a. X-linked. | ||
b. incomplete dominant. | ||
c. Y-linked. | ||
d. autosomal recessive. |
a. There is a 25% chance that Joe inherited Huntington’s disease from his father. | ||
b. There is a 50% chance that Joe inherited Huntington’s disease from his father. | ||
c. There is a 75% chance that Joe inherited Huntington’s disease from his father. | ||
d. There is a 100% chance that Joe inherited Huntington’s disease from his father. |
a. 2/3 | ||
b. 4/4 | ||
c. 3/4 | ||
d. 1/4 |
a. The genotype of all offspring is Ff Gg. | ||
b. 12.5% of the offspring has FfGg genotype. | ||
c. The offspring has parental genotype. | ||
d. The genotypic ratio is 3:1. |
a. The genotype of all offspring is Ff Gg. | ||
b. 12.5% of the offspring has FfGg genotype. | ||
c. The offspring has parental genotype. | ||
d. The genotypic ratio is 3:1. |
a. The genotypes of individuals A and B are ff, and C is Ff. | ||
b. The genotype of individual A is Ff, B is ff, and C is FF. | ||
c. The genotypes of individuals A and B are Ff, and C is ff. | ||
d. The genotype of individuals A and B are FF, and C is ff. |
a. 3/4 | ||
b. 1/8 | ||
c. 3/32 | ||
d. 7/64 |
a. 3/32 | ||
b. 1/16 | ||
c. 3/8 | ||
d. 1/4 |
a. 0% | ||
b. 25% | ||
c. 50% | ||
d. 75% |
a. The purple and the white alleles exhibit incomplete dominance, while the short allele is dominant to the tall allele. | ||
b. The purple and the short alleles are dominant; otherwise, the offspring would be white and tall. | ||
c. The pink and the white alleles are codominant; the short allele is dominant to the tall allele. | ||
d. The purple and white alleles are recessive; the tall allele is subject to X-inactivation. |
a. 1:2:1. | ||
b. 9:3:3:1. | ||
c. 3:1. | ||
d. 4:0. |
a. 100 chromosomes. | ||
b. 25 chromosomes. | ||
c. 75 chromosomes. | ||
d. 50 chromosomes. |
a. female and male gametogeneses are homologous throughout: both result in four different gametes derived from the primary gametocyte through meiotic cell division. | ||
b. female and male gametogeneses are homologous, because both result in the production of gametes through meiosis; however, while one primary spermatocyte gives rise to four sperm cells, one primary oocyte gives rise to only one oocyte. | ||
c. female and male gametogeneses are not homologous, because one primary spermatocyte gives rise to four sperm cells, while one primary oocyte gives rise to only one oocyte. | ||
d. female and male gametogeneses are not homologous, because one primary spermatocyte gives rise to four sperm cells through meiosis; one primary oocyte gives rise to the secondary oocyte through meiosis, but the oocyte is the result of mitotic cell division. |
a. Yes, it is possible, because the polar bodies and the oocyte are daughter cells that are derived from the parental primary oocyte through cell division. | ||
b. Yes, it is possible but only from the second polar body, because it is genetically identical to the oocyte. | ||
c. No, it is not possible, because the polar bodies and the oocyte are daughter cells that are derived from the parental primary oocyte through meiosis. | ||
d. No, it is not possible from the second polar body, but it is possible to determine from the first polar body. |
a. equational division | ||
b. reductional division | ||
c. kinetochore separation | ||
d. polar separation |
a. The grasshoppers, because 223 times 223 is greater than 24 times 24 | ||
b. The grasshoppers, because 223 is greater than 24 | ||
c. The grasshoppers, because 2 times 23 is greater than 2 times 4 | ||
d. The fruit flies, because they have a smaller haploid number |
a. 223 | ||
b. 232 | ||
c. 46 | ||
d. 4 |
a. Both mitosis and meiosis results in daughter cells that are identical to each other and to the parental cell; thus, there will be no unique daughter cells. | ||
b. There will be no unique daughter cells after mitosis; there will be four different daughter cells after meiosis, and these cells will also be different from the parental cell. | ||
c. There will be no unique daughter cells after meiosis; there will be eight different daughter cells after mitosis, and these cells will also be different from the parental cell. | ||
d. There will be no unique daughter cells after mitosis; there will be six different daughter cells after meiosis, and these cells will also be different from the parental cell. |
