all right good afternoon good evening good morning AP Bio this is Mr Monsour coming to you live from room 102 at tiffan beautiful at beautiful tiffan coloman high school in tiffan Ohio and this is AP Bio review number five so hopefully all of you are enjoying the review so far I'm enjoy making them of course so I always like to start off as an overview of the exam dates I know this is the fourth or fifth time but it's always good to remember that when you take the exam in May section one is 60 Questions you got an hour and a half and it's about half your score and then section two is six questions 1 hour 30 minutes and it's also 50% exam so again as we're working through this uh review I'm going to probably focus more on working on those F frqs and those task verbs and those science practi so let's warm up always good got to get that brain brain fired up there all right so in this review we're gonna be looking at the following topics topics again as I've said in my other videos I cannot review everything in unit 6 unit 6 is probably what I think is one of the more difficult units not because the content is necessarily harder but there's just so much of it this is this is all of your like molecular genetics so we'll be reviewing replication transcription and RNA processing translation regulation of gene expression mutations and also looking at the task verbs using the F frqs so this is one of my favorite things to teach but I also think sometimes students struggle because it's kind of hard sometimes to conceptualize what's occurring at this molecular level so let's let's review our task verbs again so this is the second set of task verbs I re reviewed the first eight in my previous two slides now we're going to look at these other ones so first one is identify identify is probably the simplest taser because you literally just have to say what something is right you identify or indicate or provide information about topic without any explanation I identify the variable I identify the control justify requires a bit more work you need to provide evidence and when it says provide evidence I would use data I would refer to a graph anything that you've had to construct but when you do a justify by provide evidence support qualify or defend a claim or provide reasoning to explain how evidence supports a claim maybe some of you in class have done CS claim evidence reasoning that's what I like to think of as a justification make a claim like I just said when you make a claim you're going to make an assertion based on evidence or knowledge again if it says if you're looking at evidence we had you create these beautiful graphs you put in error bars you had to put in a line of best fit make sure you refer to it you did all that work might as well use it right and then predict and make a prediction you know predict the cause or effect or change in disruption to or components of a pattern process or system so again these task verbs and F frqs they kind of give you the direction of the story to tell but we'll talk more about that as we get in these F frqs so I know you guys are all excited so we're going to move on to the next slide so let's review uh replication right DNA replication occurs during s-phase right during the cell cycle so the highlights DNA replication is semiconservative which we'll talk about you really need to know your enzymes and then one of the misconceptions that I find is the directionality of replication this 5335 Direction and so as always refer to AP daily 6.2 for more detailed information but again a little check mark you should know that DNA replication uh semiconservative and you really need to look over those enzymes but also looking at the directionality be sure you're doing those uh practice questions so let's let's look at an example here so topic 6.2 wres semiconservative replication results in a d molecule containing one original and one newly synthesized complement this was the meselson install experiment which you may have talked about but as you can see if you're going from across my screen you have the original strand it's unwinding and then you see this different color right and you see like the blue and I guess that'd be an orange and as you progress through the process as you can see at the end you have two new strands one with what would be considered the old or original and one with the new synthesize that's semiconservative right so you have part of the old and part of the new so that's important to understand what that means and that this kind of goes back into mutations as well which we'll talk about at the end of this video but any mutation that occurs in the DNA during replication will be passed on to further generations and the DNA replication enzymes enzymes enzymes I can't emphasize enough the enzymes I think are really what's important so if we talk about leading Strand and lagging strand again when DNA replicates and again I'd go back and watch the AP daily for this there's a leading Strand and a lagging strand right leading strand is continuous has to be built in a particular direction that 53 direction right the lagging strand is built but it's built in those little chunks those okazaki fragments and so but the important thing are these enzymes so helicase helicase you know I could see a question saying identify the enzyme responsible for breaking hydrogen bonds helicase would be the identification describe how DNA is uh the DNA molecule is separated well a good uh description was helicase goes through and break those bonds so seeing how they're going to apply those task verbs is important so we got helicase DNA polymerase and you got poly3 as I say in my class in poly one but poly 3 adding those nucleotides you're synthesizing that New Strand in a very specific Direction and then there's a poly3 and a poly one this one is only showing both of them and then you have your liase or ligase however your teacher wants you to say it this is the one that goes through and spots well those fragments together right and so there's some other um enzymes involved top moras and and some others gyas but again if you're doing replication you really need to kind of look at what those enzymes are what's their structure structure what's their function what happens if one of those enzymes you know has mutation no longer functions so kind of trying to tell the whole story there so let's practice practice makes perfect right so hopefully you know you're doing this practice question so 6.2 is a visual representation so in this case this skill they're going to give you a picture to look at and you're going to have to look at the picture process it