2026 AP BIOLOGY FRQ: Released and Reviewed

Added: 2026-05-07

[00:00:00]Guys, it's here. Hot off the press.

[00:00:02]College Board just released the 2026 AP BioF FRQ early this year, and we've got to talk about it.

[00:00:15]Hey guys, this is Mikey with AO Prep Academy, and I bet you thought you'd never hear that again, but here we are.

[00:00:20]So, today we're going to go through these FRQ from questions 1 through six, but I'll be upfront with you guys. I'm not going to go all deep into the biological concepts because if you're here watching this, you just took the exam on Monday and the last thing you want is a whole biology lesson. So, I'm going to sprinkle in just enough biology to make sense of these problems. I promise you. So, let's start with number one. So, here we're looking at these ducleioide polyphosphates which act as signaling molecules in plants. And we begin with your fact check question.

[00:00:52]Describe the three structural components of a nucleotide. Now this is a straightforward question with just one answer a nitrogenous base a fivecarbon sugar and a phosphate group. You could have gone and added ribos or deoxyibbal sugar and name dropped some ACGT or U but you probably didn't need to. So that's one point. So the next part of the question describes how scientists are hypothesizing that these two different ducleioide polyphosphates APA and CPC bind to a receptor called Doran one to initiate a signaling cascade that closed the stamata. They tested it and found these results. You can see that a buffer meaning that there are no APA or CPC in here is used as a baseline for a relative size measurement of the stamata and APA and CPC and even something called ABA or something that we already know to cause the stamata to close being compared. B1 identify the dependent variable. This is always the thing that we measure. So it would be simply the relative size of stamata in width and length. So that's another point. B2 describe the relative size of the stamata in CPC compared to the buffer which is definitely smaller than the size of the stamata in the buffer and this is statistically significant because the error bars don't overlap and there's a point B3 justify the use of buffer alone as a control in the experiment here we have a very standard control question where we want to say that the buffer ensured that any changes to the size of the stamata was strictly caused by the added molecules like APA and CPC rather than some other variable during the experiment. Remember that we're always trying to isolate the effect of the independent variable to the thing that we're trying to change.

[00:02:24]So there's a point. Part C shows us that they do the same experiment except they mutated the DOR one receptor. So earlier they told us that they hypothesized that these APA and CPC were engaging with the DOR1 receptor. And here's how we can actually test that idea because check it out. You can see that with a dorm one receptor being mutated, APA doesn't actually reduce the stamata size anymore. This indicates that the mutation no longer allows APA to sufficiently trigger a response. But CPC, well, it still does the whole close the stamata thing, meaning that the mutation didn't actually affect CPC's relationship with the receptor. Or perhaps CPC didn't really operate through the door one pathway. We don't really know but it doesn't really matter because C1 says justify the use of ABA.

[00:03:11]So in this question ABA acts as a positive control that shows that the closure of the stamata can occur through anticipated mechanisms. For instance, if ABA with its known mechanism failed to close the stamata and we're not just looking at the impact of dor one mutation but some other pathway that may be interlin to that to create this effect. C2 describe the difference between the effects of APA and CPC.

[00:03:35]Well, APA does not appear to make the stamata any smaller compared to the buffer, while CPC still makes the stamata smaller in plants with that mutated dor one receptor. Done. So, let's put a point up. C3. Here, we're told that dor one being activated induces some molecules that eventually leads to the stomatal closure. For the cells treated with APA, we're being asked to predict the relative production of that molecule in Dor 1 mutated cells.

[00:03:59]So definitely there's going to be less of these molecules being produced because it's clear from figure 1 that with dorm one being normal molecules were produced and that made the stamata smaller. But here in figure two with that dorm one mutation stomata do not close normally indicating a lack of those molecules or some significant decline in their production. That's another point. Part D now says that dorm one produces a signal that increases the transcription of genes involved in plant defenses. So a bit of change in our story here, but they mutate that dorm one receptor such that the intracellular or the part that's inside of the cell is mutated. D1 asks predict the effects of APA treatment on transcription of genes involved in plant defenses relative to the effects of APA on non-mutated cells.

[00:04:40]And this is taking into account that APA does in fact bind to dorm 1 from that part C. So considering the APA is known to trigger Dor1 receptors normally, you would guess that plants homozygous for the mutant dor one receptors would not initiate transcription of genes for plant defenses as much as wild type plants as a response to that APA. So there's another point. Now the justification is that the mutant dorm one receptor would not be able to convey the binding of APA1 to the intracellular side with that mutation. It will no longer begin a signaling pathway that would result in gene expression. whereas the wild type would properly respond to APA through its normal intracellular domain. So that is going to be our ninth point. Now let's move on to number two.

