Molecular Genetics — Flashcards

Flashcard 391: A 2 y/o boy comes with easy bruising. -Parent stated he develops marked bruising & open wounds following minor trauma. -Skin is difficult to suture due to its extreme fragility. -Physical exam: Hyperextensible skin, multiple ecchymoses over the forearms. - Skin biopsy: defect in extracellular processing of collagen. -Which steps of collagen synthesis is most likely impaired in this patient?

Answer: N-terminal propeptide removal or N & C terminal propeptide removal. due to defect in procollagen peptidases (defect in collagen 3 & 5) that lead to Ehlers-Danlos syndrome. (hyperextensible skin, joint laxity, tissue fragility)

Flashcard 392: How do the following mutations affect protein speed in gel electrophoresis compared to normal Hemoglobin (HbA)? - Frameshift mutation - Missense mutation - Nonsense mutation - Silent mutation - Trinucleotide expansion

Answer: - **Frameshift mutation:** Can result in shorter or longer proteins. In α-thalassemia, loss of α-chains leads to formation of **HbH (β4)**, which moves **faster** than HbA toward the anode. - **Missense mutation:** Single amino acid substitution (e.g., HbS, HbC). These lose negative charge compared to HbA and thus move **slower** toward the anode (HbA > HbS > HbC). - **Nonsense mutation:** Results in a truncated (shorter) protein, which moves **faster** due to reduced size/mass. - **Silent mutation:** No change in amino acid sequence; mobility remains the **same** as HbA. - **Trinucleotide expansion:** Results in a larger protein (e.g., Huntington disease), which moves **slower** due to increased size.

Flashcard 393: What exonuclease activities are present in DNA polymerase I and DNA polymerase III?

Answer: DNA Polymerase I: Both 5'→3' and 3'→5' exonuclease activities DNA Polymerase III: 3'→5' exonuclease activity only

Flashcard 394: An infant appears healthy at birth but develops transfusion-dependent hemolytic anemia by age 6 months. - His erythrocytes contain insoluble aggregates of hemoglobin subunits. - The child developed normally in utero because at that time he produced high quantities of:

Answer: Gamma-globin (γ-globin)

Flashcard 395: 34 y/o woman comes for breast examination. -Screening test was performed using Polymerase chain reaction (PCR) to amplify the target of exon. -Gel electrophoresis to assess the size of the exon compared to the wild type allele.What features must be known in order to perform the amplification part of this analysis?

Answer: The nucleotide sequence of the DNA regions flanking the target exon (required for primer synthesis).

Flashcard 396: In an experiment, a protein mixture obtained from a cell culture is separated by gel electrophoresis and subsequently transferred to a filter membrane. -Labeled: Double-stranded DNA probes are used to detect a specific protein of interest in the sample. -What protein is most likely to be detected by this method? a.) Ras b.) C-Jun c.) B1- adrenoreceptor d.) S-100 e.) Adenylate cyclase

Answer: C-jun Other possible answers: c-Fos, or c-Jun & c-Fos because Double stranded DNA probes were use, which means they are using Southwestern blotting technique, (DNA - binding protein).

Flashcard 397: What kind of mutation exists in the following base changes? - CUU (Leucine) → AUU (Isoleucine) - UUU (Phenylalanine) → UUC (Phenylalanine) - CAG CCC ACT → CAG TCC CAC T - CAG CAG UAA → CAG CAG CAA - UCA → UGA (Silent, Missense, Nonsense, Nonstop, Frameshift)

Answer: - **CUU (Leucine) → AUU (Isoleucine)**: Missense mutation - **UUU (Phenylalanine) → UUC (Phenylalanine)**: Silent mutation - **CAG CCC ACT → CAG TCC CAC T**: Frameshift mutation - **CAG CAG UAA → CAG CAG CAA**: Nonstop mutation (Stop codon → Amino acid) - **UCA → UGA**: Nonsense mutation (Amino acid → Stop codon)

Extra: - **Silent**: Same amino acid (often 3rd position/wobble). - **Missense**: Different amino acid. - **Nonsense**: Changes to a Stop codon (UAA, UAG, UGA). - **Nonstop**: Stop codon changes to an amino acid. - **Frameshift**: Insertion or deletion not in a multiple of 3, shifting the reading frame.

Flashcard 398: Describe these 2 genetics term? -Transition -Transversion

Answer: Transition: you replaced a purine by a purine, a pyrimidine by a pyrimidine. -Transversion: you replaced a purine by a pyrimidine. A pyrimidine by a purine.

Flashcard 399: Evaluate the following statements regarding **both** eukaryotic and prokaryotic DNA replication: - Continuous synthesis of the lagging strand - Energy-independent DNA unwinding - Multiple origins of replication - No proofreading of daughter strands - No requirement for RNA primers

Answer: All statements are **FALSE**. - **Lagging strand**: Discontinuous in both (Okazaki fragments). - **Unwinding**: Both require ATP-dependent helicase. - **Origins**: Prokaryotes typically have one; eukaryotes have multiple. - **Proofreading**: Both possess 3'→5' exonuclease activity. - **Primers**: Both require RNA primers synthesized by primase.

Flashcard 400: 15 y/o boy comes with unexplained erythrocytosis on lab analysis. -Father & sister also have high red cell levels. -Further analysis shows that this mutation impairs the ionic interaction between B-subunit & 2,3-bisphosphoglycerate. -As a result of this mutation, the patient's hemoglobin will be most similar to what hemoglobin types?

Answer: Hemoglobin F because the mutation causes low affinity of HbA to 2,3-BPG, which inhibit 2,3-BPG to bind to HbA, but will increase HbA affinity to oxygen. Thus will resemble HbF due to high affinity to oxygen. Normally, HbA has high affinity to 2,3 BPG. HbF has high affinity to oxygen.