plasmid libraries used for PACBio and Nanopore sequencing are unique in that the plasmid cloning sites have overhangs.

Answer 52

False

Question 53

An Illumina sequencing library and a BAC library differ in that the Illumina library is used for the sequencing project and then is no longer available, whereas a BAC library can be conserved and re-used.

Answer 53

True

Question 54

You are interested in aroma in mint and you know that the BAD2 gene is responsible for aroma in rice. You are wondering if there is a similar gene in mint. What can you do?

Answer 54

BLAST the BAD2 gene sequence against the mint genome sequence.

Question 55

Which of these techniques is NOT based on the principle of sequencing by DNA synthesis?

Answer 55

Nanopore

Question 56

In the PAC Bio sequencing technique:

Answer 56

The added nucleotides are fluorescence tagged from their phosphate terminal

Question 57

If your goal was to get as much long-read genome sequence as possible at the best price and in the shortest possible time for your favorite plant, which of the following methods would you choose?

Answer 57

Pacific Biosciences (PAC Bio)

Question 58

Which sequencing technique utilizes chain termination?

Answer 58

Sanger

Question 59

In PAC Bio sequencing, having the fluorescence tag attached to the terminal phosphate of the nucleotide (instead of to the base of the nucleotide) allows for shorter reads.

Answer 59

False

Question 60

Which of the sequencing techniques can, at least theoretically, easily be done in the middle of an isolated field?

Answer 60

Nanopore

Question 61

Kary Mullis, who received a Nobel prize for his development of PCR, is a direct beneficiary of royalties from Nanopore sequencing because PCR technology is essential for this sequencing technique.

Answer 61

False

Question 62

If I wanted to look for polymorphisms between two protein sequences, which would be the best platform to do this?

Answer 62

BLASTp

Question 63

If I wanted to find the DNA sequence of the Vrs1 gene in barley, which is the best platform to conduct this search?

Answer 63

NCBI Nucleotide

Question 64

In the Ilumina HiSeq, each DNA fragment results in:

Answer 64

2 short reads, one from the 3’end and another from the 5’ end

Question 65

In the PAC Bio sequencing technique:

Answer 65

The added nucleotides are fluorescence tagged from their phosphate terminal

Question 66

Which of these techniques is NOT based on the principle of sequencing by DNA synthesis?

Answer 66

Nanopore

Question 67

In PAC Bio sequencing, having the fluorescence tag attached to the terminal phosphate of the nucleotide (instead of to the base of the nucleotide) allows for shorter reads.

Answer 67

False

Question 68

In the Illumina Hiseq technique, it is essential that the DNA fragments are abundantly replicated in clusters because the fluorescence signal from one DNA molecule alone is not strong enough to be visualized. Which of the following plays an essential role in ensuring that the fragments are replicated in clusters?

Answer 68

Adaptors

Question 69

Dideoxy nucleotides are feature of which sequencing technique:

Answer 69

Sanger

Question 70

In assembling a genome based on short reads, which of the following is most true?

Answer 70

It is very useful to have a reference genome to anchor the fragments to

Question 71

You are interested in aroma in mint and you know that the BAD2 gene is responsible for aroma in rice. You are wondering if there is a similar gene in mint. What can you do?

Answer 71

BLAST the BAD2 gene sequence against the mint genome sequence.

Question 72

The principle of sequencing by ionic current modulation is mainly based on the disruption of the ionic current being independent of the shape of the molecule (A, T, C or G) translocating through the pore.

Answer 72

False

Question 73

Which of these sequencing techniques does not require PCR to be done previous to sequencing?

Answer 73

Nanopore

Question 74

In a NCBI nucleotide entry, if you click on cds you will get:

Answer 74

The sequence of a gene, including introns (if the gene has introns and the sequence is from genomic DNA).

Question 75

Which of the sequencing techniques can, at least theoretically, easily be done in the middle of an isolated field?

Answer 75

Nanopore

Question 76

In the “The gene for fragrance in rice” paper, the authors found many strong lines of evidence that the BAD2 gene is responsible for aroma in rice. However, additional experimentation is required to prove the causal effect of this gene on the phenotype. This could be done by:

Answer 76

Knocking out the BAD2 gene of a non-aromatic rice and observing a phenotypic change.

Question 77

Which of the following is FALSE about the Illumina HiSeq technique?