in your mind and come up with some kind of story or a completion of the story so figure one illustrates a model of the molecules involved in DNA replication and their placement relative to each other which of the following statement best explains okay remember explanation like explains how right how explains how topa isomerase and helicases are functionally related during the process of DNA replication how are they working this isn't an identify it's not described it's explain how so there's your topa isase there's your helicase and again this is DNA replication so going back I know my enzymes I know what helicase does I should know what topa isase does there it is and there your two enzymes so and they want you to explain that so look at our options we got a helicase adds an RNA primer at the three end of the template allowing toe ISAT to synthesize New Strand B helicase add nucleotides to the new strand of as Tope ISAT unwinds the original strands at the replication fork C hel synthesizes the New Strand as top I unwinds the original or D helicase unwinds the original strand as topa isas relieves coiling tension ahead of the replication fork all right I'm G to pretend my students are my class ABC D write down an answer if you're watching this and first one RNA primer at the three end not GNA happen first of all helicase does not add that primer right that's your that's the primase let's see so a okay let's see what our next one is B uh helicase as in the ctid the New Strand if I remember my enzymes I don't think that's correct it's not the helicases and adding the Cleo tiddes the helicase is breaking up those hydrogen bonds is it C helicase synthesizes the New Strand again if you know your enzyme you should know that helicase is not synthesizing new strands so your best answer is D give yourself a round of applause yay raise the roof whatever helicase unwinds the original strand as Tope isas relieves coiling tension and hea replication fork that's what it does it breaks those hydrogen bonds unwinds that strand you got that Tope isomerase so again if you understood what the enzymes are or their function this should not been tooo difficult of a question you know if you struggled with it a little bit go back and review those enzymes so that's kind of the replication part let's move into transcription RNA processing which I think is some of the funnest stuff but so our highlights for transcription so if you remember central dogma it goes uh you got replication but then you've got transcription translation right DNA creates that transcript that mRNA which is this next which is this part transcription R processing and then you're going to take that processed uh Mr in UK carots and produce your protein well yeah the RNA processing which we'll talk about introns and exons in a little bit so highlights genetic information flows from DNA to RNA to protein there's your central dogma and there's a series of enzyme regulated modifications occur to the MRNA transcript and UK carotic cells again enzymes are important as always look at 6.3 watch the video do the progress check do those kind of things prepare yourself for this section little check mark Make sure you do it all right all right let's look at an example so here is your flow of genetic information from DNA to RNA to protein this is your central dogma DNA right we've got a red and blue there that could probably represent you know the two different strands transcription making that mRNA that messenger RNA translation ultimately making those proteins so DNA RNA protein and if you remember uh we could you know I could see a good explanation like explain how a mutation in DNA leads to a new phenotype or describe how transcription occurs or Justify the these kind of terms that you got to think about what they're going to be asking you when they create these questions so I always like to review the types of M the excuse me I always like to review the types of RNA because I think it's important that structure of that RNA molecule determines what its function is and so you've got message RNA which carries that genetic info from the DNA of the ribosome you have Transfer RNA that TRNA that carries that specific polypeptide sequence um which helps with the production of that protein you have the ribosomal those are those units you probably heard of a small subunit large subunit you know they fit together just right to help to assemble the protein and another one that I often like to point out are Mi rnas SI rnas a lot of times if you see like an I that might mean an interference these could be small interfering rnas that help regulate gene expression or cut it up maybe when you talked about eukaryotic um processing of RNA you talked about these things called snps as I tell my students uh but those are things that'll cut up the uh that premature mRNA so I always like to point out that you know your your RNA so there's your mRNA TRNA RNA again that direction that running that five Prime to three prime or five to three but again knowing what they do will help you answer the questions better so let's look at some processing so we could identify identify the intron put an x on it we could say describe why introns are cut out or explain the importance of Exxon staying in so applying these terms here so introns or sequence of mRNA do not code for an amino acid exons are sequen of mRNA that do code for amino acids so if I look at my picture we're running 53 I got Exxon intron Exxon intron Exxon intron sounds like a song right if you see here there's my uh cap and we got a tail if you talked about that in class you got a cap and a tail because remember you have to put a cap and a tail on that primary mRNA before it goes out in the cytoplasm to kind of protect it so what we're going to do we're going to knock out those introns right we don't need them they're not going to be used so you knock out your introns whoops went a little got a little ahead of myself there we knock out those introns and what we're left there is that mature mRNA transcript so one two and three so you got the cap you got the tail so I can see a question where they gave you a sequence of DNA and they said you know each you know you've got 15 uh kilobases right and you've got two entrons each two what H how much would you have left well if you've got two entrons two * four I start out with 15 right you'd have 11 kilobases for your xon or what's going to be expressed so those kind of things but knowing that when those introns are cut out they're not going to be used and again this occurs in UK carot it's this processing part so now we can move to the next slide I got