Protein Structure & Bonds
A newly synthesized soluble protein is folding in the aqueous environment of the cytoplasm. Several nonpolar amino acid side chains become buried within the center of the protein rather than remaining exposed to water.
Which interaction is the major driving force for this process?
A. Covalent interactions
B. Hydrogen bonds
C. Hydrophobic interactions
D. Ionic interactions
E. Peptide bonds
Answer: C. Hydrophobic interactions
Explanation: Hydrophobic interactions drive nonpolar side chains away from water and into the protein interior, which is a major driving force for protein folding. You specifically annotated this as “VERY IMPORTANT” and “driving force is hydrophobic interactions
A researcher compares DNA and RNA nucleotides.
Which feature is present in RNA but not DNA?
A. Phosphate group
B. Nitrogenous base
C. 2′ hydroxyl group on the sugar
D. 5′ carbon
E. Phosphodiester bonds
Answer: C. 2′ hydroxyl group on the sugar
Explanation: RNA contains ribose, whereas DNA contains deoxyribose. Ribose has the additional 2′-OH group.
A child has recurrent painful vaso-occlusive episodes and is diagnosed with sickle cell disease.
What is the inheritance pattern?
A. Autosomal dominant
B. Autosomal recessive
C. X-linked dominant
D. X-linked recessive
E. Mitochondrial
Answer: B. Autosomal recessive
Explanation: Sickle cell disease is autosomal recessive. Rodda specifically emphasized knowing its inheritance pattern.
A cell enters the portion of interphase during which its DNA content doubles.
Which phase is it entering?
A. G1
B. S
C. G2
D. M
E. G0
Answer: B. S phase
Explanation: DNA replication occurs during S phase.
A newborn is found to have three copies of chromosome 21.
Which term best describes this numerical abnormality?
A. Polyploidy
B. Aneuploidy
C. Balanced translocation
D. Inversion
E. Mosaicism
Answer: B. Aneuploidy
Explanation: Aneuploidy is the gain or loss of individual chromosomes. Trisomy 21 is therefore an aneuploidy.
A researcher is studying a protein containing a short α-helical region. A mutation introduces an amino acid that disrupts the regular structure of the α-helix but would be well tolerated within a β-turn.
Which amino acid is most likely responsible?
A. Alanine
B. Leucine
C. Valine
D. Proline
E. Methionine
Answer: D. Proline
Explanation: Proline and glycine are poor α-helix-forming residues and are favored in β-turns. Proline is especially disruptive because its rigid cyclic structure interferes with normal α-helical geometry. Your review annotation specifically says glycine and proline = “bad for helix—they are breakers” and “amazing in beta turns.
A double-stranded DNA molecule contains 30% adenine.
Approximately what percentage of the bases are guanine?
A. 10%
B. 20%
C. 30%
D. 40%
E. 60%
Answer: B. 20%
Explanation: Chargaff's rules give A = T and G = C. If A = 30%, T = 30%, leaving 40% for G + C. Therefore, G = 20%. Your annotation specifically says to understand why Chargaff's rules are true, when they aren't, and the math using the rules.
DNA fragments are placed into an agarose gel and exposed to an electric field.
Which DNA fragments migrate the farthest through the gel?
A. Largest fragments
B. Smallest fragments
C. Most positively charged fragments
D. Most GC-rich fragments
E. All migrate the same distance
Answer: B. Smallest fragments
Explanation: Gel electrophoresis separates DNA based primarily on size. Smaller DNA fragments migrate more readily through the gel matrix. DNA travels toward the positive electrode. Your notes emphasize both points
A normal human somatic cell has completed DNA replication but has not yet entered mitosis.
Which combination correctly describes the cell?
A. 1n, 1c
B. 1n, 2c
C. 2n, 2c
D. 2n, 4c
E. 4n, 4c
Answer: D. 2n, 4c
Explanation: DNA replication doubles the amount of DNA but does not change ploidy. Therefore, after S phase the cell remains 2n, but its DNA content is 4c. Your annotations specifically say to understand how chromosome composition changes throughout the cell cycle and division
A newborn's karyotype is:
47,XX,+18
Which diagnosis is most consistent with this result?
A. Male with Down syndrome
B. Female with Edwards syndrome
C. Female with Turner syndrome
D. Male with Patau syndrome
E. Female with Klinefelter syndrome
Answer: B. Female with Edwards syndrome
Explanation: 47 = one extra chromosome, XX = female, and +18 = trisomy 18, or Edwards syndrome. Your annotations say to be able to read a karyotype, identify numerical abnormalities, and know what the notation means
A patient undergoes a permanent-wave treatment in which a chemical reducing agent is applied to the hair. The treatment alters the structure of keratin by disrupting bonds between cysteine residues.