Answer 77

Fluorescent tags are added on the terminal phosphate of each nucleotide

Question 78

Which sequencing technique produces the longest reads?

Answer 78

Nanopore

Question 79

Which sequencing technique utilizes chain termination?

Answer 79

Sanger

Question 80

If your goal was to get as much long-read genome sequence as possible at the best price and in the shortest possible time for your favorite plant, which of the following methods would you choose?

Answer 80

Pacific Biosciences (PAC Bio)

Question 81

The Illumina HiSeq is based on the principle of ionic current modulation.

Answer 81

False

Question 82

Reporter genes are included in transgenic constructs in order to

Answer 82

provide a mechanism for localizing and/or quantifying gene expression.

Question 83

A construct leading to low polyphenol oxidase activity in a diploid will be expected to show Mendelian inheritance if the transgenic plant is crossed with a variety that has high polyphenol oxidase activity.

Answer 83

True

Question 84

Genome editing is a technology that requires information about target genes in order to make specific changes in them that will result in changes in phenotype

Answer 84

True

Question 85

Introns are never included in transgene constructs because they do not have any function.

Answer 85

False

Question 86

In both hemizygotes and homozygotes for the CP4 EPSPS gene, the EPSPS enzyme (susceptible to gyphosate) is still produced, even though they are resistant to the herbicide.

Answer 86

True

Question 87

The RISC complex consists of

Answer 87

argonaute protein and a short (~ 21 nt) segment of RNA

Question 88

Which of the following is an example of a reporter gene?

Answer 88

Green fluorescent protein (GFP)

Question 89

The Tnos terminator sequence:

Answer 89

signals an end to transcription.

Question 90

An example of a constitutive promoter is one that allows a transgene to:

Answer 90

be expressed in all tissues all the time.

Question 91

In a TILLING population, the random mutations are a result of:

Answer 91

Errors in DNA replication and repair

Question 92

Coincidentally, the gene conferring resistance to glyphosate herbicide was found in the same organism that provided the gene for green florescent protein.

Answer 92

False

Question 93

Selectable markers (e.g. antibiotic resistance genes) are added to transgene constructs

Answer 93

to facilitate selection of cells that incorporated and are expressing the transgene

Question 94

In the CRISPR acronym, the “P” stands for:

Answer 94

Palindromic

Question 95

A chemical company puts the Roundup herbicide resistance gene into sugar beet. A sugar beet plant homozygous for the Roundup resistance gene cross-pollinates with an edible beet plant homozygous for the lack of the Roundup resistance gene. Assume Round-up resistance is dominant. You would expect the F2 of this cross to segregate

Answer 95

3 resistant: 1 susceptible

Question 96

The RNA primers that initiate DNA replication are encoded by the same genes encoding miRNAs

Answer 96

False

Question 97

If you use a promoter that is expressed only in floral tissue it would be best described as:

Answer 97

Tissue-specific

Question 98

In the context of RNAi, Dicer will cleave

Answer 98

Double stranded RNA

Question 99

The sentence: “In a diploid plant, once the CRISPR machinery is present in the genome, the edited sequence will be present in both homologous chromosomes.” applies:

Answer 99

Independent if the gene editing is conducted via transient or non-transient methods

Question 100

The CRISPR-cas9 system is potentially more attractive to plant geneticists that ionizing radiation or chemical mutagenesis because:

Answer 100

It is a targeted mutation system that has a lower likelihood of effects on non-target genes.

Question 101

The CRISPR-Cas9 gene editing system was developed based on a naturally occurring defense system in what type of organism:

Answer 101

Bacteria

Question 102

A key element needed for the development of glyphosate-resistant crops is that there must be sufficient expression of the CP4 EPSPS enzyme to overcome the expression of the EPSPS enzyme, because:

Answer 102

The CP4 EPSPS transgenic plant contains both the EPSPS gene as well as the CP4 EPSPS gene.

Question 103

Agrobacterium tumefaciens, it will automatically be introduced at the same position on homologous chromosomes.

Answer 103

False

Question 104

RNAi can involve a range of mechanisms, including transcriptional inhibition, mRNA cleavage, and translational inhibition.

Answer 104

True

Question 105

It has been reported in Arabdopsis that the heat shock protein (HSP) terminator increases transgene expression compared to other transgene terminators. This suggests that terminators are not generic sequences.

