Passage 8 is all about genetics, inheritance patterns, chromosomes, and much more. Follow along with the blog post for this episode.
[01:30] Passage 8
Cells of Drosophila melanogaster individuals with Fragile X syndrome were microscopically absorbed in Petri dishes. Petri dish A was stained by the lab researchers prior to microscopic observation. Cells in the interphase were shown to have heavy staining around the periphery of their nuclei. Additionally, cells in prophase, metaphase, anaphase, and telophase, were shown to have chromosomes which were heavily stained on the end of each chromosome. The arms of each chromosome were more likely stained. Petri dish B was combed to remove all cells except the ones in the mitotic G1 phase. The dish was then seeded with fluorescent oligonucleotide probes, which are specific to the lack the laci gene. In situ hybridization occurred of this gene and the oligonucleotide. After DNA supercooling occurred, fluorescence was observed on not one, but two arms of an intracellular linked chromosome pair. Petri dish C contained a cell cultured grown Drosophila cells which were genetically engineered to be incapable of producing histones or nucleosomes. It was observed that the nuclei of these cells never developed visible chromosomes, despite the fact that in theory, there should have been no impact on the structural integrity of the nuclear DNA.
[03:28] Question 40
What role do histones and nucleosomes possess relating to nuclear DNA and mitosis?
- (A) They allow visible chromosomes to form prior to mitosis by linking pairs of sister chromatids.
- (B) They cause the genetic variability that chromosomes require for mitotic splitting.
- (C) They allow small coils of euchromatin to supercoil into large coils of heterochromatin.
- (D) They facilitate mitosis by helping to supercoil nuclear DNA into chromosomes.
The correct answer here is D. Histones and nucleosomes help to pack DNA individually tight chromatin structures. There are some here that stand out as pretty wrong like B since there is variability in the DNA, not in the histones. Then for C, that’s not how it works. Since euchromatin and heterochromatin are two different structures. Euchromatin is more loosely packed chromatin and heterochromatin is more dense. Choice A talks about linking a pair of sister chromatids. It’s actually confusing that with the centromere which does that, but the histones don’t.
For you to remember this, if you can come up that works for you, then great. If not, then it could just be something that you need to remember. If you can somehow associate “Eu” with loose or like, and “hetero” as H and heavy like dense. Heterochromatin is dense and darker colored under a microscope so you can picture that.
[07:30] Question 41
Why was staining in Petri dish A only visible during mitotic stages other than interphase?
- Chromatin’s association with genetic material means that the DNA double helix must coalesce with sufficiently high density before chromatin can be microscopically visible.
- Interphase is a stage in which the proteins around which DNA is wrapped are two loosely packed to be seen with optical microscopes.
III. Chemical reactions that occur during interphase cause translucency of the chromatin structure.
- I only
- I and II only
- II and III only
- I, II, and III
The correct answer here is B. And I is one of the correct answers. Sometimes, students can be tempted to try to do everything at once. But there are negative statements so it can really get confusing. But what you can do towards the end is to ask yourself if it’s true.
III looks good as if it addresses the question but translucency of the chromatin structure is not something that happens. So III is out. And there is no II only. So it has to be I and II. II is just like the definition of interphase and it’s totally true. Interphase is a stage where DNAs wrap around proteins too loosely to be seen with microscopes and that explains why the staining was not visible during interphase.
I just a long fancy way of saying what interphase is and what chromatin is in terms of having to be dense before it’s visible. But we already know it has to be in there.
[12:52] Question 42
What component can best be expected to experience the same type of heavy staining as occurs in Petri dish A to the nuclei peripheries and chromosome ends?
- (A) The center of each chromosome pair, since most histones are located there
- (B) The arms of each chromosome pair since most histones are located there
- (C) The center of each chromosome pair since chromosomal centromeres are made of heterochromatin
- (D) The center of its chromosome pair since chromosomal centromeres are made of euchromatin
The correct answer here is C. They’re asking about staining and denser components of the chromosome are going to take up the stain and so they seem darker. The arms allude to the fact that the arms weren’t very dense already in the passage so there’s no reason for us to think they are. So ACD talk about the center. What’s key here is in the passage, they talked about the ends of a chromosome being stained, that is sort of an indirect, not explicit reference to telomeres. And telomeres are found at the ends of the chromosomes and they’re made of heterochromatin. And the other thing that’s made of heterochromatin is centromeres. So C matches that perfectly and that explains why there’s that staining in the middle.
[17:00] Question 43
Suppose the cells are either the A culture or the B culture were taken from the Drosophila individual who possessed a gene associated with hemophilia, a sex link recessive blood disorder on one and only one X chromosome. If this individual were to reproduce, what patterns of hemophilia inheritance would be manifested in offspring produced with a non-hemophiliac mate?
- (A) If the individual were female, male offspring would have a 50% chance of manifesting hemophilia and female offspring would never manifest hemophilia
- (B) If the individual were female, male offspring would never manifest hemophilia and female offspring would have a 50% chance in manifesting hemophilia
- (C) If the individual were male, the sex-linked nature of the gene would cause every male offspring to manifest hemophilia
- (D) If the individual were male, the sex-linked nature of the gene would cause no male offspring to manifest hemophilia
This is a really dense question so you may as well skip it and move on. But digesting this question, what you can do here is start at the bottom with D since male individuals are just a lot easier to deal with. But this answer is too extreme. Male offsprings always inherit X from their mothers and since the mother could be a carrier, D is wrong. And C is wrong for the same reason. A good trick with X-linked conditions is that male individuals have an X and Y. That means they must have gotten Y from their father since only other males have a Y. So any male offspring you ever see in an MCAT question must have inherited their X from their mother. So you need a lot of practice with this since it can get really confusing. So D and C are out since there’s one male with hemophiliac X but we’re talking about male offspring and they inherit it from the mother and the mother had at least one good X.
The fact that they’re non-hemophiliac mate could be a carrier is definitely a twist in this question. No, we have A and B left. Looking at B, we could get rid of this one right away because male offspring will ever inherit X-linked conditions from their mothers. And this individual who’s female, we already know this individual has hemophilia on one X chromosome so those male offspring could totally inherit that X chromosome from their mother. And finally, A is right. But it’s so easy to mess it up. If the individual were female, we have a female individual with a hemophiliac chromosome and one X chromosome, so we have a carrier. As male offspring will have 50% chance of manifesting it, you knew that was true. The hard part is the second part and female offspring would never manifest hemophilia.
We are so used to think that this non-hemophiliac mate could be a carrier because it could just have two good X chromosomes or they could be a carrier. But the non-hemphiliac mate in the case of A is a male. So it has to have only one good X chromosome. So those female offspring are always going to have at least one good X-chromosome and never going to have hemophilia manifested.
[30:50] Next Step Test Prep
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