[00:05:20]Okay, so I was a little bit surprised to see sirnas coming out again after making its appearance on the 2025 exam in questions one and two, but for those of you guys who studied that exam, this is great news cuz we already know that sirna can bind to mRNA and inhibit its translation. And in the context, they tell us that the binding of sirna to mRNA causes protein AO2 to cleave the mRNA or simply break it down. Part A, where are the ribosomes? Well, they're in the cytoplasm or on the rough ER.

[00:05:49]Now, some people online are asking if you can just say ER. And to be honest, I think you got to include the rough part because it's sort of what makes the rough ER look rough. So, there's one point. So, here's what they did in the experiment. They took two genes G&H and measured the amount of mRNA in these two genes in three different types of cells.

[00:06:07]One cell had normal AGO function and that's the homozygous wild type cell.

[00:06:11]One cell was hetererozygous, meaning that one of its chromosomes had a normal AGO gene while the other copy was a dud.

[00:06:17]And lastly, homozygous recessive cell with neither of the chromosomes producing a normal AGO2 gene product.

[00:06:23]And to be frank, the results are pretty expected. By the way, we're using relative averages again with the baseline being set for the homozygous wild type cell. So, if we begin with gene H, actually, we can see that the hetererozygous cell has more mRNA than you would expect because only one of the two Ag2 gene copies are working properly. So, you don't have as many of those AO2 products cutting up the mRNA that have the SNA attached to them. But in homozygous AGO minus, yeah, you don't have any AGOS's working and there are a ton of mRNA lying around. Now, now gene G is a little bit more surprising with those nonfunctional AGO proteins, whether it's hetererozygous or homozygous recessive, not having any impact on the degradation of gene G's mRNA. Now, B1 is graphing. This is definitely going to be a bar graph with multiple bars in each of the two categories, gan G and gene H. I'll just put my graph up here, and I think this should be about three points. B2. So for g all of the three types of cells homozygous ag plus hetererozygous and homozygous ag minus are all the same because their error bars would overlap.

[00:07:27]So that's an easy point. Part C1. What's the relationship between the number of wild type copies of Ag2 gene and the amount of gene HMRA in cells? So here we would say that the number of wild type copies of the AO2 gene relates to lower mRNA levels as having two wild type copies was lower than having one which was lower than having none. Great. So that's another point. Part C2 for G&H, calculate the percent by which the average amount of mRNA in Ag2 minus minus cells increases compared to Ag2 plus/minus. So here we're looking at a percent increase, right? So that means our final value should be three and our initial value at 2. So 3 - 2 / 2, our initial, gives us 50% increase, which makes sense because if you take something that goes from 2 to 3, then 50% of the two was added there to get us the three. So that's a 50% increase.

[00:08:18]Part D1 asks you to support the claim that SNA binding and Ag2 cleavage play a greater role in gene regulation than gene G expression. And this has been what we've been seeing in that table so far. In gene G, the genotypes of whether AO2 gene is operating normally or not has really no impact on the degradation of its mRNA as the data shows without those two functional AO2 copies of genes. So that's another point. Part D2 says that anaphase one of meiosis is blocked in AO minus minus cells. They think that this is because of a continued presence of spindle fiber stabilizing proteins. Okay, so this totally tracks. Remember that AO2 is supposed to cut mRNA and reduce the protein levels. But if it's mutated and cannot downregulate that protein expression, then too much of those spindle fiber stabilizing proteins will exist in the cell, thereby blocking proper breakdown of spindle fibers during that anaphase 1. And there is our ninth point from question number two.

[00:09:14]All right guys, so before we move on to the remainder, I just got to say thank you to all of you guys who watch my videos this year. I never thought I'd actually get mixed into Tik Tok videos with the other greats like Gabe Poser, Mr. W, and of course the queen AP Bio Penguins. So getting cut into videos within them in it felt like getting my face on Mount Rushmore of AP Bio. You know what I mean? So yeah, I love you guys. So number three is about cyanide.