Answer 105

True

Question 106

Which of the following type of RNAi is widely used experimentally to degrade specific mRNAs?

Answer 106

Small interfering (si)RNA

Question 107

When using ionizing radiation to induce mutations, it is possible to increase the frequency of mutations by manipulating the radiation dosage.

Answer 107

True

Question 108

In the CRISPR-Cas9 gene editing system, the role of the guide RNA is to:

Answer 108

Ensure that the target DNA is cut at a specific target site

Question 109

The principal drawback to the Agrobacterium method of gene transformation is that all the transformed plants also have crown gall disease.

Answer 109

False

Question 110

RNAi is triggered by the presence of ______ in the cytoplasm:

Answer 110

Doubled stranded RNA molecules

Question 111

A _____ encodes a protein (enzyme) that allows the transformed cells to grow while the growth of the non-transformed cells is inhibited.

Answer 111

Selectable marker

Question 112

Which of the following techniques leaves the gene intact and unaltered but leads to a mutant phenotype by targeting gene transcripts?

Answer 112

RNAi

Question 113

A TILLING population is a population of individuals originated from an inbred line, all containing the same mutation.

Answer 113

False

Question 114

Which method is the most specific for creating specific and targeted mutations?

Answer 114

CRISPR/Cas9

Question 115

The “Arctic” apple was developed using which one of the following techniques?

Answer 115

RNAi

Question 116

Which of the following transgenic plant examples presented in class involved multiple transgenes required to for the target phenotype?

Answer 116

EPSPS in soybean

Question 117

Which of the following are the minimal requirements for a transgene construct?

Answer 117

promoter, coding region, and terminator

Question 118

RNAi is considered a fairly precise method for gene silencing due to the partial/complete complementary base pairing between the target mRNA and the:

Answer 118

Guide RNA (single stranded miRNA or siRNA)

Question 119

You produce a test-cross generation (e.g. F1 x homozygous recessive parent) of 100 plants. The cross history is as follows: A Roundup resistant homozygote was crossed with a non-Roundup resistant homozygote to produce an F1. Assume resistance to Roundup is dominant. What ratio of living: dead do you expect in the test-cross generation?

Answer 119

50 resistant: 50 susceptible

Question 120

A genetic engineering technique that involves microprojectile bombardment of tissues with small particles coated with transgene DNA is known in the popular press as

Answer 120

the gene gun method

Question 121

Which of the following best describes the key difference between transgenics and cisgenics?

Answer 121

Cisgenics involves non-sexual transfer of genes from the same or closely related species whereas transgenics involves non-sexual transfer of genes from species that are not sexually compatible

Question 122

Reporter genes are often included in transgene constructs to:

Answer 122

Indicate the location and level of expression of the transgene

Question 123

A key difference between transgenics and genome edits is that transgenic strategies involve transfer of a gene from one organism to another whereas genome editing involves altering the DNA sequence of the target organism

Answer 123

True

Question 124

The insertion of a transgene into somatic tissues may be an interesting strategy in the varietal improvement of:

Answer 124

Bananas

Question 125

Which of the following type of RNAi is widely used experimentally to degrade specific mRNAs?

Answer 125

small interfering (si)RNA

Question 126

The CRISPR-Cas9 gene editing system can only be used to knock-out target genes; it cannot change or add functionality to existing genes.

Answer 126

False

Question 127

Inducing mutations via mutagenic treatment is a generic method in that mutations are generated randomly in the target genome. However, some mutagens tend to cause specific types of changes – for example, chemical treatment with EMS usually induces a G:C to A:T SNP.

Answer 127

True

Question 128

The nptII gene, which confers resistance to the antibiotic kanamycin, was used as a selectable marker in the EPSPS transgene construct. Thus, after transformation, EPSPS transgenic cells are selected by:

Answer 128

Kanamycin application

Question 129

In both hemizygotes and homozygotes for the CP4 EPSPS gene, the EPSPS enzyme (susceptible to gyphosate) is still produced, even though they are resistant to the herbicide.

Answer 129

True

Question 130

When inserting a transgene construct into the T-DNA of a Ti-plasmid for delivery via the Agro method, the right and left borders of the T-DNA are kept flanking the transgene construct because:

Answer 130

They contain sequences that are important for the integration of the transgene into the genome of the host plant

Question 131

Which of the following best describes a transposable element?

