Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
50 Cards in this Set
- Front
- Back
|
1. External fertilization and transparent embryos make this organism an excellent modelfor studying early embryonic development. A) Arabidopsis thaliana B) Drosophila melanogaster C) Zebra fish D) Mice E) C. elegans |
C) Zebra fish |
|
|
2. A researcher is interested in tracing the developmental fate of individual cells fromthe zygote to adulthood in real time. Which of the following model organismswould provide an opportunity to perform this analysis in real time?A) Arabidopsis thaliana B) Drosophila melanogaster C) Caenorhabditis elegans D) Zebrafish E) Mice |
C) Caenorhabditis elegans |
|
|
3. The first eukaryotic genome to be analysed completely is of A) Arabidopsis thaliana B) Human genome C) Drosophila melanogaster D) Caenorhabditis elegans E) Saccharomyces cerevisiae +10 |
E) Saccharomyces cerevisiae |
|
|
4. The most important aspect in designing a genetic screen is A) A mutation leading to a change of an amino acid B) A clear and easy phenotype to score C) Presence of dominant mutations D) Mutants with subtle phenotypes E) Large deletion mutants with no phenotype |
B) A clear and easy phenotype to score |
|
|
5. Reverse genetic analysis is A) deciphering phenotype from genotype B) deciphering genotype from phenotype C) deciphering both genotype and phenotype by comparative analysis D) deciphering the molecular basis of a mutation through mapping E) deciphering the protein encoded by a gene through computational analysis |
A) deciphering phenotype from genotype |
|
|
6. In a mutant screen, normally ----- generation is screened for identifying the recessivephenotypes A) parental B) F1 or M1 C) F2 or M2 D) F3 E) M0 mutagenised |
C) F2 or M2 (in arabidopsis, but 3 in drosophila…) From lecture 20: “Second generation - mutation will segregate into homozygous state and recessive phenotypes can be identified” |
|
|
7. Complete dominance is an example of: A) An additive interaction between different alleles at different loci B) A non-additive interaction between different alleles at different loci C) A non-additive interaction between different alleles at the same locus D) An additive interaction between different alleles at the same locus E) A multiplicative interaction between different alleles at different loci |
C) A non-additive interaction between different alleles at the same locus |
|
|
8. If an allele has the same phenotypic effect in heterozygotes as in homozygotes that is,the genotypes Aa and AA are phenotypically indistinguishable then the allele is A) Recessive B) Dominant C) Incompletely dominant D) Codominant E) Partially dominant |
B) Dominant |
|
|
9. The test to determine whether mutations are alleles of a particular gene based on thephenotypic effect of combining the mutations in the same individual is known as the A) Bradford assay B) Complementation test C) Live/Dead assay D) ELISA E) Double mutant analysis |
B) Complementation test |
|
|
10. Summer squash color is determined by the interaction of more than one gene. Thepresence of CC or Cc allele combinations produces a squash that is white in color, andthe C allele is epistatic to the G allele. The presence of GG or Gg produces a squashthat is yellow in color, and ccgg produces a squash that is green. After two heterozygoussquash are crossed one of the offspring is CcGg. What is the color of the offspringsquash? A) White B) Yellow C) Green D) Orange E) It is not possible to determine the colour based on the information given |
A) White |
|
|
11. The work of Bateson and Punnett, who studied the combs of chickens, demonstrated that: A) Two independently assorting genes can affect a trait B) Two linked genes can affect a trait C) Two independently assorting genes cannot affect a trait D) Two linked genes cannot affect a trait E) Chicken comb shape is not regulated by genetic factors |
A) Two independently assorting genes can affect a trait |
|
|
12. When two or more genes influence a trait, an allele of one of them may have anoverriding effect on the phenotype. When an allele has such an overriding effect, itis said to be: A) Dominant B) Recessive C) Penetrant D) Epistatic E) Expressive |
D) Epistatic |
|
|
13. Bateson and Punnett mated purple sweet peas with white sweet peas. F1 hybrids wereall purple in color. When the hybrids were mated a ratio of 9 purple : 7 white wereobserved in the F2 generation. Which of the following could best explain thisobservation? A) Incomplete penetrance B) Lethal alleles C) Epistasis D) Codominance E) Incomplete dominance |
