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решаем

задачи

биосинтез

белка

56789

10

типы

задач

вторая

часть

00:00:05

So guys Hello everyone,

00:00:09

such a thing happened. In short, I

00:00:12

come in before two minutes before class And

00:00:15

that the microphone doesn’t work for me, it turns out that the

00:00:17

settings are just wrong. I already thought that

00:00:19

my microphone was broken and

00:00:22

it’s good that it’s not so simple that the

00:00:25

settings are wrong, so Hello everyone.

00:00:28

Today we We will continue to solve

00:00:30

problems on protein biosynthesis. I can be clearly

00:00:32

seen. I can be heard. I am not interrupted.

00:00:34

Let’s check the connection if everyone is an

00:00:36

excellent student. Write a plus sign,

00:00:42

how do you like your homework? By the way, how did you do the first four

00:00:45

types of problems?

00:00:47

Your homework was purely based on these four

00:00:49

types. That is, what you and I solved was

00:00:56

me I haven’t done it yet Well, look Vika The main thing is

00:00:59

that you watch the last lesson

00:01:01

because it’s important now we wo

00:01:05

n’t go into detail about what points

00:01:08

we will already be at a high level,

00:01:11

that is, we won’t stop

00:01:12

there anymore why are you why and so on

00:01:15

she decides long Well, it’s clear that it’s long

00:01:17

because you’ll find it, too, you’ll have to

00:01:19

solve the problem for about 10-15 minutes in order to

00:01:22

solve it correctly,

00:01:23

and the last type seems to me the most

00:01:25

difficult Well, yes, because there you need to

00:01:28

write all kinds of what kind of replacements

00:01:31

have occurred, it’s not that it’s difficult there, it takes

00:01:34

longer this is how to solve it

00:01:38

Okay, I’m very glad that you

00:01:42

are getting it,

00:01:45

these problems seem easy to you, for

00:01:46

most people

00:01:48

today we can already solve the other types of

00:01:51

problems, look today we are solving the fifth

00:01:52

type of problems, the sixth type of problems 7 8 9 and

00:01:57

look at the 10th type of problems, unfortunately you have something in the

00:02:00

script There’s nothing wrong with it.

00:02:02

I’ll attach the

00:02:05

corrected script for problem type 10 to the recording of the lesson. Why did

00:02:09

I decide to implement it? But because

00:02:11

this type of problem is similar to 9, but there is

00:02:13

only one exception

00:02:16

and Let me tell you about this moment,

00:02:18

see 5 6 type of problem, it was encountered

00:02:20

a very long time ago the seventh task of 7 problems

00:02:24

has been encountered for the second year

00:02:26

in a row, but type 8 9 10

00:02:29

turns out to be type 8, I set it by accident,

00:02:34

and type 8 9 10 these are such new problems

00:02:37

that the guys met for the first time in the

00:02:40

twenty-second year,

00:02:43

and today you should notice that these

00:02:45

types of problems in globally they are no

00:02:46

different, only some

00:02:50

concepts are new in them. But these concepts

00:02:53

always prescribe the task. That is, you

00:02:56

just need to turn your head on and think about

00:02:58

how to solve this problem a little

00:03:00

differently,

00:03:01

but we will still consider each type of problem

00:03:03

so that you how to

00:03:06

pump up logic like this

00:03:08

Well, let's start let's start we

00:03:11

get to solve the 5th task carefully

00:03:14

read it and what we see What difference

00:03:16

do we see in comparison with previous tasks

00:03:19

in this task is different

00:03:24

I have put forward the chat so far I

00:03:28

didn’t really like solving these problems

00:03:30

I’m actually very happy what did you

00:03:32

like because in fact,

00:03:35

protein biosynthesis is considered the most complex

00:03:38

topic, so it needs to be explained to me very well so

00:03:41

that you can then read it well and competently like

00:03:45

this

00:04:00

that the complementary chains of

00:04:02

nucleic acids are antiparallel in

00:04:04

this we already know 53 ends in one chain

00:04:06

correspond only to the end of the other chain

00:04:09

synthesis of nucleic acids starts from the 5'

00:04:11

end the ribosome moves along the RNA in the

00:04:14

direction 5 to 3

00:04:16

you surprise us as if we didn’t

00:04:19

know this Yes molecules Transfer RNA

00:04:21

carrying the corresponding anticodons, the

00:04:23

ribosomes enter in the following order, that

00:04:25

is, we are given molecules here,

00:04:28

RNA molecules, but they are not given to us

00:04:32

in full, they are given only in sections,

00:04:35

and which sections of the RNA molecules are given to us

00:04:39

that carry the corresponding anticodons,

00:04:42

here the ribosomes enter in the following order, the

00:04:44

anticodons are indicated in the direction of

00:04:51

this task we are given anticodons of the tRNAs of

00:04:54

different molecules.

00:04:56

But there is one thing, there is one important

00:05:00

feature.

00:05:02

They are written here as

00:05:05

they are indicated in the direction from 5 to 3.