a. Epithelial tissue, because the tumor suppressor level is higher than the protooncogen level in this tissue | ||
b. Lymphatic cell, because the lymphatic system is not exposed to UV radiation | ||
c. Neuronal tissue, because the cell proliferation is low in the neuronal tissue | ||
d. Tendon, because oncogens cannot induce excessive cell division in this tissue due to special constraints |
a. The conversion of a tumor suppressor to an oncogene contributes to cancer development, because oncogenes turn on the cell cycle check point genes. | ||
b. The conversion of an oncogene to a proto-oncogene contributes to cancer development, because this results in excessive cell cycle control. | ||
c. The overexpression of a proto-oncogene contributes to cancer development, because proto-oncogenes turn off the cell cycle check point genes. | ||
d. The overexpression of an oncogene contributes to cancer development, because this results in excessive cell division. |
a. Crossing over occurred as Ascus B was forming but did not occur in Ascus A. | ||
b. Crossing over did not occur. | ||
c. Crossing over occurred as Ascus A was forming but did not occur in Ascus B. | ||
d. Crossing over occurred as Ascus A and Ascus B were forming. |
a. genes separate during gamete formation. | ||
b. allele pairs separate during gamete formation. | ||
c. genes mix through crossing-over. | ||
d. alleles are either recessive or dominant. |
a. Epistasis | ||
b. Epigenetics | ||
c. Dominance | ||
d. Imprinting |
a. Genetics cannot support the man’s case based on the blood types in this family: A and B alleles are dominant over O allele; thus, O blood type is a possibility for the child if both parents are heterozygotes. | ||
b. Genetics can support the man’s case based on the blood types in this family: A and B alleles are co-dominant; thus, the child should have AB blood type. O blood type is not a possibility for the child. | ||
c. Genetics cannot support the man’s case based on the blood types: blood type alleles are co-dominant; thus, the child should have ABO genotype. | ||
d. Genetics can support the man’s case based on the blood types: the inheritance of A and B alleles are incomplete dominant; thus, the child of A and B blood type parents must have AB blood type. |
a. is reset during gamete formation. | ||
b. may come from either parent by chance. | ||
c. is identical to the parental imprinted allele. | ||
d. stays imprinted during gamete formation. |
a. with an allele who exhibit a trait associated with the allele. | ||
b. with a trait that is associated with multiple genes. | ||
c. who exhibit a particular trait in a population. | ||
d. who has a particular allele in a population. |
a. epistatic, imprinted | ||
b. recessive, dominant | ||
c. epigenetic, autosomal | ||
d. multigenic, single gene |
a. Yes it can. The red-green color blindness allele is dominant, and it is on the X-chromosome; thus, little Sue’s biological father or mother must be red-green colorblind. Joe is not the biological father, because Jane is not red-green colorblind. | ||
b. No, it cannot. Red-green color blindness is a recessive autosomal condition; thus, if both parents are carriers, then there is a 50% chance that their daughter would be red-green color blind. | ||
c. No, it cannot. Red-green color blindness is maternally imprinted; thus, if Jane is a heterozygote, then there is a 50% chance that their daughter would be color blind. | ||
d. Yes, it can. Red-green color blindness allele is recessive, and it is on the X-chromosome; thus, little Sue’s biological father must be red-green colorblind. Joe is not the biological father. |
a. Epigenetic inheritance | ||
b. Mendelian inheritance | ||
c. Genomic imprinting | ||
d. Genomic silencing |
a. The altered phenotypic ratios from what was expected for dominant and/or Mendelian inheritance | ||
b. The reduced fertility of the homozygous recessive lethal mice | ||
c. The altered genotypic ratios from what was expected for dominant and/or Mendelian inheritance | ||
d. The reduced number of individuals in the laboratory mice population |
a. A is XFXf, B is XfXf, and C is XFXf. | ||
b. A is XFXF or XFXf, B is XFXf, and C is XFXF or XFXf | ||
c. A is XFXf, B is XFXf, and C is XFXF or XFXf. | ||