C) Epistasis |
|
|
14. cinnabar and scarlet mutants in Drosophila show a bright red eye phenotype. Whencrossed with each other, the scarlet mutant is able to complement the cinnabar mutant.This would indicate A) cinnabar and scarlet are allelic B) cinnabar is epistatic to scarlet C) scarlet is epistatic to cinnabar D) cinnabar and scarlet are mutations in two different genes E) cinnabar and scarlet show additive interaction |
D) cinnabar and scarlet are mutations in two different genes |
|
|
15. Mr Sweet analysed an F2 population segregating for two mutations that affectDrosophila wing shape. In the F2 population he obtained relatively fewer non-parentalphenotypes. In order to be sure that this results did not happen by chance, Mr. Sweetwould need to perform a A) Students t test B) Calculate the average wing shape C) Analysis of variance D) chi-square test E) Molecular analysis |
D) chi-square test |
|
|
16. When “albino” mutant mice were crossed with “legless” mutant mice, the F1individuals resembled wild type. When F2 progeny were analysed, there were 1/16th ofmice that were both “legless” and “albino”. The interaction between albino and leglesscan be described as A) Epistatic B) Antagonistic C) Synergistic D) Additive E) Expressive |
D) Additive According to good old wikipedia “two mutations are considered to be purely additive if the effect of the double mutation is the sum of the effects of the single mutations”. From a normal F1 cross you would expect 1/16th to be a double recessive and since 1/16th are legless and albino I would say this is just the sum of both the mutations and thus the interaction is additive? Wikipedia also adds that “when the fitness difference of the double mutant from the wild type is smaller than expected from the effects of the two single mutations, it is called antagonistic epistasis” soI don’t think this is the case here. See glossary on lecture 20. “The phenotype of the double mutant contains the effects of both single mutants in a simple additive manner.” Additive is definitely correct. |
|
|
17. “Early” mutants in peas flower 3 days earlier than the wild type peas. “Fast”mutants in peas flower 4 days earlier than wild type. When Benjamin performeddouble mutant analysis, he found the “early” “fast” double mutants were flowering20 days earlier compared to wild type. The type of genetic interaction displayed inthe double mutants is A) Epistatic B) Antagonistic C) Synergistic D) Additive E) Dominance |
C) Synergistic |
|
|
18. According to the ABC model, A activity alone specifies sepals, A together with Bspecifies petals, B together with C specifies stamens and C alone specifies carpels. Thisleads to the arrangement of sepal, petal, stamen and carpel in wild type flowers. Thismodel also suggests that A and C act against each other. What would you predict to bethe arrangement of floral organs in a A/B double mutant? A) carpel, stamen, stamen, carpel B) sepal, sepal, carpel, carpel C) stamen, stamen, stamen, stamen D) carpel, carpel, carpel, carpel E) sepal, petal, petal, sepal |
D) carpel, carpel, carpel, carpel |
|
|
19. The phenomenon in which genes on the same chromosome are separated from eachother during meiosis and new combinations of genes are formed is known as A) Linkage B) Synapsis C) Recombination D) Nondisjunction E) Disjunction |
C) Recombination |
|
|
20. Recombination is commonly the result of a process known as ---- which occurs inprophase I of Meiosis A) Crossing over B) Linkage C) Disjunction D) Non-disjunction E) Formation of the mitotic spindle |
A) Crossing over |
|
|
21. P1 blue-flowered, short-stalked plants (FFSS) and white-flowered, long-stalked plant(ffss). The resulting F1 offspring (FfSs) were intercrossed to produce the following F2progeny:400 blue, short +1400 white, long100 blue, long100 white, shortWhat is the recombination frequency, as shown in the F2 generation? A) 0.25 B) 0.20 C) 0.08 D) 0.10 E) 0.05 |
B) 0.20 |
|
|
22. Which of the following statements is false regarding linked genes? A) The tighter the genes are linked the less often they recombine B) The recombination frequency for any set of genes often exceeds 50% C) Recombination occurs frequently between genes that are loosely linked D) Linked genes do not display independent assortment E) Linked genes tend to show co-segregation |