00:05:11

But this means

00:05:13

this

00:05:16

means this means that we were shown the

00:05:20

direction of the task incorrectly.

00:05:22

Why. Because we remember that

00:05:25

anticodons

00:05:27

must be read. from 3-5 these are

00:05:34

RNA molecules that were synthesized along the

00:05:36

transcribed chain, they are read 5

00:05:38

to three. Yes, for example, codons and RNA. But

00:05:41

anticodons are anticodons, they must

00:05:43

be read antiparallel to the cadon. That

00:05:45

is, for 35 A, our problem says about

00:05:49

5 to 3. Therefore, we need to do something to

00:05:52

carry out inversion, that is, to

00:05:55

flip these anticodons in a mirror in order

00:05:58

to further solve the problem,

00:06:00

let’s get it, we’ll do it

00:06:04

like the second type of problem, only a little

00:06:06

more complicated because in the second type of

00:06:08

problem we did, we only

00:06:10

remade one, and here we’ll have to redo as many as

00:06:12

five codons, so we write

00:06:16

ordinary

00:06:19

RNA triplets Let's rewrite them as they are given

00:06:22

conditions, that is, 5 tsgu A three stroke end

00:06:26

5 A ga3 stroke end 5 Hz y three stroke

00:06:32

end 5 G a g three stroke end

00:06:37

and 5 Go 3 stroke end These are just triplets

00:06:42

and do not correspond to the anticodon But

00:06:43

they will be completely anticodons when

00:06:45

we write down their direction correctly, that

00:06:47

is, when each house is simply read in reverse,

00:06:52

we write RNA anticodons.

00:06:57

That is, if there were 5 cgu, then it will become 3 Hz, on the contrary, we

00:07:02

read like this, that is, it turns out 3

00:07:10

Next 3

00:07:12

Yeah, it does

00:07:17

n’t matter here because in the opposite

00:07:20

direction it reads the same thing further

00:07:22

3

00:07:26

Next 3 g also doesn’t change But that’s

00:07:30

okay And the last one is three a g

00:07:38

Now these are really anticodons

00:07:42

Absolutely true, but when you

00:07:46

explain this task You should say we

00:07:48

turn the chain over, this is how Ilmir writes

00:07:51

turn over

00:07:53

more precisely Anastasia Ladinova 180

00:07:57

degrees,

00:07:58

let's write this down.

00:08:01

Of course, we had to write this at the beginning.

00:08:03

Before we did it, that is,

00:08:06

write if you have already written it, you can

00:08:08

write after knowing

00:08:11

that the

00:08:12

anticodons

00:08:14

are read

00:08:16

from the 3rd stroke of the end, the

00:08:20

5th stroke of the end,

00:08:23

we convert

00:08:26

the anticodons trnk

00:08:32

or form write

00:08:39

everything we got it’s like this with you

00:08:42

Can we now write down these codes

00:08:45

in one chain? If so, how,

00:08:49

but we did this in the past In the last

00:08:52

lesson, that is, we just take it like

00:08:54

this In what sequence are the

00:08:56

sequences not written down in the

00:08:59

same sequence? We

00:09:01

write it like this, that is, three stroke end

00:09:05

ugz Just take it we rewrite it turns out

00:09:08

this is all here is how

00:09:10

we put a comma comma in one chain Next Yeah

00:09:15

comma next we have

00:09:23

then gag the

00:09:29

last one is y a g

00:09:32

You also follow me please follow me so that

00:09:35

I explain at the same time I write it down I can

00:09:38

write it down a little wrong letter

00:09:43

we get the anticodons of the RNA of

00:09:47

different molecules are written therefore separated by a

00:09:50

comma To show that these are different

00:09:53

molecules

00:09:55

like this at this stage you understand the task

00:09:59

this was the most difficult moment

00:10:07

yes Next what we do next we read the

00:10:11

conditions again but now it starts here

00:10:15

determine the sequence of the semantic

00:10:17

transcribed DNA strand RNA and

00:10:20

amino acids in the molecule of the synthesized

00:10:22

protein fragment, how will we do this?

00:10:24

We have tRNA anticodons. Can we

00:10:27

find something using them?

00:10:32

We can find a

00:10:38

whole molecule and RNA from anticodon tRNAs; a whole molecule and RNA Why?

00:10:42

Because anticodons are anti-parallel to the

00:10:44

codon, we will simply find each kadon

00:10:46

to which, accordingly, this anticodon and

00:10:48

that's it, of course, there is such a question that

00:10:52

many some guys have. Let's

00:10:54

answer it right away.

00:10:57

The recording will look at how why it

00:11:00

turns out. If we have different anticodons,

00:11:05

this is all written down in one chain with a comma.

00:11:09

How can we find it?