Answer 131

a DNA sequence that can move to a new site in the genome

Question 132

Transposable elements can play a major role in plant evolution by:

Answer 132

Creating new genes, modifying or re-programing existent genes, and even causing chromosomal arrangements

Question 133

An example of a non-autonomous transposon would be a Class 1 transposon (retrotransposon) lacking the ability to encode ______ and a Class 2 transposon (DNA transposon) lacking the ability to encode ______ .

Answer 133

Reverse transcriptase

Transposase

Question 134

In the case of codominance, heterozygous F2 plants can be phenotypically distinguished from homozygous F2 plants.

Answer 134

True

Question 135

In the case of incomplete dominance, where one parent has red flowers, the other parent has white flowers and the F1 has pink flowers what will be the expected ratio of flower color in doubled haploids derived from the F1?

Answer 135

1 red: 1 white

Question 136

Which of the following terms best describes the interaction of alleles at two loci?

Answer 136

Epistatis

Question 137

The color of grapes is an example of a phenotype influenced by transposon activity. In black grapes, the transcription factor VvMybA1 upregulates anthocyanin production. In white grapes, a retrotransposon is inserted within the transcription factor sequence, causing a loss of function.

Answer 137

True

Question 138

A ratio of 3:1 in a doubled haploid population could be diagnostic of two-locus epistasis.

Answer 138

True

Question 139

Diploid plants with the same genome sizes will all have the same numbers of chromosomes.

Answer 139

False

Question 140

Which of the following has been revealed by a deeper understanding of genome sequence and architecture?

Answer 140

More regulatory factors than anticipated

Question 141

DNA methylation:

Answer 141

Decreases gene expression

Question 142

Transposable elements in plants are

Answer 142

up to half of the DNA content in many nuclear genomes

Question 143

Fragaria vesca (woodland strawberry) is 2n = 2x = 14. This plant is reported to have 35,000 genes, and a genome size of 240 Mb. Considering these facts, one could most reasonably conclude that

Answer 143

a portion of the genome does not consist of protein coding genes

Question 144

Which of the following would be evidence of transposable element activity in a gene conferring kernel pigment color in maize?

Answer 144

colored sectors in a white kernel

Question 145

“The gene for fragrance in rice” paper describes a classic example of dominant allele conferring “more” of a phenotype, in this case fragrance. In other words, rice varieties with at least one dominant allele at the BAD2 locus are more fragrant than varieties that are homozygous recessive at the BAD2 locus.

Answer 145

False

Question 146

A _____ is an estimated DNA sequence comprised by contigs and gaps of estimated sizes.

Answer 146

Scaffold

Question 147

A fundamental expectation underlying the F1 hybrid seed industry is that the F1 of the cross of two inbred parents is expected to be 100% heterozygous at the loci where the parents have alternative alleles.

Answer 147

True

Question 148

In a perfect genetic/physical map, the number of pseudochromosomes is equal to the number of chromosomes in a given species.

Answer 148

True

Question 149

If you were interested in identifying which genes are expressed in fruits vs. roots, which of the following approached would be most useful?

Answer 149

RNA sequencing (RNAseq)

Question 150

If you were to obtain the complete nuclear genome of an organism, such as Fragaria vesca, would you expect to find genes encoding tRNAs and rRNAs? If yes, in what abundance?

Answer 150

Yes, and in abundance

Question 151

If you cross two inbred parents, both low in cyanide, and observe high cyanide in the F1 and a 9:7 ratio of high to low cyanide in the F2, you might hypothesize that which of the following is responsible?

Answer 151

Epistasis

Question 152

The C value paradox is best defined by which of the following statements?

Answer 152

Organisms of similar evolutionary complexity may vary significantly in genome size

Question 153

DNA transposons move via an RNA intermediate whereas retrotransposons move via a DNA cut and paste mechanism

Answer 153

False

Question 154

Class 1 transposable elements, also called retrotransposons, replicate their own sequence into another place in the genome, thus, leading to genome expansion. In contrast, Class 2 transposable elements move in the genome by a cut and paste mechanism: therefore, their activity does not lead to genome expansion.

Answer 154

False

Question 155

Because transposable elements in plants can eliminate or alter gene function, they are usually

Answer 155

epigenetically silenced

Question 156

In cases of incomplete dominance, heterozygous F2 plants can be phenotypically distinguished from homozygous F2 plants.