B) The recombination frequency for any set of genes often exceeds 50% |
|
|
23. A frequency of recombination that is less than 50% implies A) The genes are linked on the same chromosome B) The genes are not linked on the same chromosome C) The genes are not linked and are on different chromosomes D) The genes assort independently E) The genes are not linked and are on the same chromosome |
A) The genes are linked on the same chromosome |
|
|
24. Each crossover event at one point typically produces how many recombinants? A) 0 B) 1 C) 2 D) 3 E) 4 |
C) 2 |
|
|
25. Linked genes on a chromosome can be mapped by A) Studying how often their alleles recombine B) Studying how many individuals show the wild type phenotype C) Studying the physical structure of the chromosome D) Performing karyotyping E) Complementation testing |
A) Studying how often their alleles recombine |
|
|
26. It can be generally said that the genetic map distance is equal to A) The estimated number of crossovers that occur between two points B) The recombination frequency written as a percentage C) The non-recombination frequency written as a percentage D) The number of parental phenotypes seen in the progeny E) Number of complementation groups |
B) The recombination frequency written as a percentage |
|
|
27. Wild-type Drosophila females were mated to males homozygous for two autosomalmutations vestigial (vg), which produces short wings, and black (b), which produces ablack body. All the F1 flies had long wings and gray bodies; thus, the wild-type alleles(vg+ and b+) are dominant. The F1 females were then testcrossed to vestigial, blackmales, and the F2 progeny were sorted by phenotype and counted. Simple analysisindicates that, on average, 18 out of 100 chromosomes recovered from meiosis had acrossover between vg and b. Thus, vg and b are how far apart? A) 82 centimorgans B) 82 morgans C) 18 centimorgans D) 18 morgans E) 100 centimorgans |
C) 18 centimorgans |
|
|
28. The frequency that is often used to measure the intensity of linkage between genes isknown as the A) Linkage frequency B) Recombination frequency C) Hardy-Weinberg frequency D) Genetic frequency E) Electromagnetic frequency |
B) Recombination frequency |
|
|
29. 150 out of 1000 chromosomes are recombinant between two genes. How far apart arethey located in a genetic map? A) 150 cM B) 15 cM C) 1.5 cM D) 150 M E) 15 M |
B) 15 cM |
|
|
30. The maximum percentage of recombination one can observe in a segregating F2population is A) 25% B) 50% C) 75% D) 20% E) 100% |
B) 50% |
|
|
31. If 100 individuals of an F2 population derived from Parent A and parent B were analysed for 80 markers across the genome, the number of markers that will display heterozygosity in 50 individuals will be A) 10 B) 20 C) 40 D) 60 E) 80 |
C) 40 |
|
|
32. Polymorphic tandem repeats, which are only two to five nucleotide pairs long and areextremely valuable in the creation of high density maps of eukaryotic chromosomes areknown as: A) RFLP B) EST C) Microsatellites D) Macrosatellites E) FISH |
C) Microsatellites |
|
|
33. While generating a genetic linkage map, in an ideal situation, the number oflinkage groups will equal the number of A) Chiasmata B) Crossovers C) Recombinants D) Chromosomes E) Markers |
D) Chromosomes +7 |
|
|
34. The most common changes in human genomes are single nucleotide-pair substitutions,for example, A:T to G:C or G:C to A:T substitutions, also known as what? A) NSPs B) VNTRs C) SSLPs D) SNPs E) ESTs |
D) SNPs |
|
|
35. In an F2 population of 400 individuals Tom selected 100 flies that displayed amutant phenotype. The mutants were generated by crossing a mutant (parent A) to the wild type parent B. The F2 progeny was obtained by intercrossing wild type F1individuals. If Tom checks the genotype of a marker that is very tightly linked with the mutant phenotype in the flies that he selected, how many F2 individuals will be homozygous for parent A's marker allele?A) around 10 B) around 50 C) around 25 D) close to 100 E) None |
D) close to 100 |
|
|
36. The proportion of the total phenotypic variance that is due to genetic differences amongindividuals in a population is known as A) Narrow sense heritability B) Broad sense heritability C) Segregation distortion D) Broad sense inheritance E) Narrow sense inheritance |
B) Broad sense heritability |