00:11:11

How Yes we we can find a

00:11:14

complete RNA molecule if we look

00:11:16

at the translation process, we will see

00:11:18

that the translation process involves

00:11:20

many transport RNA molecules, but

00:11:25

if we look at both RNA and only

00:11:29

one RNA, that is, there can

00:11:31

be a lot of transport market A and there is only one RNA, that

00:11:36

’s why we will look for this one and we will look for RNA, we are

00:11:42

looking for it simply according to the principle

00:11:44

of complementarity,

00:11:46

that is, according to the principle of complementarity,

00:11:50

the principle of it is elemental

00:11:54

by the anticodon of RNA, we will find the codons of RNA

00:12:02

and Let

00:12:05

me increase the scale here, it is

00:12:08

clear that RNA will be read from 5 to

00:12:11

3

00:12:12

So Let's

00:12:15

say what acg next They

00:12:29

already know how to do this, complementarity,

00:12:35

forming new chains. I hope you have

00:12:39

already learned this,

00:12:41

so we get such a molecule and

00:12:44

RNA, we write down that these are cadons,

00:12:48

addons and RNA,

00:12:51

we got 1 2 3 4 5

00:12:56

codons,

00:12:58

then we can find from this codon

00:13:00

what

00:13:04

we can then easily find amino acids

00:13:09

Yes, using the Gene Code table, so the

00:13:12

Gene Code table is lost somewhere,

00:13:14

let’s take the next task, that

00:13:18

is, acg to us 1 goes Let’s

00:13:20

find a

00:13:24

c and g this is trionine, so we write

00:13:28

Let’s write here

00:13:31

using the

00:13:33

Gene Code table

00:13:37

by codon and RNA we’ll find

00:13:40

sequence of amino acids

00:13:47

polyp we go and 3 we have the first one then

00:13:51

we get it from Cu this we will have from C y this is the

00:13:56

sickle

00:13:58

ser

00:14:02

further

00:14:04

and g.c this we get a g c also

00:14:13

C from C this we get from him c

00:14:18

this is lei

00:14:21

and the last one we have left is the A U S

00:14:27

or

00:14:30

all of us got the

00:14:31

sequence of amino acids in a

00:14:34

polypeptide

00:14:38

like this. It

00:14:42

seems to me that this task is easier than the

00:14:44

previous one. Exactly it seems because

00:14:48

why is it so because this second lesson has

00:14:51

passed a lot of time,

00:14:53

yes, that is, you have digested

00:14:56

the information from the past a little classes and now it

00:14:58

seems to you that it’s all simpler because

00:15:01

he said that these tasks, as it were, will

00:15:03

not become more complicated, the techniques

00:15:05

that we use these tasks they

00:15:08

used in the past

00:15:12

If I told you all 10 types of tasks in the last

00:15:15

lesson it would also be

00:15:16

difficult, so time must pass, and

00:15:18

then over time, before the Unified State Exam, you

00:15:20

will click these tasks like seeds,

00:15:23

what else do we have to find in these in this

00:15:26

task, we just need to find another DNA

00:15:28

semantic transcribed chain, how

00:15:31

will we search for these chains by which chain

00:15:34

Which chain do we begin with, we will find

00:15:36

transcribed or sense and by which

00:15:38

molecule let us answer this question,

00:15:41

in this case we have presented

00:15:45

we have codons and RNA there are amino acids

00:15:49

we are at the beginning Here you are Absolutely correct in what you write,

00:15:52

we will find at the beginning the template chain It is also

00:15:56

transcribed from

00:15:59

messenger RNA

00:16:01

and then we’ll find the semantic one according to the

00:16:04

transcribed perfectly.

00:16:06

Let’s do it

00:16:12

this way

00:16:19

according to the principle of complementarity,

00:16:26

according to the principle of complementarity

00:16:30

about the codon and RNA, we’ll find

00:16:34

the sequence of

00:16:37

nucleotides

00:16:40

in the

00:16:43

template DNA chain.