a little too excited I'm just so amped up today all right so let's let's practice a little bit all right visual representation again again this skill you're looking at a picture right so I tell you look at the picture first so we've got Direction and syn is we've got two strands we've got 53 all right so this is transcription you know how I know it's transcription because I've got urasil there being produced on mRNA so transcription is a process involved in gene expression and includes the use of enzymes we've already talked about that figure one represents a transcription process which of the following statement best explains how how does this occur explains how RNA polymerase is involved in the transcription process now I did not mention RNA polymerase in the previous slide but hopefully that's one of those enzymes that you remember right RNA polymerase essentially does all of the jobs right so look at our picture okay I'm seeing I got 53 35 there's my direction of synthesis so here are my options a RNA pimas synthesize a new DNA strand in a 53 Direction B RNA pyase synthesizes mRNA by base pairing with a coding strand C RNA polymerase copies the gene by base pairing with The non-coding Strand or D RNA polymerase copies the gene by synthesizing TRNA from The antient Strand now they've used some terms like coding non-coding and anti-sense I don't have time for it in this video but I would encourage you to go back and look at how those terms are used so all right class let's pretend you're here if you're at home write down an answer so first thing I want to point out there's that 53 directionality which is important so how is RNA Pim Ras involved well we know it's not a let me I kind of let me go back look at a RNA p r syn new DNA strand RNA P Ras is not making DNA RNA polymerase is producing mRNA B is not correct RNA polymerase syn mRNA by Bas pairing with the coding strand it's not the coding strand D RNA Pates copies the gene by synthesizing TRNA from an antient strand again we're not making TRNA we're not making DNA we're making mRNA however the best answer here is that RNA polymerase copies the gene by Bas pairing with The non-coding Strand and so the key there is the non-coding Strand all right so the difference between B and C is difference between coding and non-coding and so so just so you know the non-coding Strand is used as a template for m RNA but again that's one of those things you can eliminate down to the two coding versus non-coding but again using that picture will help you all right here's another skill representation so oxin are plant hormones that coordinate the growth and development indol 33 atic acid IAA is an oxin that is usually synthesized from the amino acid tryptophan figure one gene trip- T and codes and enzyme that converts tryptophan to endol 33 pyruvic acid I3 paa which is then converted to IAA by an enzyme encoded by the gene YC or yuck first things first don't get all caught up in these IAS and the trips and the yucks and all those kind of things just read it first don't don't get caught up on like what is indol 33 atic acid just realize that it's an important part of the process we have figure one over here model of two-step enzymatic plant pathway for synthesis of IAA from tryptophan so if you look you've got Gene trip t gen yuck you've got mRNA then you have your enzymes all right okay I know what those are got I know what a gene is I know what mRNA is I know what an enzyme is okay so we must be doing central dogma mind blown right then you've got tryptophan with an arrow I3 paa with an arrow and I a so you know you're talking about a visual representation here so there's your figure one a model of Step enzymatic plant pathway for synthesis of IIA from trip toin which I already said so this is a circle and identify right so Circle one arrow that represents transcription on the template pathway so you're going to actually have to Circle something and identify the molecule that would be absent if enzyme yuck is nonfunctional so I'm sitting here thinking okay I got to Circle something well I know that transcription right I have to go from DNA to RNA right and so I know either one of those that's DNA RNA so I could Circle either one of those it says Circle one Arrow Arrow you wouldn't have to do both I'm just circling both of them for you so I Circle that one right there's my arrows and then identify the molecule that'd be absent if enzyme yuck is nonfunctional it's probably not yuck I probably shouldn't say it's probably y see but so if we look at that enzyme y see right there it is if I put an X over it all right so if I put an X over that pathway I'm probably not going to produce IAA anymore and so the molecule that would be absent if the enzyme is nonfunctional is the IAA so in this question you had to do two things you had to circle simply Circle and I had to identify and simply identifying is simply stating and so we knock out the enzyme knocked out the IAA and then there's our answer we got our Circle we got our Circle we identified we've got our two points there now this is again an F frq so that would been two points now we're going to B predict how the deletion of of one base pair in the fourth codeon of the coding region of Gene trpt would most likely affect the production of IAA justify your prediction now we've got two points here we have to make a prediction and you have to justify it with a j ification there's not necessar any numbers here for you to use but you should probably try to use something in that pathway so predict how a deletion I know if I do a deletion in a codon I'm probably gonna change a structure change of function I may not have a functional protein and so an acceptable prediction would be a reduction in IA production or no production of IA that's a prediction I know that if I remove a code on I may cause some kind of mutation whether it's a frame shift right and so I've now probably or more than likely I'm going to have a reduction uction in the production of IIA or I'm going to have no production at all my justification mutation will result in the translation of an inactive trpt enzyme so again applying what I know about DNA if I have a mutation in the DNA how it will affect the enzyme and so there's your prediction there's your justification I should have gave you time to write that down four points right so hopefully that helps but trying to like dissect the frq all right here's another visual representation so figure one represents part of a process that occurs in UK carotic cells there are untranslated regions utr in this sequence so across the top we got our five to three I got Exxon intron