Biology 200 Exam 3 Questions with
Complete Answers
Law of Independent Assortment (Ch. 14) - ANSWER-When two or more
characteristics are inherited , individual hereditary factors assort independently
during gamete production, giving different traits an equal opportunity of occurring
together
Multiplication Rule (Ch. 14) - ANSWER-To determine the probability of two or more
traits together multiply the probability of one event by the probability of the other
Addition Rule (Ch. 14) - ANSWER-To determine the probability of at least one event
will occur add the probability of each event occurring together
Incomplete Dominance (Ch. 14) - ANSWER-Mixture of both possible outcomes;
blending
Codominance (Ch. 14) - ANSWER-Fully both phenotypes
Epistasis (Ch. 14) - ANSWER-Second gene determines whether or not the first gene
is allowed to be expressed, controls phenotype
Polygenic Inheritance (Ch. 14) - ANSWER-Many alleles influence; range of possible
phenotypes
Pleiotropy (Ch. 14) - ANSWER-One gene has more than one effect
Barr Body (Ch. 15) - ANSWER-Inactive X chromosome in females; random chance
Wild-Type (Ch. 15) - ANSWER-Original trait/gene; most common in the wild
Mutant (Ch. 15) - ANSWER-Variation observed rarely in wild, or deliberately
changed
Linked Genes (Ch. 15) - ANSWER-Two traits appear together when on the same
chromosome
Chi-Square (Ch. 15) - ANSWER-The test used to verify if genes are linked.
Genomic Imprinting (Ch. 15) - ANSWER-Phenotype varies depending on whether
the allele came from the mother or father.
Organelle Genes (Ch. 15) - ANSWER-Mitochondria and chloroplast genes can only
come from eggs not sperm.
, Deletion (Ch. 15) - ANSWER-Removes a chromosomal segment.
Duplication (Ch. 15) - ANSWER-Repeats a chromosomal segment.
Inversion (Ch. 15) - ANSWER-Reverses a segment within a chromosome.
Translocation (Ch. 15) - ANSWER-Moves a segment from one chromosome to a
nonhomologous chromosome.
Transformation (Ch. 16) - ANSWER-Some chemical in dead cells is picked up by
living cells.
Semiconservative DNA Replication (Ch. 16) - ANSWER-Forms new strands
complementary to the template strands made by separating the parental molecule.
DNA Proofreading and Repair (Ch. 16) - ANSWER-Initial errors are detected by
polymerase and replaced. Later errors are cut by nuclease, corrected by DNA
polymerase, and sealed by ligase.
Telomeres (Ch. 16) - ANSWER-Extra repetitions of non-gene DNA at the five prime
end to protect the genes.
Chromatin (Ch. 16) - ANSWER-The combination of DNA and histones.
Transcription (Ch. 17 and Ch. 18) - ANSWER-Makes a RNA copy of the DNA code.
Translation (Ch. 17 and Ch. 18) - ANSWER-Translates from nucleic acid language to
protein language.
Codon (Ch. 17 and Ch. 18) - ANSWER-Codes for amino acids made from three
nucleotides. Start codon is AUG and there are three different stop codons.
Steps of Transcription (Ch. 17 and Ch. 18) - ANSWER-1) Initiation: Binding to
promoter
2)Elongation: RNA polymerase
3) Termination: Prokaryotes- termination sequence and polymerase detaches
Eukaryotes- signal then pre-mRNA cut free by proteins
Chaperone Protein (Ch. 17 and Ch. 18) - ANSWER-Protein that helps with correct
polypeptide folding.
Regulating Cell Activity (Ch. 17 and Ch. 18) - ANSWER-Can turn enzymes on or off.
Gene Regulation (Ch. 17 and Ch. 18) - ANSWER-Long term regulation of processes
by changing enzyme amounts.
Silent Mutation (Ch. 17 and Ch. 18) - ANSWER-No effect on amino acid sequence.