00:16:53

Well, let’s look now

00:16:56

at the codons and RNA and put a three

00:17:03

so here you have to be as

00:17:04

careful as possible yes that is, first we write

00:17:07

tgc

00:17:09

turns Next Yeah cg

00:17:15

so cg

00:17:18

and t.a.g

00:17:26

we found it, it turns out the matrix chain It is

00:17:29

transcribed now we will also

00:17:30

find the

00:17:35

semantic chain according to the principle of

00:17:37

complementarity until the bottom Already remove to by

00:17:40

triplets of the

00:17:44

template

00:17:48

DNA chain we will find the sequence of

00:17:50

nucleotides in the semantic chain

00:17:54

semantic chain

00:17:56

Let's

00:17:59

semantic chain is read from 5 so

00:18:02

we write 5 Reeve end Let's go further

00:18:07

it will be a cg

00:18:11

and

00:18:13

C

00:18:14

and ATC three stroke end

00:18:21

the problem is solved only we spent time on this task

00:18:25

we spent 15 minutes

00:18:28

Well, okay This is with explanations And you

00:18:31

probably minutes if you learn to solve

00:18:33

You can solve them even in five minutes

00:18:34

ideally

00:18:37

The most time consuming

00:18:39

tasks like this are to build chains and not

00:18:42

get confused on the contrary And this is you on the contrary and

00:18:45

so on

00:18:48

this is normal

00:18:52

Rate this task by complexity from 1 to

00:18:55

10 Let’s

00:18:58

estimate from 0 to 10

00:19:02

so

00:19:04

47

00:19:05

considers the problem difficult

00:19:08

and Mira 9 how is it

00:19:11

two three five but again yes you won’t

00:19:14

understand the problems that are difficult and not difficult until you

00:19:16

solve them yourself

00:19:18

so two three five but the most correct

00:19:22

answer is 5 because you haven’t

00:19:25

solved them yet You yourself doesn’t know this

00:19:28

if, in principle, there are no questions about this

00:19:32

problem Write the advantages we are moving on

00:19:35

Type 6 is such a spoiler it is even simpler

00:19:51

but the

00:19:52

complexity in the sixth type of problems is included

00:19:55

in the fact that here you need to know something

00:20:00

has changed in the sixth type of problems it is written

00:20:05

about viruses Yes viruses have not yet passed

00:20:07

viruses we will pass at the end of general

00:20:10

biology when we come with you mitosis we are

00:20:12

its santhogenesis and so on here it is written

00:20:14

that some viruses There and so on

00:20:16

Okay, so let's, before

00:20:19

solving such a problem, I will explain to you a little

00:20:20

about viruses, brief information

00:20:23

which will help you solve such problems

00:20:24

and then we will go through them globally on a

00:20:27

separate topic,

00:20:28

you don’t have to write this down now,

00:20:31

you just listen to me and understand the

00:20:36

most important thing, understand look, we

00:20:38

get viruses, who are these viruses,

00:20:41

this is an

00:20:43

organism that separates the kingdom of viruses into a separate

00:20:46

kingdom

00:20:47

and there are two

00:20:50

types of viruses: DNA-containing viruses and RNA-

00:20:53

containing

00:20:56

DNA-containing

00:20:59

viruses and RNA-containing

00:21:04

viruses; if we talk about the structure of

00:21:06

viruses, it is very simple; essentially it is a

00:21:10

nucleic acid. That is, it is either DNA

00:21:13

or RNA plus protein and the protein here

00:21:16

performs a protective function. That is, it

00:21:19

hides just the same

00:21:22

nucleic acid underneath,

00:21:24

you can just guess the same function,

00:21:29

preservation, transmission and implementation of

00:21:32

hereditary information. You

00:21:34

see, we have a bacteriophage that

00:21:37

you will later go through, and here the

00:21:39

bacteriophages are located inside this

00:21:41

protein shell called

00:21:43

capseit this is DNA

00:21:46

This is a bacteriophage DNA containing a virus

00:21:49

there is RNA containing viruses for example the

00:21:52

tobacco mosaic virus And also the

00:21:55

human immunodeficiency virus, that is, HIV, it

00:21:57

also contains RNA and it is also structured in the

00:22:01

same way

00:22:03

inside there is a nucleic acid the

00:22:06

nucleic acid is protected by what kind of

00:22:08

protein shell

00:22:10

what else is there Super capsid we will

00:22:13

go through later this is the main thing

00:22:17

And therefore the viruses we have

00:22:20

differ in nucleic acids either

00:22:23

DNA or RNA

00:22:25

Therefore the cycle of these viruses, that is,

00:22:28

their life cycle will also

00:22:30

differ, DNA containing viruses

00:22:32

will have their own life cycle and the market

00:22:35

containing the same cycle will change if

00:22:38

it changes life cycle, then

00:22:40

our

00:22:42

reproduction of these viruses will change,

00:22:46

and you should also know that

00:22:49

our viruses are intracellular

00:22:51

parasites,

00:22:53

intracellular parasites, that is, in

00:22:57

order to multiply, they must

00:23:00

get into another cell

00:23:03

and multiply there,

00:23:07

let’s look at this process using the

00:23:10

example of RNA containing viruses,

00:23:11

only they still need to contain

00:23:13

the virus for our tasks,

00:23:16

let's look at their development cycle, here is the

00:23:19

picture shown. That is,

00:23:22

look, we have a cell

00:23:24

Yes, this is a cell, for example, a host, a

00:23:27

human cell, this is HIV viruses, it is for the deficiency of the

00:23:31

person in which it is located RNA