Exxon intron Exxon right say that 12 times now if you can see I Circle those intron if you remember from a previous slide a few slides back those introns are cut out they're not any good we got our cap we got a tail down there in the bottom and so figure one represents part of a process that occurs in UK carotic cells which of the following statements best explains again why this process occurs so why is this occurring why are we cutting out the introns why are we putting a cap and a tail so that's what they're asking you here for so I'm going to read through the uh options you feel free to you know kind of read ahead of me pause it and see what you do here a introns must be removed from the mature transcript because they do not code for amino acids and only sequence coding for amino acids are sent to ribosomes B introns and exons must be separated into different mature transcripts because ribosomes can only accept transcripts that are either introns or exons not both C introns are removed because the inclusion of introns makes the transcript structure too big and this will prevent the transcript from exiting the nucleus D introns are not retained in the mature transcript because including intron sequences will result in proteins with too many amino acids all right so pause the video talk to your friends I'm gonna pretend my students are here they're all looking at me like I'm crazy as usual okay so we know that we got to cut out those introns right so we know that they're processed out so we could probably get rid of B introns and exons must be separated to different mature transcripts they're not separated into different mature transcripts the introns are cut out the exons are left in C is not a good answer introns are removed because the inclusion of introns make the transcript structure too big has nothing to do with how big the transcript is it's not going to prevent it from exiting the nucleus and then D is not the best answer introns are not retain the mature transcript because including intron singles result in proteins with too many amino acids that's not it the best answer is a as I said previously slide introns are removed from the mature transcript they don't code for anything and only the sequences that code which are your exons will stay in so again looking at that picture having some of the previous knowledge will be very helpful all right let's go into translation I know I'm moving quick here pause the video take a break if you need to hopefully you got some snacks or something so translation highlights translation of mRNA generates polypeptides translation occurs in three main steps we got initiation elongation and termination and then near all organisms use the same genetic code and this is evidence of common ancestry among all living things I like to think of when we see everything used in same genetic code is a conserved process right this is a process that's conserved through multiple um groups of living things as always go to AP daily 6.4 for more detailed information and let's move on all right so let's look at an example here so translation of mRNA generates polypeptides right so translation occurs on the ribosome procaryotes only have those cytosolic ribosomes they're going to be present in outside or in the cytoplasm UK carots have both cytosolic ribosomes and ribosomes bound to the ruar so you heard about free and rough so again with protein synthesis DNA right you got your transcription and you got your translation so we're going to focus on translation now so translation here's your three steps and in procaryotes translation occurs while the RNA is being transcribed so they occur at the same time in UK cariot they're a separate process but we have initiation right initiation is the start so they could ID they could ask you identify the first step in Translation well that's initiation right and so your first uh code on Aug right hopefully you know how to use your code on chart right that's initiation right you got the the large subunit and the small subunit come together elongation is just the building of that polypeptide you're adding those amino acids right and so it's getting longer and longer you're creating those bonds and then termination right it's that process that where it stops and so it gets to a point in that sequence where okay we've built our polypeptide it's now time for it to be to stop and for it to go and go off to where it needs to go so that's initiation elongation termination so hopefully the chart you you'll have access to an amino acid sequence chart on the test I know some teachers use a wheel and a chart I would encourage you to get some practice with this chart and so I've got a series of amino acids here right there's your SE or there's your chart and then I have some codons aug CCU GGG GCA Gua cgu UA right alphabet soup but really that does it will tell us a message so if you want to pause the video here and just as a quick review go ahead and convert those codons into their amino acids I'm gonna kind of give you the answer here in a second but couple things I want to point out a lot of the or several of these amino acids will have multiple codons so if I look at Leu it has uua uu CU and so on so forth so there's multiple um codon sequences for specific amino acids I always like to point out Aug which is also start and I like to point out your stop codons because those those are those are important when you're going through and you're building that poly peptide so hopefully by now you've unpaused me and you're ready to see what your sequence is so hopefully you were able to come up with me P gly ala V A RG and stop and so that's just good practice um going through and using that that chart so again I would encourage you use this chart there is a wheel the wheel is great but you will be you're not going to have to memorize this chart for the test but I think it's good that you get practice using this chart has first SEC first letter second letter third letter anyway so that there there's that idea of taking that mRNA convert it into your sequence there so let's practice some so argumentation this is requires a little bit of more thought more processing taking ideas from multiple um multiple units or multiple ideas so figure one representing the process of translation so we've got mRNA we've got TRNA going to the ribosome and then we have a polypeptide right messenger RNA I know that's what that is TRNA I know what that is it's carry amino acids they're going to meet at the ribosome ribosome is going to do its you know initiation elongation termination and then we have a polypeptide so the process of