Answer 156

True

Question 157

You are a geneticist working on okra. Knowing the Haldane cM distance between two loci in an ancestral diploid species and in a polyploid relative is most useful for potentially understanding which of the following?

Answer 157

Homoeology and synteny

Question 158

Locus A determines flower color. The dominant allele produces an enzyme critical for pigment formation, which gives a red color. The recessive allele is non-functional and gives no color (white flowers). Knowing that the two alleles show partial dominance, what phenotypic ratio do you expect in the F2 from a cross between a plant with white flowers and one with red flowers, both completely homozygous?

Answer 158

1 red : 2 pink : 1 white

Question 159

A common feature of the F2 epistatic ratios presented for two unlinked loci is that they all add up to

Answer 159

16

Question 160

According to the classical “overdominance theory” for heterosis, it should be possible to develop an inbred with performance equal to the F1.

Answer 160

False

Question 161

The number of transposable elements in and around genes is most likely to

Answer 161

Increase with time

Question 162

In general, if genome sizes of flowering plants are compared to those of other organisms,

Answer 162

flowering plants show the greatest range in genome sizes

Question 163

Which type of heterosis for grain yield in maize is most likely to be preferred in the hybrid seed business?

Answer 163

High parent heterosis

Question 164

Epigenetic changes are heritable. They are, however, potentially reversible. That’s why it is very useful to know if a phenotype is due to epigenetics, but could be risky to depend on epigenetics for a phenotype because the phenotype could revert at some point in the future.

Answer 164

True

Question 165

After years of research, Barbara McClintock discovered the role of transposons in colored kernels in maize. A small pigmented spot in a white kernel suggests that:

Answer 165

A transposon was excised late in kernel development

Question 166

Which of the following plants would you expect to suffer the most from inbreeding depression?

Answer 166

Maize (cross pollinated)

Question 167

What is the approximate number of genes one might expect in a diploid plant? For this question use the classical definition of a gene: a DNA sequence that codes for a mRNA that is translated to a protein, or that codes for a functional RNA (tRNA or rRNA).

Answer 167

30,000

Question 168

There are two classical theories for the genetic basis of hybrid vigor. These are:

Answer 168

Overdominance and dispersed dominant genes

Question 169

Flow cytometry is a technique used to:

Answer 169

Measure genome size

Question 170

In the Fragaria genome sequencing paper, the authors

Answer 170

Chose a wild diploid species because of the small genome size

Question 171

If a species is 2n = 2x = 14 and there is complete genome sequence information available, how many pseudochromosomes would you expect?

Answer 171

7

Question 172

In addition to being cold tolerant, facultative type barleys do not require vernalization to flower because:

Answer 172

The VRN-H2 gene is completely deleted.

Question 173

The two classical theories of heterosis are mutually exclusive in that according to one of the theories the F1 will always be superior to the parents and according to the other theory it is theoretically possible to accumulate all possible favorable alleles in a single homozygous plant

Answer 173

True

Question 174

In the reading on sequencing the Fragaria genome, the authors show a diagram demonstrating that different types of genes are expressed in fruits and roots. (Technically speaking the part of the strawberry we usually eat is neither a fruit nor a berry, but that is a matter for another class). This means that if you eat a strawberry “fruit” you are eating the same genes (as DNA) that are present in a root, or for that matter, a leaf.

Answer 174

True

Question 175

Transposable elements and their proliferation may help to explain the C value paradox

Answer 175

True

Question 176

Gene ontology is a representation of:

Answer 176

Gene and gene product

Question 177

Genes can be regulated:

Answer 177

All the answers are correct

Question 178

We can always predict a phenotype directly from the coding sequence of gene.

Answer 178

False

Question 179

Transposable elements are always very small and never exceed 21 nucleotides in length

Answer 179

False

Question 180

Which of the following is true about the C-value?

Answer 180

C-value refers to a constant value. The term came up when scientists observed that within a species the genome size is quite constant.