00:23:33

it is an intracellular parasite

00:23:35

in order for

00:23:38

this virus to multiply it needs to

00:23:40

get into the cell how it gets into the

00:23:42

cell it approaches the cell it merges with

00:23:45

the membrane of the host cell

00:23:49

the most important thing is that it gives into the cell it

00:23:53

gives nucleic acid

00:23:56

further this RNA begins to move towards the

00:24:00

nucleus that's

00:24:05

what happens Further further in

00:24:10

this virus there is a special enzyme,

00:24:12

it is called reverse transcriptase,

00:24:16

which ensures the process of reverse

00:24:19

transcription What is reverse

00:24:21

transcription, let us remember What

00:24:23

is transcription in general, this is when there is a

00:24:26

DNA molecule and it is translated into an

00:24:28

RNA molecule using a DNA template, RNA is formed

00:24:32

reverse transcription When

00:24:35

the opposite happens When we have RNA

00:24:37

and RNA, we form DNA,

00:24:41

and only viruses can do this because they contain RNA,

00:24:47

thanks to their special enzyme,

00:24:49

which contains reverse transcriptase in these viruses,

00:24:51

so the viral

00:24:53

RNA interacts with this enzyme,

00:24:56

reverse transcriptase, and we

00:24:58

form a viral DNA molecule, here is a

00:25:02

double-stranded viral DNA DNA is a

00:25:05

double-stranded viral DNA that is introduced

00:25:07

into the genome of the host cell, that

00:25:11

is, such a genome. Well, it turns out

00:25:14

that the host has DNA, this DNA

00:25:18

is simply introduced into the host’s DNA, if

00:25:20

simply, and then the usual process already occurs there,

00:25:23

that is, the DNA of the cell

00:25:27

begins the process of transcription and is formed

00:25:33

this is already normal RNA and the RNA is

00:25:35

already a little different than it used to be

00:25:38

because this RNA contains Genes

00:25:42

that will synthesize viral

00:25:45

proteins

00:25:47

further, this RNA comes out and goes

00:25:49

to the translation process and there

00:25:52

Protein synthesis occurs, the translation process is what is

00:25:57

called

00:25:59

further these viral proteins

00:26:01

accumulates accumulates and

00:26:03

new viruses are assembled, which is called

00:26:06

HIV, these HIV viruses accumulate and

00:26:10

leave the cell and then go to

00:26:13

some other cell.

00:26:15

Of course, we will go through this process in more

00:26:18

detail, we will

00:26:19

describe it so that it is remembered

00:26:21

now, the most important thing for you to understand What

00:26:24

is reverse transcription? This is when

00:26:25

RNA turns into DNA.

00:26:28

That’s the fact that

00:26:31

RNA-containing viruses are capable of this. At

00:26:37

this point, you understand that this is

00:26:40

happening

00:26:43

now. Let’s understand the task even better.

00:26:51

Read the task. Now some viruses

00:26:54

carry RNA as Greek material.

00:26:56

But we already know this, yes, this is RNA

00:26:57

containing viruses. Which viruses, having infected

00:27:00

a cell, insert a DNA copy of their

00:27:02

genome; the

00:27:06

genome of the host cell has entered the cell;

00:27:09

viral RNA has entered the cell; the following

00:27:10

sequence is the

00:27:12

sequence of viral RNA;

00:27:24

RNA Write the sequence of

00:27:27

two flower fragments of DNA Indicate the 5th

00:27:29

and 3rd stroke ends

00:27:34

Well, let's do what first Let

00:27:36

's just take it and write out the viral RNA

00:27:42

so the first point is

00:27:44

viral RNA like 5

00:27:50

we write out a y

00:27:54

Hz y

00:27:58

and the

00:28:01

end stroke

00:28:04

Ok what we do next

00:28:09

we do next remember the cycle of viruses

00:28:12

that explained what it turns out, the viral

00:28:15

RNA enters the

00:28:18

cell then what happens Next, with

00:28:21

this viral RNA that we just

00:28:23

wrote down, the reverse transcript enzyme will interact

00:28:25

and we will have

00:28:28

the process of reverse transcription, that is,

00:28:31

DNA will be synthesized, it is important for us to

00:28:34

understand

00:28:35

what kind of chain it will be

00:28:38

synthesize The first semantic strand

00:28:40

or transcribed one is synthesized,

00:28:46

it is logical to assume that if the viral RNA

00:28:49

is read from 5 to 3, then of course the

00:28:54

transcribed target will be synthesized first because there are

00:28:56

three to five and the semantic one

00:28:59

will then be read from the

00:29:00

transcribed one.

00:29:05

Yes, you correctly said what is

00:29:07

being transcribed to you in condition, in

00:29:10

fact, there is a hint, see Determine

00:29:12

what the sequence of the

00:29:13

viral protein will be if the template for

00:29:16

synthesis and RNA serves as a chain

00:29:19

chain What is

00:29:20

the complementary viral RNA, that is,

00:29:23

Let this goal be like X as here is the

00:29:30

viral protein if the template for synthesis

00:29:33

and RNA

00:29:34

serves as the matrix for synthesis and market

00:29:38

there must be RNA and Let's

00:29:41

just remember from the past tasks

00:29:44

who has always served as a matrix for synthesis and

00:29:46

RNA

00:29:49

Matrix

00:29:52

Therefore this is X this is the

00:29:54

transcribed DNA chain the

00:29:58

transcribed DNA chain

00:30:01

let's find it

00:30:04

and then using this transcribed alcohol we

00:30:07

'll find the RNA which here in

00:30:10

the picture was the one

00:30:12

that then came out of the nucleus.