translation involves RNA molecules figure one shows a basic model of the translation process which of the following statements best represents the relationship between the RNA molecule evolved in Translation so represents okay got to I'm argumenting I got to make a I got to make a reason why this image is the best one a TRNA reads mRNA and rrna adds Amo acids to the growing polypeptide B rrna reads mRNA and TRNA adds amino acids to the growing polypeptide C mRNA reads TRNA and RNA adds amino acids to Growing growing polypeptide or D TRNA reads rrna and mRNA adds amino acids to Growing polypeptide again you have to know what they do so go ahead pause the video like okay I think it's a b c d talk to your friends so if we look here though you're looking for the best representation of how these things work together so we know it's not b r RNA does not read the MRNA right and does and TRNA does not add the amino acids C is not correct Mr mRNA reads TRNA and RNA adds amino acids again knowing what these do so now we're down to a or D so a or D anybody half and half okay and I'm pretending my students are here so D is not the best answer either TRNA reads um RNA and mRNA adds amino acids that's not correct the correct answer is the TRNA reads mRNA you have this anti-codon codon right they come together and then the rrna adds amino acids to the growing polypeptide and so what's going to happen is you've got the messenger RNA you've got the transfer RNA the transar can amino acid and then the ribosomal RNA kind of creates or kind of builds that growing polypeptide so again if you knew you what your um types mRNA did you should be good all right we got another one argumentation an mRNA molecule was synthesized incorrectly and does not contain a stop codon sequence which the following best predicts again a prediction the effect on translation if the MRNA molecule is sent to the ribosome again we're looking at a prediction here so okay we don't have a stop cat on so what's going to happen so again incorrectly there's a mutation so we got a transfer RNA will bring the wrong amino acids and the polypeptide will have an incorrect amino acid sequence B the ribosomal subunits will not attach to the MRNA no protein will be synthesized C the start codon will also serve as the stop codon and protein production will continue as normal or D Transfer RNA will continue to bring amino acids and the polypeptide will be longer than normal um ABC or D pause the video jot down your answer talk to your friends we know it's not a transfer RNA will bring the wrong amino acid the polypeptide will have an incorrect amino acid sequence it's not like those transfer rnas can just swap out and just because there's there's no stop Cod on B is not correct the ribosomal Subs will not attach mRNA no protein will be synthesized you're still going to synthesize a protein it's just the fact that there's no stop codon next one the start codon will serve as a stop it doesn't work that way it just can't like go from the beginning to the end so our best answer is D Transfer RNA will continue to bring amino acids and the polypeptide will be longer than normal it's just going to keep reading that sequence it has to get to a stop codeon so again knowing like what a codeon is right uh understanding what the different mRNA and TRNA and RNA is do but also making a prediction I had to think about it for a second okay if I knock this out I add these in right hope you like my sound effects all right moving on another one I think this is our last one for 6.4 I promise bacteria were culture in ass system allowed for the continual addition of fresh nutrients and removal waste products a bacteria Fage which is a virus were added at the time shown the following population changes were observed all right describe how the Genome of a retrovirus like HIV uh human immuno deficiency virus becomes incorporated into the Genome of the host cell the giveaway here is retrovirus right and so when we look we talk about a retrovirus there's they have this special enzyme called reverse transcript days right and so reverse transcript a is an enzyme that copies the viral genome into the DNA essentially what it does is it causes the central dogma to kind of go in reverse it doesn't go DNA to RNA that reverse transcript transcript a what happens is it causes the causes viral RNA to be copied into DNA then that DNA getss integrated in hosome BEC transcribed and you start making new viruses so describe how the Genome of a retrovirus like HIV becomes Incorporated genome Host this enzyme reverse transcript as copies viral RNA into DNA and that would be your description so again if you look at this picture here there's your virus added right it's the virus is added you can see uh where the bacteria Decline and the viruses increase because the viruses have essentially taken over the cell Machinery so that's just another way of looking at how viruses can incorporate mutation or differences into cell all right let's move on to 6.5 regulation gene expression there's a lot of of reading on this one so highlights regulatory sequences are stretches of DNA that interact with proteins epigenetic changes can affect gene expression and in procaryotes groups of genes called operons are transcribed in a single mRNA and again I would encourage you to watch 6.5 Gene Express regulation gene expression AP daily this one is really important so again all of this is important but I think just even with work with my own students this is one that we kind of struggle with a little bit so let's look at an example so here is some regulatory sequences or stretches of DNA that could be used to promote or inhibit protein synthesis so we got sequence one you have that regulatory Gene producing that mRNA to that regulatory protein you got sequence two operator Gene a Gene B Gene C you have some structural genes there then you have mRNA and you have your protein so where that Gene a Gene B and Gene C is at your DNA transcription translation I bring it together so again there's your regulator Gene right which is going to produce that regulatory protein you have your operator which is like the switch right that's what's going to determine the operator is what will allow transcription to happen or not then your regulatory protein again is what regulates this process so let's look at some other examples here so what we also have to understand is that some the same transcription Factor May regulate the transcription of multiple genes and exhibit different regulatory effects on different genes so I've got Gene a transcribed into mRNA and then we have these transcription factors so on Gene