Which type of interaction is directly disrupted?
A. Hydrogen bond
B. Ionic interaction
C. Hydrophobic interaction
D. Van der Waals interaction
E. Covalent bond
Answer: E. Covalent bond
Explanation: Two cysteine residues can form a disulfide bond, and disulfide bonds are covalent bonds. Reducing agents break these disulfide bonds. Rodda actually used keratin/perms as the example in the review, and you annotated “disulfide bonds = covalent bonds.
A pathologist examines chromatin during interphase and identifies regions that are loosely packed and distributed throughout the nucleus.
These regions are most consistent with:
A. Heterochromatin
B. Euchromatin
C. Centromeric DNA
D. Telomeric DNA
E. Mitotic chromosomes
Answer: B. Euchromatin
Explanation: Euchromatin is relatively open chromatin and is generally associated with active gene expression, whereas heterochromatin is more condensed. Your annotations emphasize knowing the distinction between these forms of chromatin
Two restriction enzymes cut different DNA recognition sequences but leave single-stranded overhangs.
Which finding indicates that their products can be ligated together?
A. Both enzymes recognize six-base sequences
B. Both enzymes cut DNA at the same chromosome
C. Their resulting overhangs are complementary
D. Both enzymes generate blunt ends
E. Their recognition sequences have identical GC content
Answer: C. Their resulting overhangs are complementary
Explanation: Compatible sticky ends require complementary overhangs. Your annotation says to look at the position of the arrows and the internal four bases when determining compatibility
During gametogenesis, a cell undergoes its first meiotic division.
Which structures normally separate during this division?
A. Sister chromatids
B. Homologous chromosomes
C. Individual DNA strands
D. Centromeres only
E. Nucleosomes
Answer: B. Homologous chromosomes
Explanation: Meiosis I separates homologous chromosomes. Meiosis II separates sister chromatids. Your review emphasizes knowing the differences between these divisions
A healthy woman undergoes genetic testing after several miscarriages. She is found to carry a reciprocal translocation but has no loss or gain of genetic material.
Which term best describes her chromosomal rearrangement?
A. Unbalanced translocation
B. Balanced translocation
C. Aneuploidy
D. Polyploidy
E. Deletion
Answer: B. Balanced translocation
Explanation: A balanced rearrangement changes the organization of chromosome material without a net gain or loss. Therefore, the carrier can be phenotypically normal. Your annotations emphasize knowing the difference between balanced and unbalanced rearrangement
A researcher isolates a protein composed of several polypeptide subunits connected by disulfide bonds. The sample is boiled, treated with SDS and a reducing agent, and then subjected to SDS-PAGE.
Which property will primarily determine how far each resulting polypeptide migrates through the gel?
A. Native electrical charge
B. Number of disulfide bonds
C. Molecular size
D. Ligand-binding affinity
E. Hydrophobicity
Answer: C. Molecular size
Explanation: SDS unfolds the protein and coats it with negative charge, removing native charge as the major variable. The reducing agent breaks disulfide bonds, allowing disulfide-linked subunits to separate. Therefore, SDS-PAGE separates the resulting polypeptides primarily according to size. Your annotation is essentially the test-taking shortcut: “If run protein prep on SDS Page gel → get size only.”
A tumor suppressor gene develops extensive methylation of CpG islands within its regulatory region.
Which outcome is most likely?
A. Increased transcription
B. Formation of more euchromatin
C. Increased accessibility to transcription factors
D. Gene silencing with heterochromatin formation
E. Removal of histones from DNA
Answer: D. Gene silencing with heterochromatin formation
Explanation: DNA methylation is associated with heterochromatin formation and transcriptional repression. Your annotations specifically connect methylation with heterochromatin and emphasize understanding its mechanism
RNA is isolated from two tissues and analyzed for expression of a specific gene. Tissue A produces a much darker band than Tissue B.
Which technique was most likely used?
A. Southern blot
B. Northern blot
C. Western blot
D. Karyotyping
E. SDS-PAGE
Answer: B. Northern blot
Explanation: Northern blotting analyzes RNA, allowing comparison of gene expression between samples. Southern blotting analyzes DNA.
Two siblings inherit very different combinations of maternal and paternal alleles despite having the same parents.
Which meiotic mechanisms contribute most directly to this genetic diversity?
A. DNA replication and transcription
B. Crossing over and independent segregation of homologous chromosomes
C. Cytokinesis and DNA repair
D. Mitosis and translation
E. Chromosome condensation and decondensation
Answer: B. Crossing over and independent segregation of homologous chromosomes
Explanation: Meiosis generates genetic diversity through homologous recombination/crossing over and the independent assortment/segregation of homologous chromosomes.