00:30:16

So let me not write now based on the

00:30:20

principle of complementarity, let’s find

00:30:21

the sequence. It’s already clear that

00:30:24

we need to write just straight away the

00:30:27

DNA that is being edited,

00:30:31

we write down three strokes, the end of

00:30:38

cga

00:30:43

and cg,

00:30:47

so I hope I wrote everything down correctly, it’s

00:30:51

transcribed. Now let’s find the DNA

00:30:53

and RNA

00:30:55

according to the Matrix of the transcribed chain,

00:30:59

again we need to write there according to the principle

00:31:00

of complementarity according to the matrix

00:31:01

transcribed, we’ll find the RNA and write it down,

00:31:05

we won’t waste time because we

00:31:07

still have a lot of types to sort out,

00:31:09

so we’ll write the 53rd end

00:31:13

and at g

00:31:16

gsu

00:31:18

so on I don’t see here let me go down a little bit

00:31:21

So with gcu

00:31:24

[music]

00:31:25

and gca there are

00:31:28

three strokes at the end, we get this and the

00:31:32

RNA is already, as it were, the host’s non-viral one

00:31:35

that was synthesized then from the nucleus

00:31:38

came out of the nucleus I mean

00:31:41

Okay, what can we do

00:31:44

next using the table of non-human code

00:31:47

you can easily find

00:31:50

[music]

00:31:52

just our polypeptide

00:31:55

Let's find it This is how it turns out

00:31:57

Methionine

00:32:00

I won't even look at the Gene Code table

00:32:02

because I have to remember the topics What do

00:32:06

we have G C

00:32:10

and

00:32:15

Then three

00:32:20

is a hairdryer

00:32:26

a And the last one is gca

00:32:31

gc a this is Alla

00:32:37

This is what we have got as a peptide field,

00:32:40

now what other tasks do we need to find

00:32:44

the sequence of a two-stranded

00:32:46

DNA fragment, but we only have one

00:32:49

chain found, we also need to find a semantic

00:32:51

chain

00:32:53

Well, let’s do this, that was the third point

00:32:57

4 point find means DNA is already a completely

00:33:01

semantic goal search using

00:33:03

transcribed DNA so we start 53

00:33:07

end So you AC means here atg

00:33:11

then we will have

00:33:17

and

00:33:28

the problem is solved

00:33:35

first we need to find transcribed DNA

00:33:37

only then RNA Well, how

00:33:39

else will you find RNA

00:33:42

Well, you won’t find it any other way directly

00:33:45

with the viral you won’t find it because but

00:33:47

it’s impossible, that’s what

00:33:52

we’ve just decided, yes,

00:33:54

these are all four points, they’re all kind of

00:33:58

depicted in this diagram, in fact,

00:34:07

this is the task,

00:34:15

if there are no questions, put a plus Let’s

00:34:17

move on, so the last semantic DNA, yes,

00:34:20

let’s write it down this is a semantic DNA

00:34:33

cool problem Well, I wish everyone this on the Unified

00:34:37

State Exam

00:34:39

No, they’re all easy, it’s just a

00:34:41

matter of practice

00:34:47

How many hours will it be in two hours?

00:34:51

Only two types were sorted out in half an hour,

00:34:53

everything is clear here, let’s go to the seventh type of

00:34:56

problems,

00:34:57

here we actually have a little

00:35:01

geometry, so those who doesn’t like geometry

00:35:07

But why geometry because here you

00:35:09

need to solve such a problem through

00:35:10

an assumption Why so give it to

00:35:13

me yourself Tell me what But what makes this

00:35:17

problem different from all previous problems

00:35:19

especially from problems of the first five types of the

00:35:21

first four

00:35:23

like Dan given DNA molecules 2 chains Well,

00:35:29

as if Okay, it will fly there, it starts to

00:35:32

be changed Okay, there was no first, like

00:35:35

there is a GIS

00:35:37

only in this difference or something

00:35:39

else,

00:35:41

Regina Sultanova Well done, it is not

00:35:45

written here where What kind of DNA is there, that is, the

00:35:47

task was always written there The

00:35:49

lower goal is transcribed The upper

00:35:52

semantic A this one was not indicated here,

00:35:59

and here is just a new

00:36:02

new condition, look, it is known that this

00:36:06

policy begins with

00:36:07

at least four amino acids, that is, we

00:36:10

must solve the problem here and at the end

00:36:12

compare whether our solution coincides with the

00:36:16

data in the conditions, if it does not coincide,

00:36:19

then we have solved the problem wrong if

00:36:20

everything matches then we are great and we can

00:36:24

move on to the next task in Kimi

00:36:28

like this

00:36:34

Therefore, if the task is not indicated

00:36:36

which chain Which one should we assume

00:36:39

but look you always thought always in the

00:36:44

problem they said that the Lower one is

00:36:45

transcribed let us assume the

00:36:47

Lower goal is transcribed What if we are

00:36:50

lucky we write first point, let’s assume

00:36:57

that I’m writing this now, it really needs to be

00:36:59

written, then for the decision, let’s

00:37:02

assume the

00:37:04

lower chain is

00:37:14

transcribed,

00:37:21

let’s write it out, I won’t write it out,

00:37:24

I’ll just take it and copy it, as I

00:37:26

always like to do, so I’ll take it,

00:37:29

copy it, cut it like this

00:37:34

[music]