[00:09:35]Basically the text tells us that low concentration of cyanide slightly increases the cytochrome C oxidase activity. That should correlate with slightly greater ATP production while high cyanide greatly decreases its activity resulting in very little to no ATP being produced. So here they put mamalian cells into medium containing cyanide of varying concentrations or no cyanide and varying concentrations of medium. The data table is a little bit convoluted but it's still asking about cyanide concentrations with the medium concentrations. A ask describe the advantage of aerobic respiration versus glycolysis. The answer is that aerobic respiration produces 30 to 32 ATPs while glycolysis alone produces two. B asked identify two treatment groups that served as controls and that would be our groups three and four that did not include cyanide to show that changes in ATP production would be linked to cyanide and nothing else. Now part C says predict the group that would have the highest ATP production and that would be group two with a little bit of cyanide because earlier they told us that a little bit of cyanide slightly increase the activity of the CCO which should yield more ATP. Now part D we need to support that when CCO is inhibited we should see lactic acid production. Now this one might be tricky but when CCO is inhibited the ETC doesn't work properly and the electrons do not flow through normally. This would not allow NADH or FADH to be regenerated into NAD+ or FAD+ thereby stopping the KB cycle and eventually pyrovate oxidation and as such only glycolysis can occur as part of that lactic acid fermentation process. Moving on to number four. Now, this was a little bit unexpected. They haven't done a cell division or a reduction problem in a while, but here we are. So, a describe one characteristic of chromosome movement during meiosis one. So, I'm guessing here that they're talking about how homologous chromosomes separate during anaphase 1 and move toward the motic poles because we're specifically talking about meiosis 1. B. Explain why chromosomes are more visible during mitosis or meiosis rather than at any other times. And this is because during prophase or prophase 1, we see the condensation of that chromatin into chromosomes so that they can facilitate the proper segregation of DNA. Done.

[00:11:43]Part C. Here we see a diagram of non-isjunction where inevitably two of the gameamtes would have what we call M plus1 number of chromosomes. Now ultimately if this game were to form a zygote, it would have a tricomeme which would overproduce that mRNA because it has three copies of genes rather than the normal two. Now on the other hand the N minus one gametes will result in a monostomy and in those gameamtes they would produce only half the amount of gene products from the chromosome that they would have gotten from just the other parent.

[00:12:21]All right. So question five. I feel like this one should be okay for most students given that you're just working with maps and symbols. A asks describe the role that changes in abiotic factors have in natural selection. So yeah, selection could come from other living things like predators, but abiotic features like temperature and precipitation can certainly act as a force of selection. Changes in abiotic factors would positively select members of the population with traits that are better suited to those changes and those traits would accumulate in the population over time through selection.

[00:12:50]Done. B. So storm intensity is shown as these dark shapes. So, it appears that the greater storm intensity is correlated with larger and larger toe pads, and that aligns with what they tell us that these animals do, which is just like clinging onto trees and stuff.

[00:13:04]C. Number four, experienced the least storm intensity in 2019 because it's these colors right there. And lastly, D.

[00:13:12]So, speciation question, it can definitely result if these different populations under different selectional forces do not interbreed. And these selectional changes would accumulate as differential genetic changes in different populations would accumulate until after a long period of time they wouldn't be able to reproduce and produce viable fertile baby anolles anymore. All right. So what's that like 30 points so far, right? So four more to go. And number six has that nasty box and whisker plot which has come out before, but it's not so bad. Just know that the whiskers refer to the range and the boxes as quartortile and the most important part that middle line as the median or the middle number in the data set. So they're looking at these three regions where you have protected areas with no farming activity and unprotected areas with farming activity. And we're interested in changes to raptor populations, which by the way are not these raptors, but these raptors, just birds of prey. Keep in mind that the graph shows annual percent changes in population size in both directions, positive and negative, because that's important. A says, identify the median annual percent change in populations in the unprotected areas of region 2. And that is just around -3%, I want to say.

[00:14:20]Now B says, identify the region and the area that experience the greatest median annual percent change in population size. So here is why these negative numbers are important because greatest percent change can go in either directions. Now, of course, in this case, the greatest percent change was into the negatives in region 1, where we see like -4.2ish change in population size in that unprotected area. Now, some people thought that the question was asking about the differences between the protected and the unprotected area within each region. But seeing the y-axis label and knowing that these two areas are actually different areas, not like it was protected and is now unprotected or anything like that. I think the biggest change is simply just the right answer here with whatever that is the biggest change on the y- axis. C says that in region 2, if we eliminated farming, there wouldn't be a decline in population size, and we're told to evaluate that claim. So, a lot of people online are comparing protected versus unprotected, arguing whether a box and whisker plots can tell you about whether a significant difference exists. But if you pay careful attention to the wording, it asks whether this specific protected area would experience a decline in the population size. And if you look at the median, it's set at zero, meaning that more than half of the protected sites either had no change or an increase in the population. So to me, that sounds like it's a pretty good argument for the idea that eliminating farming would in fact increase the population size. And I think I'm on the right track with this one, guys. And lastly, we're asked to explain why ecosystem resilience in unprotected areas may decline over time. And this is a call back to the fact that these raptors, no, not those, these raptors are a keystone species that stabilize the ecosystem in its community by playing a large role in population control. So, with the decline in raptor populations, instability may occur. So, final thoughts. I thought it was okay.

[00:16:06]This year leaned a little bit more heavily on gene expression and less on cell signal transduction, which was a bit of a surprise. But there it is. So, thanks for watching and we'll see you guys around.