d. A and C are XfXf, and B is XFXf. |
a. Dominant | ||
b. Mitochondrial | ||
c. Recessive | ||
d. X-linked |
a. A, B, and O | ||
b. AB only | ||
c. A and B only | ||
d. A, B, AB, and O |
a. it takes place during gamete formation. | ||
b. kinetochore microtubules pull the sister chromatids away from each other very fast. | ||
c. the exchanged chromosome segments are identical. | ||
d. homologous non-sister chromatids exchange segments. |
a. Yes, we can conclude from the recombination frequencies that the order of the genes is ACB. | ||
b. Yes, we can conclude from the recombination frequencies that the order of the genes is CAB. | ||
c. No, there is not enough data available; the order of the genes can be either ACB or CAB. | ||
d. No, there is not enough data available; the order of the genes can be either ACB or BCA. |
a. 4.8 cM | ||
b. 47.5 cM | ||
c. 0.9 cM | ||
d. 23.7 cM |
a. 4.8 cM | ||
b. 47.5 cM | ||
c. 41.2 cM | ||
d. 2.4 cM |
a. 25 cM | ||
b. 20 cM | ||
c. 10 cM | ||
d. 12.5 cM |
a. 10 cM | ||
b. 6 cM | ||
c. 12 cM | ||
d. 20 cM |
a. Crossing over | ||
b. Epistasis | ||
c. Crosslinking | ||
d. Epigenetics |
a. All crossing over is genetic recombination. | ||
b. All genetic recombination is crossing over. | ||
c. The appearance of non-parental phenotype combinations is always the result of crossing over. | ||
d. The disappearance of non-parental phenotype combinations is always the result of crossing over. |
a. The recombination frequency is higher between genes that are close to each other during meiosis. | ||
b. The recombination frequency is higher in prokaryotes than in eukaryotes during meiosis. | ||
c. The recombination frequency is random throughout the genome during meiosis. | ||
d. The recombination frequency is more frequent in certain regions of the genomes during meiosis. |
a. A D C B or B C D A | ||
b. A C B D or D B C A | ||
c. A D B C | ||
d. D B C A |
a. H, B, A, G | ||
b. B, H, G, A | ||
c. A, B, H, G | ||
d. B, H, A, G |
a. triploid. | ||
b. diploid. | ||
c. haploid. | ||
d. aneuploid. |
a. anaphase, telophase | ||
b. prointerphase, interphase | ||
c. metaphase, prometaphase | ||
d. cytokinesis, telophase |
a. A | ||
b. B | ||
c. C | ||
d. D |
a. 47 | ||
b. 43 | ||
c. 46 | ||
d. 44 |
a. Exchange | ||
b. Inverse translocation | ||
c. Reciprocal translocation | ||
d. Transposition |
a. Duplication | ||
b. Inversion | ||
c. Inversion | ||
d. Mutation |
a. 1. deletion, 2. duplication, and 3. inversion | ||
b. 1. addition, 2. replication, 3. crossing | ||
c. 1. loss, 2. insertion, 3. translocation | ||
d. 1. elimination, 2. doubling, 3. reciprocal inversion |
a. 1. inverse deletion and 2. translocation | ||
b. 1. translocation and 2. reciprocal translocation | ||
c. 1. transposition and 2. exchange | ||
d. 1. elimination and 2. reciprocal inversion |
a. Klinefelter syndrome | ||
b. Edward’s syndrome | ||
c. Patau syndrome | ||
d. Down syndrome |
a. Duplication | ||
b. Deletion | ||
c. Inversion | ||
d. Aneuploidy |
a. Nondisjunction during meiosis | ||
b. Nondisjunction during mitosis | ||
c. Folic acid deficiency | ||
d. Maternal allele inheritance |
a. 25 times more | ||
b. 225 times more | ||
c. 252 times more | ||
d. 50 times more |
a. The in vivo process always takes place in a cell; the in vitro process always takes place in a thermocycler. | ||
b. The in vitro process may involve thermostable DNA polymerase; the in vivo process always utilizes DNA polymerase isolated from Thermus aquaticus. | ||
c. The in vivo process uses ribonucleotide primer; the in vitro process uses oligonucleotide primer for DNA polymerase. | ||
d. The in vivo process does not use dNTPs; the in vitro process uses dNTPs. |
a. it is technically difficult to determine the phenotypes. | ||
b. the structure of the dsDNA was not known yet. | ||
c. these elements were jumping on the chromosomes. | ||
d. Ds and Ac cancelled each other out. |
a. they never contain mistakes. | ||
b. they are highly variable. | ||
c. they are short. | ||
d. they are long. |
a. To identify point mutations | ||
b. To analyze DNA fingerprints | ||
c. To amplify genomic DNA | ||
d. To analyze differential gene expression in tissues |
a. Plasmids, Nuclear | ||
b. Mitochondrials, Nuclear | ||
c. Organelles, Chromosomal | ||
d. Plastids, Chromosomal |