00:37:38

but you can don’t write it down on the sheet

00:37:42

there is

00:37:44

So what next

00:37:48

We need to ultimately find the peptide field

00:37:50

Therefore, from this

00:37:52

transcribed chain we know what we will

00:37:54

do to look for and RNA then through RNA we will

00:37:57

find it turns out our peptide field

00:38:00

seems everything is easy yes in fact

00:38:02

Let’s check

00:38:04

so we

00:38:06

will find and RNA according to the principle of

00:38:09

complementarity,

00:38:13

I will now write in abbreviated form these

00:38:16

phrases, in fact, we should have written it,

00:38:18

let’s find and write the market for complementarity

00:38:23

along the transcribed DNA chain,

00:38:28

it turns out 5

00:38:34

5 So you Yeah, it turns out with then

00:38:38

GHz

00:38:41

and

00:38:45

so if

00:38:50

a and t AC So we have auc ggc and y a is

00:38:56

then we have

00:38:59

C

00:39:02

Hz y

00:39:04

and

00:39:06

AC it will be a y g

00:39:12

I hope everything is written down correctly

00:39:14

and RNA

00:39:20

further here is said

00:39:25

in a fragment of the policy encoded by this

00:39:27

section of DNA if it is known that it will

00:39:29

fly starts with an amino acid,

00:39:32

you see amino acids

00:39:35

we need What to do now we need

00:39:37

to find the Gene Code table Here it is, this is the

00:39:41

Gene Code table, let’s also

00:39:44

copy it so that it does

00:39:48

n’t loom constantly

00:39:57

[music]

00:40:00

so here is our our table

00:40:05

and we need to find it turns out GIS in the

00:40:08

table Gene Code where is it But here it

00:40:11

turns out here it

00:40:15

turns out

00:40:16

GIS we have What codons this corresponds to

00:40:20

Ca and C a c

00:40:23

we write About the table The gene code to

00:40:36

the amino acid GIS corresponds to

00:40:41

the codon and

00:40:47

as we did in the first type of problems in the

00:40:49

first type of problem we get there

00:40:51

we had aug mine they were looking for simply in the

00:40:53

chain Let's find here this one

00:40:58

can really exist because we

00:41:02

can’t take the beginning Because it’s not According to the

00:41:04

condition it should be 1

00:41:06

so Well, let’s look at cg No cg doesn’t

00:41:11

fit

00:41:12

But we

00:41:14

found it So maybe there’s something else there

00:41:19

but it does

00:41:21

n’t fit yet

00:41:24

Oh we found it’s cool that is, we have a

00:41:26

GIS Therefore, next we will have agts

00:41:30

and y a y

00:41:33

Let's look at

00:41:35

the action Au this is GIS up from the table Gene code

00:41:39

we can immediately find further A g.ts here and here

00:41:43

G And this is

00:41:46

Sergei C this is this is the

00:41:50

server

00:42:01

And the last thing we have left with

00:42:07

this is a shooting gallery

00:42:10

Yeah It’s just like, well, it doesn’t fit because

00:42:15

giving some amino acid once it

00:42:18

turns out that we got 3

00:42:20

amino acids in total,

00:42:22

then we solved the problem correctly or not

00:42:31

if we read the conditions further, it

00:42:35

says it has a length of at least four

00:42:38

amino acids And we only get 3

00:42:40

it turns out to be a bummer Yes, we solved the problem

00:42:43

incorrectly. We incorrectly assumed

00:42:45

that the Bottom chain is transcribed. What does it

00:42:48

mean to do next? imagine, in

00:42:50

reality, you will come to the Unified State Exam, you are

00:42:52

sitting there, worried and you did not know about the

00:42:55

assumption about these types of problems; you

00:42:58

solve,

00:42:59

notice what you don’t have here it’s written

00:43:01

what the goal is What Such a

00:43:03

damn thing and yes this is a bit of prank you can

00:43:09

really say you were pranked

00:43:12

in fact no one pranked you

00:43:13

because this is a separate type of task

00:43:15

What we do next means how we get out of this

00:43:17

situation we must say we

00:43:23

must draw a conclusion in the end we

00:43:26

got in field peptide

00:43:30

amino acids

00:43:33

a According to the condition

00:43:36

According to the condition there must be at least 4

00:43:42

therefore the

00:43:47

Bottom strand

00:43:51

is not transcribed, we're

00:43:59

all

00:44:02

done here if the Bottom strand

00:44:05

is transcribed, then it's logical

00:44:07

that the Top target will be

00:44:08

transcribed let's write this down in the

00:44:10

second paragraph

00:44:14

Top strand

00:44:17

suppose let's

00:44:20

assume that the Top strand is

00:44:23

transcribed

00:44:28

So what we have to do now is The top

00:44:32

target is read from 5 to 3 to this is how we

00:44:36

what are we going to do now

00:44:43

yes Again turn it over because we

00:44:47

have no right to say that the

00:44:48

transcribed strand is read from 5 to

00:44:51

3 because but this is wrong not according to the

00:44:52

rules we are we must take and

00:44:56

turn the chain back over, that is, read it

00:44:58

backwards 3gta this is

00:45:06

how we did it, it turns out in the fifth

00:45:09

type of problem Yes, only we were already working

00:45:12

with

00:45:14

RNA there, and here we are working with DNA

00:45:18

We are working within the same chain Therefore,

00:45:21

let’s turn it over, we just

00:45:25

write knowing that the

00:45:29

transcribed the

00:45:35

transcribed chain is read from 3 to

00:45:41

5,

00:45:43

we will turn the

00:45:46

top chain 180 degrees

00:45:54

Well, just take it and

00:45:56

rewrite it the other way around.