Question 181

In the woodland strawberry genome sequencing project, before extracting total RNA from roots and fruits and characterizing the transcriptomes of these tissues, the authors subjected plants to a wide array of environmental conditions in order to:

Answer 181

Maximize the number of mRNAs transcribed

Question 182

In the case of complete dominance, the homozygous recessive condition always leads to less of a phenotype (e.g. less color, less aroma, fewer fertile florets per rachis node)

Answer 182

False

Question 183

In the woodland strawberry genome sequencing project, some scaffolds were anchored to bins. Imagine two given scaffolds A and B that anchored to Bin X, which is between markers 1 and 2. One can say:

Answer 183

That scaffolds A and B are located somewhere between markers 1 and 2.

Question 184

Which of the following is the best definition of a transcription factor, such as the functional allele (Nud) at the nud locus?

Answer 184

A gene that encodes a protein that binds to DNA, thus regulating gene expression.

Question 185

Match the conceptual model with the intra-locus interaction:
The dominant allele is functional and overrules the effect of the recessive allele, which is non-functional

The dominant allele is functional and does not overrule the effect of the recessive allele, which is non-functional

Both dominant and recessive alleles are functional and encode different functions

Answer 185

Complete dominance

Incomplete (partial) dominance

Codominance

Question 186

Which of the following is most directly related to genes in chromatin being “switched on”?

Answer 186

More acetylation of histones

Question 187

When alleles at a locus show overdominance:

Answer 187

The phenotypic value of the heterozygote is superior to both homozygous states.

Question 188

What statement is FALSE about transcription factors?

Answer 188

They can only enhance the expression of another gene; they cannot repress expression of another gene.

Question 189

Heterosis is most commonly observed in crosses involving parents that are

Answer 189

inbreds

Question 190

Epistasis can be observed when considering alleles at just one locus.

Answer 190

False

Question 191

In the context of genome sequencing, more coverage means:

Answer 191

More reads representing each part of the genome

Question 192

Cyanide concentration in clover is an example of duplicate recessive epistasis. If a low cyanide plant AAbb is crossed with another low cyanide plant aaBB, and doubled haploids are produced from the F1, what is the phenotypic ration in the doubled haploid population?

Answer 192

1 high cyanide : 3 low cyanide

Question 193

In hazelnuts, the S locus determines self-incompatibility and the alleles at the S locus usually show:

Answer 193

Codominance

Question 194

Which of the following type of molecular marker will allow you to identify heterozygotes?

Answer 194

codominant

Question 195

In the woodland strawberry genome sequencing project, the authors concluded that the species’ small genome size was, in large part, due to:

Answer 195

The absence of large duplications and the limited number of transposable elements

Question 196

In the context of allelic interactions within a locus, a ___________ allele is often the result of mutations involving deletion, altered transcription (for example, a mutation in the promoter), or altered translation (for example, a premature stop codon).

Answer 196

Non-functional

Question 197

Why was the woodland strawberry chosen over the cultivated strawberry as the representative of the genus Fragaria for genome sequencing?

Answer 197

It has a smaller genome, it is diploid and amenable to transformation, and it has synteny with other species of the Rosaceae family

Question 198

In the woodland strawberry genome sequencing project, the authors observed strong syntenic relationships between woodland strawberry and Prunus. This suggests that woodland strawberry and Prunus:

Answer 198

Diverged at similar evolutionary time from a common ancestor.

Question 199

Which statement below is FALSE about the dispersed dominant genes theory:

Answer 199

It implies that it is not possible to have inbreds with the same phenotypic value as the F1.

Question 200

In domesticated barley varieties, vernalization requirement can be explained by a 2-locus epistatic model. According to the model presented in class, a winter type barley will flower when:

Answer 200

The VRN-H2 gene is not expressed, and, as a consequence, VRN-H1 gene determining meristem differentiation is expressed.

Question 201

The metabolic pathway model explains the duplicate recessive epistasis observed in clover. According to the model, the alleles at both genes need to be functional because:

Answer 201

The enzyme produced by one gene is a precursor to the enzyme produced by the other gene.

Question 202

In the woodland strawberry genome sequencing project, short reads were assembled de novo because there was not a reference genome for the Fragaria genus at the time. This type of assembly:

Answer 202

Requires the reads to overlap

Question 203

possible to identify the allelic state of all individuals with a dominant marker.

Answer 203

False

Question 204

A phenotype can be a result of all of the following: RNAi, chromatin remodeling, inter-locus interactions, intra-locus interactions, promoters, transcription factors, mRNA stability, protein modification, and possibly much more.

