Mendelian Genetics Quiz

Multiple Choice Questions

• Question 1: Monohybrid cross vs. dihybrid cross: A monohybrid cross examines one trait, whereas a dihybrid cross examines two traits in an organism that is heterozygous for both traits.

• Question 2: F1 offspring of Mendel's pea cross similarities: The F1 offspring resemble one of the parental varieties due to complete dominance of one phenotype over another.

• Question 3: Number of unique gametes produced: An individual with the genotype AaBbCCDdEE can produce 16 unique gametes through independent assortment.

• Question 4: Importance of F2 generation in Mendel's experiments: Examining the F2 generation was crucial because it revealed the reappearance of traits not seen in the F1 generation, supporting the law of segregation and independent assortment.

• Question 5: Genotypes corresponding to heterozygous plants: In Figure 14.1, boxes 2 and 3 represent the genotypes corresponding to heterozygous plants with a heterozygous genotype.

• Question 6: False statement about the law of segregation: The false statement is that the method cannot be used to determine the number of chromosomes in a plant.

• Question 7: Pea plant traits and independent assortment: The fact that all seven pea plant traits obeyed independent assortment likely signifies that all the genes controlling these traits are located on different chromosomes within the pea plant itself.

• Question 8: Black fur and long tails in mice progeny: The cross BbTt × BBtt will produce 3/8 offspring with black fur and long tails.

• Question 9: Red terminal flowers from pea plants: Approximately 190 of 1000 F2 offspring are expected to have red, terminal flowers.

• Question 10: Resulting progeny from a true-breeding sharp-spined vs. spineless cactus cross: The cross yields 25% sharp-spined, 50% dull-spined, and 25% spineless progeny.

• Question 11: Description of cystic fibrosis: Cystic fibrosis is characterized by pleiotropy as it affects multiple organ systems in the human body.

• Question 12: Example of polygenic inheritance: Skin pigmentation in humans is an example of polygenic inheritance.

• Question 13: Likelihood of wooly hair in progeny of IV-3 and IV-4: The likelihood of wooly hair in the progeny of individuals IV-3 and IV-4 is 50%.

• Question 14: Explanation for male-specific white-eyed flies in F2 generation: The gene responsible for eye color (red or white) is located on the X chromosome.

• Question 15: Description of SRY: SRY is a gene region on the Y chromosome that triggers male development in humans.

• Question 16: Observation of recombination frequency: A 0.1% recombination frequency implies that the two genes involved are located very close to each other on the same chromosome.

• Question 17: Mechanism for genetic recombinants: Crossing over and independent assortment are the main mechanisms for the production of genetic recombinants.

• Question 18: Order of genes on chromosome map: The order of the genes on the chromosome map is rb-cn-vg-b based on recombination frequencies.

• Question 19: Interpretation of Barr bodies: Three Barr bodies in a human interphase nucleus indicate the presence of four X chromosomes.

• Question 20: Chromosome number after nondisjunction: When homologous chromosomes fail to separate during anaphase I, the resulting gametes will have either n+1 or n-1 chromosomes, whereas the other two gametes will have the normal haploid number.

• Question 21: Chromosomal alteration due to fragment joining: Joining a fragment to a nonhomologous chromosome results in a translocation.

• Question 22: Requirement for chromosomal inversion/translocation: Chromosome breakage and rejoining is required for chromosomal inversion or translocation. 

• Question 23: Griffith's observation in transformation experiments: Mixing a heat-killed pathogenic strain of bacteria with a live non-pathogenic strain can transform some living cells into the pathogenic form.

• Question 24: Key fact utilized by Hershey and Chase: DNA contains phosphorus, whereas protein does not.

• Question 25: Significance of Chargaff's DNA base composition analysis: Chargaff determined that the amount of adenine always equals thymine (A=T), and cytosine always equals guanine (C=G).

• Question 26: Uniform diameter of the DNA double helix: Purines pair with pyrimidines, which maintains a uniform diameter of the DNA double helix.

• Question 27: Enzyme for DNA strand elongation: DNA polymerase III.

• Question 28: Telomerase's method of solving replication at chromosome ends: Telomerase adds numerous repetitive sequences to the ends of chromosomes to prevent loss of DNA during replication. 

• Question 29: Significance of highly conserved telomere DNA: The high conservation of telomere DNA reflects a critical function that is crucial and must be maintained.

• Question 30: RNA primer sequence: The RNA primer sequence is 5'ACGUUAGG3'.

• Question 31: Enzyme that unwinds DNA strands: Helicase.

• Question 32: Difference between leading and lagging strands: The leading strand is synthesized continuously in the same direction as the replication fork, whereas the lagging strand is synthesized discontinuously in fragments (Okazaki fragments).

• Question 33: Function of topoisomerase: Topoisomerase relieves the strain in the DNA ahead of the replication fork.

• Question 34: Role of DNA ligase in lagging strand synthesis: DNA ligase joins together Okazaki fragments.

• Question 35: Proteins that hold DNA strands apart: Single-strand binding proteins (SSBPs) bind to and stabilize single-stranded DNA.

• Question 36: Impact of telomerase deficiency: Cells lacking telomerase are more likely to become cancerous, which is related to a reduction in chromosome length.

• Question 37: Description of histones: Histones are proteins that are positively charged and tightly bind to negatively charged DNA.

• Question 38: Reason for tight histone-DNA binding: The positive charge on histones and the negative charge on DNA lead to the tight binding of those two molecules.

• Question 39: Nitrogenous base found in various groups: Adenine is found in the groups of ATP, proteins, RNA, and DNA.

• Question 40: Possible template strand DNA sequence: The template strand DNA sequence is 3' AAA-GAA-TAA-CAA 5'.

• Question 41: Peptide sequence from mRNA: The peptide sequence is phe-leu-ile-val.

• Question 42: Reason for large transcription units: Transcription units are large because the mRNA has many noncoding stretches of nucleotides.

• Question 43: Actions of eukaryotic mRNA after transcription: Eukaryotic eukaryotic mRNA undergoes substantial alteration, including excision of introns.

• Question 44: Biological significance of introns: Introns allow for exon shuffling, increasing the possibilities of different protein combinations during evolution.

• Question 45: Description of snRNPs: snRNPs join together to form a large structure called a spliceosome, that assists in removing introns from the primary transcript.

• Question 46: Components of eukaryotic mRNA: Eukaryotic mRNA includes 5' and 3' untranslated regions (UTRs) located at the end of the coding sequence.

• Question 47: mRNA codon for phenylalanine: The mRNA codon for phenylalanine is UUU or UUC.

• Question 48: Most abundant type of RNA: rRNA.

• Question 49: Correct sequence of protein synthesis events: 4,1,3,2,5

• Question 50: Events during ribosome translocation: The tRNA in the A site moves into the P site, and the tRNA that was in the P site moves into the E site, and is released.

• Question 51: Products from reaction stop at stop codon: An assembled ribosome, a polypeptide, and free tRNA in the P site would be isolated.

• Question 52: Impact of point mutation on protein activity: A point mutation might substitute an amino acid in the active site of an enzyme, causing a significant change in the protein's activity, as well as alter the folding.

• Question 53: Cause of a frameshift mutation: A frameshift mutation results from either an insertion or deletion of a base.