X Gene X on what on that one there can produce protein T if it's on transcription Factor on Gene y may turn it off no protein made or transcription Factor on Gene z may turn it on and produce protein s so these different genes depending on the MRNA if they turn different genes on or off can lead to the production of different proteins and so if we see here here's the MRNA we've got different transcription factors and depending on what the transcription factor is Gene X may be on Gene z may be on Gene y may be off and depending if that Gene's off you're not going to produce anything but then you can get protein T and protein s so understanding that there's multiple ways that this gene expression can be regulated I would pause this picture or this slide and maybe just kind of try to process it here but just understanding that different transcription factors can regulate different pathways epigenetics all right or uh this is this is always a cool one epigenetic chains involve reversible modification of DNA or histone so I have a model here and so this is chromatin right and then this is eukariotic these white things here these are histones all right and so this is what the genetic material is going to wrap around I've got two genes here one in blue one in red and so we talk about acetylation and methylation so if this is all scrunched up I'm going to ask ask you a question you can jot down the answer so this is if this material is all scrunched up like this are red and blue available for um expression right can they be transcribed hopefully he said no if I stretch these apart though right hopefully you can see that my red and my blue if these get stretched apart you can probably see yeah you know th those genes are accessible for expression so we talk about epigenetic change these are changes that deal with the DNA and how in a mod ification so there's a term called methylation if I methylate this DNA I get these little methyl markers and they cause that DNA to scrunch up those genes are not expressable they're not not able to be expressed I don't know if expressable is a word Google it I'm not sure if there's another term called A cation if I acetate it I can separate those out and so those genes are available for expression so this is just showing you that there are multiple ways that eukaryotic genome can be regulated so I always like to show this again this is a model all models are wrong I don't have the right number of hones right but I just wanted to show you not accessible right it's all buried on the inside these ones are accessible so let's talk more about the operon though so this is another so now we're going to go to procaryotes my histones that was UK carots now we're going to go into procaryotes in procaryotes groups of genes are called operons they're transcribed in a single mRNA so you've got as you can see in the lack operons considered in inducible because it's usually turned off I'm not going to get into all of the biology about inducible but you've got regulatory sequence one regulatory Gene regory sequence 2 operator lack Z lack y lack a and so you've got your mRNA you've got your alil lactose which binds the repressor it's an active form of the regulatory protein what happens is that alil lactose will bind into that repressor right and it activates it so the AL Alo lactose acts as a repressor when it's not attached it can attach that operator and block RNA so here's my best operon I can make right so I've got my your regulatory sequence I've got operator and then I've got all my Lac my Lac genes and then this would be my RNA pimas so as my RNA plase goes down it'll attach here it can read this sequence of genes and produce the transcript this right here this binder clip I know really high tech models here this can act as a repressor and so what'll happen is or reg if my regory protein makes this if I attach it right here at my operator right what's going to happen is when my RNA polymerase goes to attach or try to you know read those genes it gets blocked so this is how op this is in procaryotes operon blocking the operator I know this pretty fancy I try so reatory Gene there's your RNA polymerase there's your operator right there's the operator and so that is kind of your switch there so if we block the operator we can't read the rest of the gene or produce that mRNA so let's do some practice here so we got a lot of reading here some more argumentation all right so we got figure one a marine stickle back figure two freshwater stickleback the three spine stickleback I'm not going to try to read that scientific name is a small fish found in both saltwater and freshwater environments saltwater stickleback populations consists mostly of individuals with pronounced pelvic spines as shown in figure one so as you see figure one pronounced spines individuals in freshwater stickleback populations on the other hand have reduced pelvic spines so you can see that those pelvic spines are small as shown in figure two which of the following claims here we go a claim best accounts for the difference in stickleback population phenotypes shown in figure one and figure to we're asking about a claim right so again this is going to require more information it's argumentation I would read the question again and I would look at the image okay so the pelvic spine I've got these sequences pelvic spine on the freshwat I got these sequences but there's an X over The hind limb one says it's disabled due to mutation I'm going to read through the um answers a b c and d and then we'll kind of go back to it so a they're asking you what claim best accounts for the difference a the pitx1 gene does not independently assort in the cells of saltwater sticklebacks but does independently assort in the cells of freshwater sticklebacks B the genetic code of the pitx1 gene is translated differently in males and females C expression of the pitx1 gene is affected at mutations at different location or different genetic locations Loi or D the pitx1 gene is transcribed in the cells of freshwater sticklebacks but not transcribed in the cells of saltwater sticklebacks H again great place to pause the video you have to take this in you have to look at the picture you have to read the questions so again we're looking for a claim so if I look over here there's my pelvic spine on on the Marine pelvic spine on the freshwater all right I got pituitary I got jaw I got hin limb I got no hind limb okay I got an X over it mutation it's not working I have pituitary and I have jaw all right so there was a mutation and so I have to think about there's something happened to the gene well we know it's