Mendelian Genetics Quiz

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Questions and Answers

What is the difference between a monohybrid cross and a dihybrid cross?

A dihybrid cross involves organisms that are heterozygous for two characters and a monohybrid only one. (correct)

A monohybrid cross results in a 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1 ratio.

A monohybrid cross produces a single progeny, whereas a dihybrid cross produces two progeny.

A monohybrid cross is performed for one generation, whereas a dihybrid cross is performed for two generations.

A monohybrid cross involves a single parent, whereas a dihybrid cross involves two parents.

The F1 offspring of Mendel's classic pea cross always looked like one of the two parental varieties because

the traits blended together during fertilization.

one phenotype was completely dominant over another. (correct)

no genes interacted to produce the parental phenotype.

each allele affected phenotypic expression.

different genes interacted to produce the parental phenotype.

How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?

32 (correct)

It was important that Mendel examined not just the F1 generation in his breeding experiments, but the F2 generation as well, because

parental traits that were not observed in the F1 reappeared in the F2. Signup and view all the answers

Which of the boxes correspond to plants with a heterozygous genotype?

2 and 3 Signup and view all the answers

Which of the following about the law of segregation is false?

It is a method that can be used to determine the number of chromosomes in a plant. Signup and view all the answers

The fact that all seven of the pea plant traits studied by Mendel obeyed the principle of independent assortment most probably indicates which of the following?

All of the genes controlling the traits behaved as if they were on different chromosomes. Signup and view all the answers

Black fur in mice (B) is dominant to brown fur (b). Short tails (T) are dominant to long tails (t). What fraction of the progeny of the cross BbTt × BBtt will have black fur and long tails?

1/2 Signup and view all the answers

Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all F1 individuals have red, axial flowers. The genes for flower color and location assort independently. If 1,000 F2 offspring resulted from the cross, approximately how many of them would you expect to have red, terminal flowers?

190 Signup and view all the answers

A cross between a true-breeding sharp-spined cactus and a spineless cactus would produce

all sharp-spined progeny. Signup and view all the answers

Cystic fibrosis affects the lungs, the pancreas, the digestive system, and other organs, resulting in symptoms ranging from breathing difficulties to recurrent infections. Which of the following terms best describes this?

Pleiotropy Signup and view all the answers

Which of the following is an example of polygenic inheritance?

Skin pigmentation in humans Signup and view all the answers

What is the likelihood that the progeny of IV-3 and IV-4 will have wooly hair?

50% Signup and view all the answers

When Thomas Hunt Morgan crossed his red-eyed F1 generation flies to each other, the F2 generation included both red- and white-eyed flies. Remarkably, all the white-eyed flies were male. What was the explanation for this result?

The gene involved is on the X chromosome. Signup and view all the answers

SRY is best described in which of the following ways?

A gene region present on the Y chromosome that triggers male development Signup and view all the answers

A 0.1% frequency of recombination is observed

in genes located very close to one another on the same chromosome. Signup and view all the answers

What is the mechanism for the production of genetic recombinants?

Crossing over and independent assortment Signup and view all the answers

In a series of mapping experiments, the recombination frequencies for four different linked genes of Drosophila were determined as shown in the figure. What is the order of these genes on a chromosome map?

b-rb-cn-vg Signup and view all the answers

If a human interphase nucleus contains three Barr bodies, it can be assumed that the person

has four X chromosomes. Signup and view all the answers

If a pair of homologous chromosomes fails to separate during anaphase of meiosis I, what will be the chromosome number of the four resulting gametes with respect to the normal haploid number (n)?

n + 1; n + 1; n - 1; n - 1 Signup and view all the answers

One possible result of chromosomal breakage is for a fragment to join a nonhomologous chromosome. What is this alteration called?

Translocation Signup and view all the answers

In order for chromosomes to undergo inversion or translocation, which of the following is required?

Chromosome breakage and rejoining Signup and view all the answers

In his transformation experiments, what did Griffith observe?

Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form. Signup and view all the answers

In trying to determine whether DNA or protein is the genetic material, Hershey and Chase made use of which of the following facts?

DNA contains phosphorus, but protein does not. Signup and view all the answers

Chargaff's analysis of the relative base composition of DNA was significant because he was able to show that

the amount of A is always equivalent to T, and C to G. Signup and view all the answers

Why does the DNA double helix have a uniform diameter?

Purines pair with pyrimidines. Signup and view all the answers

Which enzyme catalyzes the elongation of a DNA strand in the 5' → 3' direction?

DNA polymerase III Signup and view all the answers

The enzyme telomerase solves the problem of replication at the ends of linear chromosomes by which method?

adding numerous short DNA sequences such as TTAGGG, which form a hairpin turn Signup and view all the answers

The DNA of telomeres has been found to be highly conserved throughout the evolution of eukaryotes. What does this most probably reflect?

that the critical function of telomeres must be maintained Signup and view all the answers

At a specific area of a chromosome, the sequence of nucleotides below is present where the chain opens to form a replication fork: 3'CCTAGGCTGCAATCC 5' An RNA primer is formed starting at the underlined T (T) of the template. Which of the following represents the primer sequence?

5'ACGUUAGG3' Signup and view all the answers

Which of the following separates the DNA strands during replication?

helicase Signup and view all the answers

The leading and the lagging strands differ in that

the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction. Signup and view all the answers

What is the function of topoisomerase?

relieving strain in the DNA ahead of the replication fork Signup and view all the answers

What is the role of DNA ligase in the elongation of the lagging strand during DNA replication?

join Okazaki fragments together Signup and view all the answers

Which of the following help to hold the DNA strands apart while they are being replicated?

single-strand binding proteins Signup and view all the answers

Which would you expect of a eukaryotic cell lacking telomerase?

a reduction in chromosome length Signup and view all the answers

Which of the following statements describes histones?

Histone H1 is not present in the nucleosome bead; instead it is involved in the formation of higher-level chromatin structures. Signup and view all the answers

Why do histones bind tightly to DNA?

Histones are positively charged, and DNA is negatively charged. Signup and view all the answers

The nitrogenous base adenine is found in all members of which group?

ATP, RNA, and DNA Signup and view all the answers

A possible sequence of nucleotides in the template strand of DNA that would code for the polypeptide sequence phe-leu-ile-val would be

3' AAA-GAA-TAA-CAA 5' Signup and view all the answers

What is the sequence of a peptide based on the following mRNA sequence? 5'... UUUUCUUAUUGUCUU 3'

phe-ser-tyr-cys-leu Signup and view all the answers

A transcription unit that is 8,000 nucleotides long may use 1,200 nucleotides to make a protein consisting of approximately 400 amino acids. This is best explained by the fact that

many noncoding stretches of nucleotides are present in mRNA. Signup and view all the answers

Once transcribed, eukaryotic mRNA typically undergoes substantial alteration that includes

excision of introns. Signup and view all the answers

Introns are significant to biological evolution because

their presence allows exons to be shuffled. Signup and view all the answers

Which of the following is (are) true of snRNPs?

They join together to form a large structure called the spliceosome. Signup and view all the answers

Study Notes

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.

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Description

Test your knowledge on Mendelian genetics with this quiz focused on the principles of monohybrid and dihybrid crosses, F1 and F2 generations, and gamete production. Explore key concepts from Mendel's experiments and deepen your understanding of inheritance patterns.