Answer 204

True

Question 205

Consider two loci. One locus (C) controls fruit shape. A second locus (F) determines texture. CC and Cc = round; cc = spherical. FF and Ff = mushy; ff = firm. One parent is CCFF and the other parent is ccff. You score the two traits in 100 doubled haploid progeny derived from the F1 of this cross and record the following data:

Answer 205

90

Question 206

Given the following data from a doubled haploid population derived from the F1 of the cross between parents QQZZ x qqzz, the two loci are definitely on the same chromosome.

QQZZ

QQzz

qqZZ

qqzz

40

10

10

40

Answer 206

True

Question 207

The co-segregation of markers in a linkage map is usually caused by small population size (due to rare recombinants within the region of the markers).

Answer 207

True

Question 208

Linkage maps can be useful for plant geneticists as a starting point for cloning genes.

Answer 208

True

Question 209

In linkage maps based on different crosses within the same species, locus orders will usually be consistent but estimates of distance may vary.

Answer 209

True

Question 210

If you read that two loci are 10 cM apart, this means that in every individual heterozygous at these loci, there will be 10 crossovers between the two loci.

Answer 210

False

Question 211

Alleles at loci on the same chromosome will never show independent assortment

Answer 211

False

Question 212

You are building a linkage map for your favorite plant species based on a doubled haploid population, and you come up with the linkage data below. What is your diagnosis of the data point in bold (marker allele in position 15 cM)?

Chromosome

Marker position (cM)

Marker allele

1

1

A

1

5

A

1

10

A

1

15

B

1

20

A

1

25

A

1

30

A

Answer 212

It indicates a double crossover, which is rare within a 5 cM distance, thus this data point can considered missing data.

Question 213

You read in a journal article that two loci are on the same chromosome according to a map made in a doubled haploid population. But when you calculate the % recombination between the two loci in an F2 population derived from the same cross it is 50%. Which of the following is the most likely explanation?

Answer 213

The authors of the journal article probably had more markers available to make their map than you did.

Question 214

You observe that in a very large (n = 1000) F2 population derived from the cross of two completely homozygous parents, two specific combinations of traits are always inherited together: all purple-flowered progeny are susceptible to a fungal disease and all white-flowered progeny are resistant to the fungal disease. This is most likely a case of:

Answer 214

Pleiotropy

Question 215

Alleles at loci on different chromosomes should always show independent assortment.

Answer 215

True

Question 216

A linkage group is ~ 150 centiMorgans long. Loci at the opposite ends of this linkage group are expected to show close to what % recombination?

Answer 216

50%

Question 217

If you calculate a recombination value of 30% between two loci and convert this value to centiMorgans (Haldane or Kosambi)

Answer 217

The cM value will be larger than the % recombination value

Question 218

Which meiotic phenomenon would you hypothesize to be most related to the synaptonemal complex?

Answer 218

Crossing over

Question 219

Assuming you have sufficient markers to generate a medium density linkage map with an average spacing of 3 cM in a species that is 2n = 2x = 30, how many linkage groups would you expect?

Answer 219

15

Question 220

Mutation is the source of new alleles whereas recombination between linked loci, and independent assortment of alleles at un-linked loci, are sources of new combinations of alleles.

Answer 220

15

Question 221

Crossovers occur only in the “dark matter” between protein coding genes, never within genes.

Answer 221

False

Question 222

If you had sufficient SNP markers to provide complete genome coverage of all chromosomes - at an average density of 2 cM - in an allopolyploid species that is 2n = 6x = 42, how many linkage groups would you expect?

Answer 222

21

Question 223

Loci “far enough” apart on same chromosome show independent assortment – due to “sufficient” crossovers between the loci in a population of individuals

Answer 223

True

Question 224

An advantage of higher density linkage maps is that you can more accurately convert cM to Mb.

Answer 224

False

Question 225

The Oregon Wolfe Barley doubled haploid population is a great choice for linkage mapping because:

Answer 225

It is very diverse (which means many recombination events can be identified ) and there are abundant genotype data.

Question 226

You are offered a job as a tomato breeder, with primary responsibility for breeding a new variety of tomato with great flavor and long shelf life. Experience and data confirm that poor flavor and long shelf life are associated. Which of the following genetic phenomenon would offer you the greatest chance of success in your new job?