not d the pitx1 Gen is transcribed in the cells of freshwater SLE backs but not transcribed in the cells of saltwater there's nothing in there that would tell you that the pitx gene is is present right B the genetic code of the pitx1 gene is translated different to males or females there's nothing in there that even gives you any idea that there's a difference between um male or female uh or a the pitex one gene does not independently assort in the cells of uh sticklebacks but doesn't sort the cells of freshwater CLE backs they're not referring to like independent assortment we're not looking at any kind of of of Al the best answer there is C expression of the pitx gene is affected by mutations at other gentic Loy it talks to you there's a mutation there so whatever happens that mutation it causes reduction in that pelvic sign I would encourage you look back to see what enhancers are um but in this case looking at the picture making great argument all right got another one here so we've got a table you've got an operon secr so the diagram above represents a segment of the eoli chromosome that contains the Lac I Gene and part of the Lac oper Accord accordin regulated set of genes that are required for the metabolism of lactose the presence of lactose which causes the repressor to be released from the operator results an increased transcription of lack opon again what students often need to do is make sure you go back up and read what all of the structures and functions are I'm not going to read those to you because I don't want this to turn into an hour and a half so which of the following is the most likely consequence of a mutation at the operator this is your operator right so this is where this will bind Locus that prevents binding of the repressor protein a expression of the structural genes will be repressed even in the presence of lactose RNA polymerase will attach at the PAC Locus but transcription will be locked beta galacto galactose I butchered that I'm sorry will be produced even in the absence of lactose the operator Locus will code for a different protein thereby prevent transcription of the structural gene a lot of information there again I would pause the video read through it process it because I'm just going to move on sorry so we've got some information there's your regulatory Gene remember the regulatory Gene are the things that produce these repressors you got your promoter you got your operator you got your genes I like to remember promoter operator and genes is POG promoter operator Gene P so again the presence of lactose will cause the repressor to be released from the operator results in increased transcription so you what happens is if you have lactose the repressor is going to be released however you got a consequence of a mutation where it does not bind so most likely consequence it's not going to be a expression of the structural genes will be repressed even in the presence of lactose it has nothing to do with what happens if there's a consequence of mutation RNA Pat will attach at the PAC Locus but transcription will be blocked if RNA polymerase can attach right at the at the location and it's transcription not going be blocked D the operator Locust will code for different protein and thereby prevent transcription the structural gene that oper operator isn't coding for the protein the operator is a structural part so your best answer is C of course the enzyme that I cannot pronounce correctly beta galacto oidas I know I butchered it I'll be get will be produced even in the absence of lactose so again looking at the image looking at the chart looking at the functions will help you answer that question and another one for 6.5 argumentation in animals the Hawks genes and code of family transcription factors are important for proper Vel embrionic segments and are widely conserving organisms the figure below shows the embryotic segments in which one such Gene Haw six is expressed in the Embry of a mouse a chick and a goose embryonic segments are counted from anterior end during formation of vertebrae the most anterior embryonic segment that expresses Hawk six marks the end of the cervic cervical vertebrae which is your neck and the beginning of the thoracic vertebrae all mammals have seven cervical vertebrae so again we're looking at the image we got Mouse chick Goose right no Hawks is expressed in white Hawks uh is expressed in the dark and then we've got our um labels there so which of the following statements is most likely true so first of all you got there's your anterior end which is important I drew a line there to show you where it's expressed where it's not expressed so which of the following stat is most likely true a the chick and the goose have the same number of thoracic vertebrae the Mo B the most anterior expression of Haw six is the eighth vertebrae in the mammal C hawk six is expressed in the same embryonic segments in birds and mammals or D Hawk sixs expressed in the same vertebrae at the anterior end of all anterior end of all bird embryos well looking at the picture again there's our question which one's true I drew my line and so again pause the video we already know it's not a the chick and the goose have the same number of thoracic vertebrae if you look at the chick and the goose you just got to count there they do not D Hawk express the same vertebrae at the anterior end of all bird embryos again it is not so you can see a difference between the white boxes and the gray boxes now we're down to B andc okay B andc anybody any body well it's not C hawk six is expressed in the same embrionic segments in birds and mammals that's not the case your best answer is B the most anterior expression of Hawk six is the eighth vertebrae in mammals I drew the line in there so you can see where it was all right let's talk about mutations 6.7 this will be our last topic so mutations uh changes in genotype can result in changes in phenotype alterations in a DNA sequence can lead to changes in the type or amount of protein produced and the consequence of phenotype and errors in mitosis or meosis can result in change in phenotype so this is essentially change in structure change in function and as always check out 6.7 for uh more information so let's do a little practice here lots of reading on these argumentations so just bear with me if you don't want to hear me pause the video a stero AAR plate uh one sorry that should be one is stre with a pure culture bacteria by means of aseptic technique paper disc treated with the antibiotics a and penicillin are placed at opposite sides of the plates as shown in the diagram the plate is examined