Question

2. A dominant allele H reduces the number of body bristles that Drosophila flies have, giving rise to a “hairless” phenotype. In the homozygous condition, H is lethal. An independently assorting dominant allele S has no effect on bristle number except in the presence of H, in which case a single dose of S suppresses the hairless phenotype, thus restoring the "hairy" phenotype. However, S also is lethal in the homozygous (S/S) condition. What ratio of hairy to hairless flies would you find in the live progeny of a cross between two hairy flies both carrying H in the suppressed condition?

a) 7 hairy: 2 hairless

b) 2 hairy: 1 hairless

c) 9 hairy: 7 hairless

d) 4 hairy: 3 hairless

e) 1 hairy: 1 hairless

4. MYB31 is a transcription factor that promotes expression of the gene encoding the enzyme chalcone synthase (CHS). CHS is necessary to make purple pigments in plants known as anthocyanins. CHS mRNA abundance is reduced by 90% in the recessive null myb31 mutant, resulting in low anthocyanin levels and lilac (light purple)-colored flowers. The recessive null chs mutant has white flowers. What phenotypic ratios do you expect among the progeny of a cross between two dihybrid (e.g. myb31/+ chs/+) individuals?

a) 9 purple: 3 lilac: 4 white

b) 9 purple: 4 lilac: 3 white

c) 9 purple: 7 white

d) 15 purple: 1 white

e) 12 purple: 1 lilac: 3 white

5. You suspect that one or more transcription factors may promote CHS expression. Which of the following strategies would not be a way to find candidate gene(s) that may encode these transcription factor(s)?

a) mutagenize myb31 and screen for white-flowered mutants

b) mutagenize myb31 and screen for purple-flowered mutants

c) mutagenize chs and screen for lilac- or purple-flowered mutants

d) search a genome database for genes encoding proteins with similar sequences to MYB31

e) mutagenize wildtype and screen for additional mutants with lighter flower pigmentation

6. You discover another mutant allele of MYB31 that unexpectedly has nearly white flowers and almost no CHS transcripts. Which term best describes the effect of this allele?

a) amorph

b) antimorph

c) hypermorph

d) hypomorph

e) neomorph

14. To deduce the possible functions of the ~6000 open reading frames (ORFs) in Saccharomyces cerevisiae, a collection of null alleles in each ORF was systematically generated. 15% of the ORFs are lethal when knocked out, 25% show a mutant phenotype, and 60% show no detectable mutant phenotype. This could imply that 60% of yeast ORFs have no function, but that's quite unlikely. What is a better explanation for a knockout without a phenotype?

a) there is functional redundancy among homologous (similar) ORFs

b) an independent, parallel pathway does the same job as the pathway the ORF is in

c) the knockout mutation produces a subtle phenotype that went unnoticed

d) the knockout has a phenotype only under a very specific condition that has not been tested

e) all of the above

15. Why is the most likely reason exome sequencing would fail to identify a disease-causing mutation in a person affected with a rare disorder?

a) the affected gene is not expressed in the tissue sample from which the nucleic acids are collected

b) the mutation lies in an intergenic region

c) the mutation is an indel rather than a SNP

d) the mutation causes a conservative rather than non-conservative amino acid substitution

16. A scaffold is a collection of contigs that are known to be physically near each other on a chromosome but are separated by gaps of unknown sequence.

a) True

b) False

17. Paired-end sequencing of long DNA fragments (e.g. 10 kb) can be used to link two contigs together on a scaffold. In this process, the entire DNA fragment is amplified on the flow cell, followed by sequencing in a limited distance from each end.

a) True

b) False

18. Chromatin immunoprecipitation is a method to determine _________ interactions.

a) protein-protein

b) protein-DNA

19. Affinity purification can be used to isolate ______

a) specific nucleic acids

b) specific proteins

c) both nucleic acids and proteins

20. Which of the following statements is not true? In RNA-seq, coverage

a) is correlated with the abundance of a given transcript

b) indicates how many times a specific base in a gene is represented among a collection of reads

c) is essentially absent in the introns and UTRs

d) is typically uneven across a gene

e) is affected by alternative splicing

21. De novo genome assemblies from next-generation shotgun genome sequencing (i.e. sequencing of billions of random genomic DNA fragments) usually fail to produce full-length sequences of chromosomes at first. Which of the following is not a reason why assemblies can fail?