A phenotypically normal man carries a balanced chromosomal translocation. His partner becomes pregnant with a fetus that inherited an unbalanced product of this rearrangement.
Which abnormality is the fetus most likely to have?
A. No change in gene dosage
B. Complete tetraploidy
C. Partial monosomy and/or partial trisomy
D. Only mitochondrial DNA abnormalities
E. Loss of all homologous chromosomes
Answer: C. Partial monosomy and/or partial trisomy
Explanation: A balanced parent may be clinically normal but can produce unbalanced gametes. The offspring can therefore inherit excess material from one chromosome and deficient material from another—producing partial trisomy and partial monosomy. You specifically wrote this implication in your review notes
A pharmacologist studies the binding of two experimental drugs to the same receptor. Drug X has a dissociation constant (Kd) of 2 nM, whereas Drug Y has a Kd of 200 nM.
Which statement best describes these drugs?
A. Drug Y has greater affinity because it has the larger Kd
B. Drug X has greater affinity because less ligand is required to achieve half-maximal binding
C. Drug X has lower affinity because it dissociates more readily from the receptor
D. Both drugs have equal affinity because Kd does not describe ligand binding
E. Their relative affinities cannot be determined without calculating receptor concentration
Answer: B. Drug X has greater affinity because less ligand is required to achieve half-maximal binding
Explanation: Kd describes protein–ligand affinity. A smaller Kd means tighter binding and therefore higher affinity. Drug X's Kd of 2 nM is much smaller than Drug Y's 200 nM, so Drug X has the higher affinity.
This matches your Rodda annotation almost word-for-word: “Kd → describe affinity between protein and ligand → small number tighter binding/higher affinity.” You also wrote “no calculations,” so the important skill is interpreting the relationship, not doing math.
Following DNA replication, the parental strand of a CpG-containing region remains methylated, while the newly synthesized complementary strand is initially unmethylated.
How is the methylation pattern maintained?
A. Histone acetyltransferase methylates the new strand
B. A maintenance methyltransferase recognizes the hemimethylated DNA and methylates the new strand
C. DNA polymerase transfers methyl groups during replication
D. The methylated parental strand is replaced
E. RNA polymerase methylates both strands during transcription
Answer: B. A maintenance methyltransferase recognizes the hemimethylated DNA and methylates the new strand
Explanation: The existing methylation pattern on the parental strand directs methylation of the complementary new strand, restoring the symmetrical pattern. This is exactly the mechanism your annotation tells you to understand
A mutation causing HbS eliminates a restriction enzyme recognition site present in the normal HbA allele. DNA from a patient is digested with this enzyme.
Compared with the HbA allele, which result is expected from the HbS allele?
A. More restriction fragments because additional sites are created
B. Smaller fragments because HbS is digested more frequently
C. A larger fragment because the DNA is cut fewer times
D. No DNA fragments because HbS cannot enter the gel
E. Identical fragments because a single-base mutation cannot affect restriction digestion
Answer: C. A larger fragment because the DNA is cut fewer times
Explanation: Loss of a restriction site means the enzyme makes one fewer cut, producing a larger fragment. Your annotations say to approach these by analyzing the map, comparing the two alleles, and predicting the resulting fragment sizes; sickle cell is the emphasized example
Genetic analysis of an abnormal ovum demonstrates that it contains two different homologous copies of maternal chromosome 18 rather than two copies derived from the same homolog.
At which stage did nondisjunction most likely occur?
A. Maternal meiosis I
B. Maternal meiosis II
C. Paternal meiosis I
D. Paternal meiosis II
E. Postzygotic mitosis
Answer: A. Maternal meiosis I
Explanation: In meiosis I nondisjunction, homologous chromosomes fail to separate, so the gamete can receive both different maternal homologs. In meiosis II nondisjunction, sister chromatids of one homolog fail to separate.
A phenotypically normal adult undergoes karyotyping and is found to have 45 chromosomes due to fusion of the long arms of two acrocentric chromosomes.
Which statement best explains this finding?
A. The individual must have monosomy and therefore cannot be phenotypically normal
B. The individual has a balanced Robertsonian translocation and can be phenotypically normal despite having 45 chromosomes
C. The individual has an unbalanced reciprocal translocation
D. The individual necessarily has Turner syndrome
E. The individual has triploidy
Answer: B. The individual has a balanced Robertsonian translocation and can be phenotypically normal despite having 45 chromosomes
Explanation: A balanced Robertsonian translocation carrier can have 45 chromosomes because two acrocentric chromosomes have effectively become one chromosome without clinically significant loss of essential long-arm genetic material.