after 20 24hour incubation period and a clear ring is discovered around disk a but not around disk P within the clear ring around dis a a single bacterial Colony with physical characteristics like those of the pure culture is observed a second sterile agar plate two is then stri with this single colony and also incubated with antibiotics which of the following would most likely be observed what can you see in plate two after 24 hours a a clear ring that a clear ring larger than that around dis a in plate one would appear around dis a only b a clear ring larger than that around dis a and plate one would appear around dis P only c a clear ring smaller than that around dis a and pl one would appear around dis P only or D there would be no there would not be a clear ring around either dis a or dis P lots of words again pause the video to trying to read those but what we've given you here is you've got the plates there's that little Colony there if you notice that colony is growing within that zone right so those antibiotics have probably diffused out there should be no growth in that zone we took that little dot we put another plate and now the plate is covered I drew that plate it's not part of the question so I drew that plate in for you to see so let's look what our options are it would not be c a clear ring smaller than that around dis a would appear around dis P there's going to be no clear ring so a clear ring larger than that around dis a and plate one would appear around dis a only again uh no ring is going to appear and then a clear ring larger than that around dis a and plate one would appear around disp again you're not going to have any of those clear Rings because a mutation has occurred that had that that small group of bacteria in that area they are resistant so they have they have a selective Advantage there there would not be a clear ring around either dis a or dis P that's why I put that there because those bacteria are resistant they have a mutation that confirmed an advantage and they're going to be able to grow on that plate so all right so hopefully that kind of made sense we've got the small little dot onto the plate it grows and so that's why D is the best answer okay another one here so gibber so another argumentation so gibberellin is the primary plant hormone that promotes stem elongation ga3 beta hydroxy g3h is an enzyme that catalyzes the reaction that converts a precursor of gibberellin to the active form of gibberellin a mutation the g3h gene results in a short plant phenotype when a pure breeding tall plant is crossed with a pure breeding short plant all Offspring and F1 are when the F1 plants are crossed with each other 75% of plants and the F2 are tall and 25% of the plants are short so question a says the wild type Alo encodes a ga3h enzyme with alanine ala a non-polar amino acid at position 229 the mutant Alo encodes a g3h enzyme with th or polar amino acid at position 229 here's your this is an F frq this require more work describe the effect of the mutation on the enzyme and provide reasoning to support how this mutation results in short plant phenotype type in homozygous recessive plants a lot of information there again pause the video read it work it out but again I'm GNA move ahead so with our description describe the effect of mutation on the enzyme well for one point a good description would be the amino acid substitution changes the shape structure function of protein that's a description mutations change the shape change of structure change of function now provide evidence provide reasoning excuse me to support how this mutation results in short PL well the mutation decreases or eliminates the production of the gibberellin right change in structure change in function I change the structure of that amino acid right substitution ultimately leads to a change in that gibberellin which would change its ability to do its job and then for Part B using the code on chart provided predict the change in the codeon sequence that resulted in the substitution of alanine for 39 amino acid position 229 so again if you look at your chart okay I've got ala I'm changing it trr so it's a prediction a a g got changed to an A in the first position of the code on that's just a prediction simply this got changed to this and you did that from your chart so again these longer questions probably good to pause and try to work them out all right so what should we take away there was a lot in this video I know if you've made it this whole way I'm very proud of you I'd give you gold stars and anything else so takeaways review the suggest AP daily videos that I said again I could not cover all of unit six go back and review take the progress checks practice F frqs focus on those task verbs don't cram have a plan and ATP answer the prompt but I always but one takeaway that I always like to do at the end of my videos is about the frq so we talked about F frq1 we're talking about F frq let's talk about F frq 3 and four today so fq3 it's a free response question it's a four-point question that prevents you with a lab investigation and you're going to be assessed in three areas part a describe a concept biological concept again describe Part B identify experimental procedures part C predict some results and Part D justify again telling a story so again pause these videos but they this fq3 is always is gonna be a four-pointer describe identify predict justify same thing with fq4 remember there's six of them so we're going to do two this section session and then two my last session conceptual analysis is a four-point question that prevents you with a scenario describing some kind of Phenomenon with A disruption right you know biology like life is full of disruptions right and so this is your ability to do the following things describe a concept explain the concept or process predict the cause or effect in that change and justify a prediction so again describe explain how or why predict what that cause did and justify your prediction all right so as always our next AP live will be natural selection part one with the always wonderful Margaret Evans she's awesome she's amazing she's probably one of my favorite people and again I just want to say thank you again this video probably ran a little bit longer than my first two lot of information again AP teachers thank you for working so hard AP students uh thank you for watching these I know I mispronounce things but yeah it is what it is so again thank you and again this is uh Mr Monsour signing off from beautiful room 102 on Tiff's beautiful Southside in northwest Ohio have a great day and keep it classy AP Bio for