Answer 226

Flavor and shelf life are controlled by two tightly linked genes

Question 227

What is the main purpose for applying a mapping function such as the Haldane or Kosambi function?

Answer 227

To adjust observed recombination fractions for double crossovers, so that calculated map distances (cM) are additive along the chromosome.

Question 228

You have two different linkage mapping populations in the common bean. Which of the following is most likely (assuming that your data are excellent for both populations)?

Answer 228

The loci in the two maps will be in the same order for a given linkage group but distances between loci may vary

Question 229

If a plant is 2n = 6x = 24 and there are sufficient polymorphic loci to make complete linkage maps of all chromosomes, there will be how many linkage groups?

Answer 229

12

Question 230

The co-segregation of markers in a linkage map is usually caused by small population size (due to rare recombinants within the region of the markers).

Answer 230

True

Question 231

Alleles at loci on the same chromosome will never show independent assortment

Answer 231

False

Question 232

Linkage map data can be used in gene discovery. In QTL analysis, for example, marker positions obtained from linkage maps are associated with traits. This is often the first step in determining the location of a gene associated with a trait.

Answer 232

True

Question 233

Crossing over can (but does not always) lead to observable recombination. A crossover will not lead to observable recombination when it occurs_____. (Mark all that apply)

Answer 233

Between two sister chromatids in a 100% heterozygous plant
Between two sister chromatids in a doubled haploid plant
Between two non-sister chromatids in a doubled haploid plant

Question 234

You are building a linkage map for your favorite plant species based on a doubled haploid population, and you come up with the linkage data below. What is your diagnosis of the data point in bold (marker allele in position 15 cM)?

Chromosome

Marker position (cM)

Marker allele

1

1

A

1

5

A

1

10

A

1

15

B

1

20

A

1

25

A

1

30

A

Answer 234

It indicates a double crossover, which is rare within a 5 cM distance, thus this data point can considered missing data.

Question 235

Crossing over is the basis of linkage mapping. Which type of crossover will be most informative for map construction?

Answer 235

Crossovers between polymorphic loci

Question 236

Linkage maps can be useful for plant geneticists as a starting point for cloning genes.

Answer 236

True

Question 237

Locus A determines tomato color and locus B determines tomato flavor. AA=red, aa=pink. BB=mild flavor, bb=strong flavor. The A and B loci are in a repulsion linkage, and have a recombination frequency of 1%. Color is easy to measure. Flavor is difficult and expensive to measure. What would the minimum (and affordable) doubled haploid population size you would need to have a reasonable probability of recovering 5 plants with red color and strong flavor?

Answer 237

500 plants

Question 238

Consider two loci. One locus (Y) controls flower color. A second locus (F) determines flavor. YY and Yy = yellow; yy = white. FF and Ff = no flavor; ff = flavor. One parent is YYFF and the other parent is yyff. You cross the two parents, produce 100 doubled haploids from the F1 and score the two traits in the doubled haploid progeny. You record the following data:

Yellow and no flavor: 46 individuals

Yellow and flavor: 6 individuals

White and no flavor: 4 individuals

White and flavor: 44 individuals

How many non-parental (aka recombinant) individuals are in the doubled haploid population?

Answer 238

10

Question 239

You observe that in a very large (n = 1000) F2 population derived from the cross of two completely homozygous parents, two specific combinations of traits are always inherited together: all purple-flowered progeny are susceptible to a fungal disease and all white-flowered progeny are resistant to the fungal disease. This is most likely a case of:

Answer 239

Pleiotropy

Question 240

Mutation is the source of new alleles and recombination is the source of new combinations of alleles.

Answer 240

True

Question 241

Regions around the centromere are generally heterochromatic and therefore will have, on average, more Mb/cM than euchromatic regions with higher recombination rates.

Answer 241

True

Question 242

In comparing linkage maps of the same chromosome in the same diploid species based on two different samples of 100 random F2 indviduals from the same cross, which of the following is the most likely?

Answer 242

Locus orders will the same but cM distances will vary

Question 243

Alleles at loci on different chromosomes

Answer 243

will always show independent assortment

Question 244

If there is 14% recombination between the N locus and the L locus; there is 39% recombination between the W locus and the N locus; and there is 48% recombination between the W and L loci, what is the most likely order of these loci in a linkage map?

Answer 244

WNL