Molecular Genetics — Flashcards

Flashcard 381: 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 382: 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 383: A genetic researcher is comparing the DNA replication process of prokaryotic and eukaryotic cells. - E. coli and human cells are cultured in separate media containing tagged nucleotides and their rates of DNA replication are determined. - Eukaryotic DNA is much larger and more complex, but replication still occurs in a timely manner. - Which feature of eukaryotic replication best explains this observation?

Answer: Multiple origins of replication

Extra: Eukaryotic chromosomes are large and linear with multiple origins of replication, allowing simultaneous replication at many points. In contrast, prokaryotes (like E. coli) typically have a single circular chromosome with a single origin of replication (oriC). Despite prokaryotic DNA polymerases being faster, the multiple origins in eukaryotes allow for faster overall genome replication.

Flashcard 384: 24 y/o woman is diagnosed with gestational diabetes mellitus during her 1st pregnancy. -Mother & younger sister had high blood surgars during pregnancy. -If this patient's gestational hyperglycemia is genetically predisposed, she is most likely to have decreased activity in what enzyme?

Answer: Glucokinase Glucokinase need to phosphorylate glucose to breakdown in glycolysis in order to release ATP to close down the K channel, which lead to influx of Ca that causes release of insulin. due to Maturity Onset Diabetes of the Young (MODY) (hereditary mutation of glucokinase gene) Key: Her mother & her younger sister had diabetes too during their pregnancy, very young.

Flashcard 385: 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 386: HbC disease is caused by a single amino acid substitution (Glutamate → Lysine) at position 6 of the β-globin chain. Patients homozygous for HbC (HbCC) have mild chronic hemolytic anemia. Patients with HbS (Sickle Cell Disease) generally have a more severe condition. Which property of HbS best explains why it is associated with more pronounced clinical manifestations than HbC?

Answer: HbS involves a substitution of Glutamate (polar) with **Valine (nonpolar)**, which allows for **hydrophobic interactions** and polymerization.

Extra: In HbS, the nonpolar valine residue on the surface of the β-globin chain creates a 'sticky' hydrophobic patch. In the deoxygenated state, these patches interact, causing HbS molecules to polymerize into long fibers that distort the RBC (sickling). In contrast, HbC involves a substitution with **Lysine**, which is polar/positively charged and does not promote the same hydrophobic polymerization seen in HbS.

Flashcard 387: Identify the location or function of the following molecular markers: - CCA sequence - Poly-A tail - 7-methylguanosine cap - TATA box

Answer: - **CCA sequence**: Located at the 3' end of **tRNA**; the site where amino acids bind ("**C**an **C**arry **A**mino acids"). - **Poly-A tail**: Added to the 3' end of **mRNA** after transcription (post-transcriptional modification). - **7-methylguanosine cap**: Located at the 5' end of **mRNA** (post-transcriptional modification). - **TATA box**: An upstream eukaryotic **DNA promoter** sequence that binds TATA-binding protein (TBP) to initiate transcription.

Flashcard 388: Describe the characteristics or mutations associated with the following Hemoglobins: - Hemoglobin C - Hemoglobin F - Hemoglobin H - Hemoglobin S

Answer: - **HbC**: Glutamate → **Lysine** mutation at position 6 of the β-globin chain. - **HbF**: Fetal hemoglobin (α2γ2); has higher O2 affinity than HbA. - **HbH**: β4 tetramers; occurs in **3-gene deletion** α-thalassemia. - **HbS**: Glutamate → **Valine** mutation at position 6 of the β-globin chain.

Extra: Note: **Hb Barts** (γ4) occurs in **4-gene deletion** α-thalassemia (Hydrops fetalis). Both HbH and Hb Barts have very high oxygen affinity and are poor at oxygen delivery. Mnemonic for β6 mutations: - Hb**S**: **V**aline (**S**ickle) - Hb**C**: **L**ysine (Ly-**C**-ine)

Flashcard 389: 65 y/o woman with chronic obstructive pulmonary disease & type 2 DM. -She's hospitalized with septicemia. -Cultures shows rapidly dividing gram-negative bacteria. -Replication of theses microbial cells requires synthesis of two daughter strands of DNA using the parents strands as templates. -Which processes will differ the most between the 2 dauther strands formed at each replication fork?

Answer: Joining of DNA fragments by ligase which is specific for lagging strands.

Flashcard 390: In which strands will the followings occurs? (leading or lagging strands) -Enzymatic function of DNA helicase -Interaction with single stranded DNA binding proteins -Joining of DNA fragments by ligase -Proofreading of the newly synthesized DNA -Relief of supercoils by topoisomerase

Answer: They all will occur in both leading and lagging strands except for Joining of DNA fragments by ligase which only occurs in the lagging strands.