a. Polymerization, denaturation, melting | ||
b. Melting, base-pairing, elongation | ||
c. Annealing, melting, polymerization | ||
d. Denaturation, annealing, elongation |
a. Cisgenesis only | ||
b. Transgenesis only | ||
c. Transgenesis and cisgenesis | ||
d. Transgenesis and conventional breeding |
a. B shows a transposon, and A shows a retrotransposon. | ||
b. A shows a transposon, and B shows a retrotransposon. | ||
c. A shows a transposon, and B shows a duplicating transposon. | ||
d. B shows a duplicating transposon, and A shows a retrotransposon. |
a. Interchromatine | ||
b. Chromosome | ||
c. Nucleolus | ||
d. Plasmid |
a. Recombinant DNA always combines DNA from at least two sources. | ||
b. Recombinant DNA is the end-result of every cloning process. | ||
c. Recombinant DNA always uses a restriction endonuclease and a ligase. | ||
d. Recombinant DNA is always made in the test tube, not in a living cell. |
a. Class II transposons are segments of DNA that move from one place to another through the cut-and-paste mechanism. | ||
b. Class II transposons are characterized by their short lengths and stretch of 15 base pairs that occur at each end of the transposon. | ||
c. Class II transposons are part of a highly unique group that forms large portions of the genomes of eukaryotes. | ||
d. Class II transposons are incorporated through retrotransposition. |
a. It only takes one mutation of one base pair to make a large difference in protein production. | ||
b. In order to detect a mutation, at least three base pairs must change at the same time. | ||
c. Populations are often filled with individuals that have a mutation in the first base of the triplet codon, because this does not make much of a difference in how the protein is made. | ||
d. Silent mutations most often happen in the second base of the triplet codon. |
a. each gene codes for one characteristic. | ||
b. every gene can be correlated to a trait that it impacts. | ||
c. each gene codes for the production of a specific enzyme. | ||
d. every gene codes for an enzyme that will impact only one trait. |
a. viral transduction. | ||
b. transformation. | ||
c. R-factor. | ||
d. F-factor. |
a. merozygote. | ||
b. homologous pair. | ||
c. heterozygote. | ||
d. disomy. |
a. a host DNA. | ||
b. a bacteriophage. | ||
c. a nuclease. | ||
d. a virulent strain. |
a. Transfection | ||
b. Induction | ||
c. Transformation | ||
d. Transposition |
a. Transfection | ||
b. Transduction | ||
c. Induction | ||
d. Transposition |
a. Conjugation | ||
b. Transformation | ||
c. Transduction | ||
d. Mating |
a. This is the Trp operon, which is repressed by tryptophan (5). Tryptophan inactivates the RNA polymerase (2) by removing it from the DNA and leaving the transcription factor (3) behind. | ||
b. This is the Trp operon (4), which is repressed by tryptophan (5). Tryptophan inactivates the RNA polymerase (2) by removing it from the DNA and leaving the transcription factors (3 and 4) behind. | ||
c. This is the Lac operon, which is induced by allolactose (5). Allolactose inactivates the repressor (2) and allows the RNA polymerase (3) to transcribe the genes. | ||
d. This is the Lac operon (4), which is induced by lactose (5). Lactose inactivates the repressor (2) and allows the RNA polymerase (3) to transcribe the genes. |
a. Transfection | ||
b. Transduction | ||
c. Induction | ||
d. Conjugation |
a. Introns are removed from the transcript before translation begins. | ||
b. Genes that play a role in the same metabolic pathway are organized into an operator. | ||
c. The expression of genes that play a role in the same metabolic pathway is orchestrated by upstream enhancers. | ||
d. Genes that play a role in the same metabolic pathway are organized into an operon. |
a. The Lac operator | ||
b. Positive feedback | ||
c. Negative feedback | ||
d. The Lac operon |
a. produces overexpression mutants and is also a tool to analyze the natural expression of that gene if the transposon has a reporter gene. | ||
b. generates jumping genes and mosaic yeast cell; this is similar to the transposon effect in corn that makes color-colorless strikes in seeds. | ||
c. generates transgenic yeast that glows in the dark; thus, it attracts great interest in the bakeries. | ||