|
|
37. If the broad sense heritability is close to 0 for a population then A) Much of the observed variability in the population is due to genetic differences B) Much of the observed variability in the population is due to heritable traits C) Little of the observed variability is due to genetic differences D) Little of the observed variability is due to environmental differences E) All of the observed variability is due to genetic similarities |
C) Little of the observed variability is due to genetic differences |
|
|
38. The additive genetic variance, Va as a fraction of the total phenotypic variance, is called A) Total phenotypic variance B) Broad sense heritability C) Narrow sense heritability D) Broad sense inheritance E) Narrow sense inheritance |
C) Narrow sense heritability |
|
|
39. The difference between the mean of the selected parents and the mean of the populationfrom which they are selected is known as A) Selection quotient B) Selection differential C) Selection additive D) Response to selection E) Response differential |
B) Selection differential |
|
|
40. Which of the following is a component of narrow sense heritability A) Additive effects of alleles B) Dominance interactions among alleles C) Environmental effects D) Epistatic relationships among alleles E) Synergistic relationships among alleles |
A) Additive effects of alleles |
|
|
41. Which of the following is not a component of broad sense heritability A) Additive effects of alleles B) Dominance interactions among alleles C) Environmental effects D) Epistatic relationships among alleles E) Total genetic variability |
C) Environmental effects |
|
|
42. Variations in the lengths of DNA fragments produced by digestion by restrictionenzymes are known as A) RFLPs B) ESTs C) PCRs D) VNTRs E) STRs |
A) RFLPs |
|
|
43. PCR refers to A) Polymerase chain reaction B) Polymerase charge reaction C) Polymerase clear reaction D) Powerful chain reaction E) Powerful charge reaction |
A) Polymerase chain reaction |
|
|
44. QTL stands for A) Quantitative Trait Lesion B) Quantitative Trait Locus C) Quantitative Trait Leverage D) Quantitative Trait Loss E) Qualitative Trait Locus |
B) Quantitative Trait Locus |
|
|
45. Genes underlying QTL are A) impossible to identify B) typically microRNAs C) typically transcription factors D) typically no genes E) increasingly becoming possible to identify, although it is hard |
E) increasingly becoming possible to identify, although it is hard |
|
|
qualitative (simple) and another trait is quantitative (complex). Sally is likely tofind A) continuous segregation of both traits B) discrete segregation of both traits C) continuous segregation of quantitative trait and discrete segregation of qualitative trait D) discrete segregation of quantitative trait and continuous segregation ofqualitative trait E) No phenotypic segregation at all |
C) continuous segregation of quantitative trait and discrete segregation of qualitative trait |
|
|
47. Four of us were climbing the K2 peak (Himalayas). As we go higher up we found it difficult to cope with the altitude. However, all of us were not getting affected to the same level. This is due to A) Genetic differences only B) Environmental differences only C) Genotype x environment interaction D) Environment x environment interaction E) Genotype x genotype interaction |
C) Genotype x environment interaction |
|
|
48. Which of the following statements is false? A) Simple traits can also display environmental effect B) Complex traits are often controlled by several genes C) Whether a trait is a simple trait or complex trait depends both on the genetic architecture for the trait as well as the way in which the trait is measured D) Genotype x environment interactions do not contribute to total phenotypic variation E) QTL analysis is used for analyzing the genetic basis of complex traits |
D) Genotype x environment interactions do not contribute to total phenotypic variation |
|
|
49. Huntington’s disease is due to a A) point mutation B) frameshift mutation C) silent mutation D) repeat expansion E) nonsense mutation |
D) repeat expansion |
|
|
50. RFLP refers to A) Restriction Fragment Locus Polymorphism B) Redirected Fragment Length Polymorphism C) Restriction Fragment Length Polymorphism D) Restriction Feature Length Polymorphism E) Restriction Fragment Long Polymorphism |
C) Restriction Fragment Length Polymorphism |