Complete Answers
Law of Independent Assortment (Ch. 14) - ANSWER-When two or more
characteristics are inherited , individual hereditary factors assort independently
during gamete production, giving different traits an equal opportunity of occurring
together
Multiplication Rule (Ch. 14) - ANSWER-To determine the probability of two or more
traits together multiply the probability of one event by the probability of the other
Addition Rule (Ch. 14) - ANSWER-To determine the probability of at least one event
will occur add the probability of each event occurring together
Incomplete Dominance (Ch. 14) - ANSWER-Mixture of both possible outcomes;
blending
Codominance (Ch. 14) - ANSWER-Fully both phenotypes
Epistasis (Ch. 14) - ANSWER-Second gene determines whether or not the first gene
is allowed to be expressed, controls phenotype
Polygenic Inheritance (Ch. 14) - ANSWER-Many alleles influence; range of possible
phenotypes
Pleiotropy (Ch. 14) - ANSWER-One gene has more than one effect
Barr Body (Ch. 15) - ANSWER-Inactive X chromosome in females; random chance
Wild-Type (Ch. 15) - ANSWER-Original trait/gene; most common in the wild
Mutant (Ch. 15) - ANSWER-Variation observed rarely in wild, or deliberately
changed
Linked Genes (Ch. 15) - ANSWER-Two traits appear together when on the same
chromosome
Chi-Square (Ch. 15) - ANSWER-The test used to verify if genes are linked.
Genomic Imprinting (Ch. 15) - ANSWER-Phenotype varies depending on whether
the allele came from the mother or father.
Organelle Genes (Ch. 15) - ANSWER-Mitochondria and chloroplast genes can only
come from eggs not sperm.
, Deletion (Ch. 15) - ANSWER-Removes a chromosomal segment.
Duplication (Ch. 15) - ANSWER-Repeats a chromosomal segment.
Inversion (Ch. 15) - ANSWER-Reverses a segment within a chromosome.
Translocation (Ch. 15) - ANSWER-Moves a segment from one chromosome to a
nonhomologous chromosome.
Transformation (Ch. 16) - ANSWER-Some chemical in dead cells is picked up by
living cells.
Semiconservative DNA Replication (Ch. 16) - ANSWER-Forms new strands
complementary to the template strands made by separating the parental molecule.
DNA Proofreading and Repair (Ch. 16) - ANSWER-Initial errors are detected by
polymerase and replaced. Later errors are cut by nuclease, corrected by DNA
polymerase, and sealed by ligase.
Telomeres (Ch. 16) - ANSWER-Extra repetitions of non-gene DNA at the five prime
end to protect the genes.
Chromatin (Ch. 16) - ANSWER-The combination of DNA and histones.
Transcription (Ch. 17 and Ch. 18) - ANSWER-Makes a RNA copy of the DNA code.
Translation (Ch. 17 and Ch. 18) - ANSWER-Translates from nucleic acid language to
protein language.
Codon (Ch. 17 and Ch. 18) - ANSWER-Codes for amino acids made from three
nucleotides. Start codon is AUG and there are three different stop codons.
Steps of Transcription (Ch. 17 and Ch. 18) - ANSWER-1) Initiation: Binding to
promoter
2)Elongation: RNA polymerase
3) Termination: Prokaryotes- termination sequence and polymerase detaches
Eukaryotes- signal then pre-mRNA cut free by proteins
Chaperone Protein (Ch. 17 and Ch. 18) - ANSWER-Protein that helps with correct
polypeptide folding.
Regulating Cell Activity (Ch. 17 and Ch. 18) - ANSWER-Can turn enzymes on or off.
Gene Regulation (Ch. 17 and Ch. 18) - ANSWER-Long term regulation of processes
by changing enzyme amounts.
Silent Mutation (Ch. 17 and Ch. 18) - ANSWER-No effect on amino acid sequence.