d. produces gene disruption mutants and is also a tool to analyze the expression of that gene if the transposon has a reporter gene. |
a. Y chromosomes. | ||
b. X chromosomes. | ||
c. somatic chromosomes. | ||
d. autosomes. |
a. they have physically seen the crossover events on electron micrographs. | ||
b. they found genes in the chloroplast that were known to be mitochondrial. | ||
c. the endosymbiosis hypothesis predicts such gene transfers. | ||
d. proper organellar function requires nuclear gene products. |
a. as the body of an organism becomes more complex, the size of the genome will increase. | ||
b. there can be a large difference in the size of the genome even between organisms that are similar in body pattern and size. | ||
c. as the body of an organism becomes more complex, the size of the genome will decrease. | ||
d. prokaryotic genome is not always smaller than the eukaryotic genome. |
a. only the regulatory regions of the genes differ in animals; all animals have the same number of genes. | ||
b. the fugu genome has less exon and less intergene space. | ||
c. the fugu genome has smaller introns, less intergene space, and fewer repeats. | ||
d. the fugu genome has no intergene space and no repeats. |
a. Yeast two-hybrid system | ||
b. Eukaryotic UAS-Gal4 control | ||
c. Bacterial transcriptional activation | ||
d. Plant gene induction |
a. Series of bacterial conjugation | ||
b. A theory for the origin of eukaryotes | ||
c. A theory for the origin of prokaryotes | ||
d. Plasmid multiplication in fused bacteria |
a. Approximately 50% in flies and 99% in mice | ||
b. 0% in flies and 75% in mice | ||
c. Approximately 10% in flies and 75% in mice | ||
d. 100% in both flies and mice |
a. A locus refers to the location on the chromosome, whereas an allele is used to refer to the alternative forms that a gene comes in. | ||
b. An allele refers to the location on the chromosome, whereas the term locus refers to the alternative forms of the gene. | ||
c. A locus refers to the phenotype of the organism, whereas an allele refers to the genotype. | ||
d. None of these answers |
a. GloFish | ||
b. Glowing bacteria | ||
c. Glow in the dark cats | ||
d. Glowing mice |
a. it spends most of its life cycle in diploid condition. | ||
b. it spends most of its life cycle in the haploid condition. | ||
c. when the zygote undergoes mitosis or meiosis, it spreads. | ||
d. its genome is sequenced. |
a. They will not change color. | ||
b. They will turn pink. | ||
c. They will turn brown. | ||
d. None of these answers |
a. these regions commonly show evidence of horizontal gene transfer; thus, they simplify the tree construction. | ||
b. the repeats in these regions simplify the in silico sequence analysis. | ||
c. there are more non-coding regions than coding regions; thus, this approach offers more data to work with. | ||
d. mutations in these regions are less influenced by natural selection. |
a. harsh weather that no other crop could survive. | ||
b. endotoxin toxicity. | ||
c. the low diversity in the population. | ||
d. the sudden diversity increase in the population. |
a. genetic drift. | ||
b. the gene flow. | ||
c. growing human populations. | ||
d. nonrandom mating. |
a. small, the founder effect | ||
b. small, nonrandom mating | ||
c. large, the bottleneck effect | ||
d. large, genetic drift |
a. Mutation | ||
b. Emigration | ||
c. The founder effect | ||
d. The bottle neck effect |
a. 25% for the A allele and 75% for the a allele | ||
b. 37.5% for the A allele and 62.5% for the a allele | ||
c. 62.5% for the A allele and 37.5% for the a allele | ||
d. 75% for the A allele and 25% for the a allele |
a. p + q = 1 | ||
b. 2pq = 1 | ||
c. p – q = 0 | ||
d. p + 2pq + q = 0 |
a. Stabilizing selection | ||
b. Bidirectional selection | ||
c. Disruptive selection | ||
d. Directional selection |
a. Genotype frequency is always the same as phenotypic frequency. | ||
b. Genotype frequency and phenotypic frequency are different. | ||
c. Allele frequency always determines gene frequency. | ||
d. Allele frequency always determines phenotypic frequency. |
a. Founder effects | ||
b. Nonrandom mating | ||
c. Bottleneck effect | ||
d. Genetic drift |
a. the population is evolving. | ||
b. the population is not evolving. | ||
c. the population is large. | ||
d. there is more than one species in the population. |