a) insufficient coverage

b) repetitive sequences

c) short read lengths

d) reads have high error rates

e) genes with low expression levels are not well represented in the dataset

23. Forward genetics refers to the process of identifying an interesting mutant phenotype, then tracking down the location of the mutation. In reverse genetics, a mutation is created/identified and then a mutant phenotype is searched for.

a) True

b) False

24. What advantage does insertional mutagenesis (i.e. with T-DNA or mobile DNA/transposons) have over chemical mutagenesis as a method to generate mutant populations for reverse genetic studies?

a) the T-DNA or mobile DNA have known sequences, enabling their locations in individual lines to be easily determined by PCR methods that can capture flanking DNA sequences

b) large insertions of DNA into a gene are more likely to cause a loss-of-function than point mutations

c) only one or a few mutations are typically introduced in each line, rather than hundreds or thousands, making it easier to attribute a phenotype to a specific mutation

d) all of the above

e) none of the above

25. Selectable markers offer a clever way to identify cells or organisms that have undergone rare transformation events. When creating targeted gene knockouts in mice, a population of embryonic stem (ES) cells are transformed with a vector containing a neomycin-resistance gene flanked by DNA sequences that match the targeted gene. This provides selection for cells that have integrated the neomycin-resistance gene into a chromosome. A tk+ gene is also included in the vector, but it is outside the regions with homology to the targeted gene. The purpose of the tk+ gene is to select against ectopic insertions, as well as to select against cells in which _________ crossovers occurred between the vector and the targeted gene.

a) single

b) double

26. In the example of a targeted mouse gene knockout we described in class, embryonic stem (ES) cells were derived from agouti mice (A/A) and implanted into embryos from mice with black fur (a/a), producing chimeric mice that were mated to mice with black fur. In the next generation (F1), agouti mice were selected for further work. Why?

a) agouti fur indicates that the ES cell lineage contributed to the germ line of the chimera

b) agouti fur indicates the mice that carry the knockout gene allele

c) both of the above

d) none of the above

27. Creating a transgenic mouse can be as simple as injecting DNA into a single-cell mouse embryo, followed by implantation. However, the ectopic insertions that would be present in resulting transgenic lines have drawbacks. What of the following is not a potential problem?

a) because insertions are random, the transgene may disrupt another gene

b) the location of the transgene insertion can influence how it is expressed

c) the selectable marker (e.g. antibiotic resistance gene) can be picked up by bacterial symbionts

d) transgenes often insert as integrated arrays, which can vary in size and structure

e) integrated transgene arrays can undergo rearrangements

28. The protospacer adjacent motif (PAM) requirement for Cas9 activity prevents cleavage of the CRISPR repeat arrays in the bacterial genome.

a) True

b) False

29. Cas9 cleaves DNA, but does not introduce mutations itself. Why does CRISPR-Cas9 activity typically lead to mutations at targeted sites?

a) the nonhomologous end-joining repair mechanism is error-prone

b) error-free repair of the target site remakes the sequence that Cas9 recognizes and cleaves

c) nonhomologous end-joining is more commonly used by cells than homology-directed repair

d) all of the above

30. CRISPR-Cas9 offers a great deal of potential for gene editing. However, most scientists were deeply concerned by the recent report of gene editing in humans. Societal ethics aside, what are safety issues of concern?

a) Cas9 may cleave DNA sequences that are similar, but not exact matches, to the guide RNA

b) double-stranded breaks can cause chromosomal rearrangements

c) although the mutation location is easily controlled, the exact mutation produced is not

d) all of the above

31. (10 points) In yeast, you have sequenced a piece of wild-type DNA and it clearly contains a gene, but you do not know what the gene does. To investigate further, you would like to find out its mutant phenotype. How would you used the cloned wild-type gene to do so? Draw in all key features of the yeast integrative plasmid you would create to accomplish this.

Describe what you would do to isolate the mutant.

32. The cystic fibrosis gene was cloned in 1989, before the human genome project had even started. It was found by mapping the cystic fibrosis mutation to chromosome 7. The researchers were eventually able to clone a portion of an exon of the putative cystic fibrosis gene, but they were unable to clone the entire coding sequence from genomic DNA (note that introns in humans can be very large). In as much detail as you can, describe how you would have identified a full-length coding sequence for the cystic fibrosis gene in an era before nextgeneration sequencing and a human genome assembly. Assume you have the ability to extract nucleic acids from a biopsy of any human tissue you require.

Having found the coding sequence for the putative CF gene, how would you go about confirming that you had found the correct gene? (Transformation of humans is unethical and not allowed here.)

Answer 1

2 ___ A

4