00:45:59

Let's

00:46:01

get it like this: 3 GTA

00:46:04

is

00:46:06

G

00:46:08

cg ATA

00:46:10

cg and csta

00:46:14

GTA cg a

00:46:18

t a cg and ctr

00:46:25

everything

00:46:27

seems to be correct,

00:46:29

so we transfer it here

00:46:32

and now according to the principle of complementarity

00:46:34

we are looking for and RNA

00:46:39

about the principle of

00:46:41

complementarity

00:46:42

we are finding and RNA

00:46:47

Let's 53 end

00:46:50

5 stroke end and in

00:46:53

agc

00:47:03

[music]

00:47:10

this is RNA

00:47:14

So explain why you can’t understand by the

00:47:15

stroke of the end Where is which chain, otherwise

00:47:17

I don’t see the meaning of the problem at all Well, look,

00:47:21

you understand in such in problems

00:47:24

they can specifically write the

00:47:27

chain incorrectly, but at the same time in the first type of

00:47:30

problem they always always clarified it for

00:47:33

a reason, they clarify it and say that the

00:47:36

Upper chain is like this, the Bottom chain is like this,

00:47:39

but in this problem they didn’t clarify this for you

00:47:41

because in general any chain How

00:47:44

can any word be written in

00:47:46

I even explained in different ways there is a word, for example Arthur, yes it

00:47:49

can be written this way correctly. But you can write it

00:47:53

differently, you can write

00:47:55

it like this, yes. And then this word will

00:48:00

lose its meaning and will no longer be the same as the

00:48:02

first. Here the same thing can be written

00:48:05

according to the normal semantic goal As it should

00:48:07

be in the direction of three five Well, semantic

00:48:11

35 to semantic 5, I can write it

00:48:14

the other way around. Just read it the other way around. Then there

00:48:17

will be no meaning. This is exactly what the task is about.

00:48:19

You found the correct chain,

00:48:28

but now let’s further, according to the principle

00:48:31

of complementarity, find the RNA, here is Coy

00:48:33

agc

00:48:35

further

00:48:36

according to the Gene Code table

00:48:39

amino acid git corresponds to the codon Co

00:48:42

Let's find them,

00:48:44

but here it is Co, we found it right away Let's

00:48:48

see, maybe there are some more,

00:48:51

but there aren't any more,

00:48:54

but we found y

00:48:59

Now let's build a polypeptide, this

00:49:03

will be later

00:49:06

So agc we will have a

00:49:10

server

00:49:13

then

00:49:18

and g. ts we will have Alla

00:49:24

it will be

00:49:30

and with the condition Everything fits That is, we have

00:49:33

more than four amino acids more precisely How are

00:49:36

the conditions no less, we should have

00:49:39

no less than five,

00:49:47

so Regina Sultanova correctly wrote

00:49:50

what they can give, give strokes the ends as

00:49:52

anyone wants sequences

00:49:55

want any sequence,

00:50:05

all the problem is solved

00:50:07

Well, we can also draw the following conclusion and

00:50:09

write it because

00:50:12

since we ended up with at least 4

00:50:17

amino acids

00:50:20

in the peptide field The

00:50:23

upper chain is transcribed,

00:50:37

so it turns out in terms of speed. By the way, we

00:50:40

’re going quickly, we can even do it early, we

00:50:44

solved as many as three problems in 50 minutes

00:50:49

8 9 10 8 in general we will quickly solve

00:50:53

a 9 10 They are similar 9 I

00:50:56

will explain the maximum in detail 10

00:51:00

So you want to take a break

00:51:04

easily and so as not to forget centuries You need to

00:51:07

decide

00:51:08

You can anyone You can

00:51:11

forget anything if this is not addressed do not

00:51:14

repeat and so on then

00:51:16

you can forget the multiplication table if you

00:51:18

don’t count money every day

00:51:21

So let’s

00:51:23

create a poll Let’s

00:51:26

give you a break,

00:51:29

just for a break, of course

00:51:37

we can for about 5 minutes even make an explosion

00:51:46

so Olesya may not notice because she’s

00:51:49

such a flooder if there were more, then there

00:51:52

were still, I just don’t understand the question can be

00:51:55

more detailed

00:52:00

like this And for you and for me I have a task What

00:52:04

task didn’t come across I passed it in

00:52:08

2019, we didn’t even have a direction at all,

00:52:12

we were given just letters for us, we didn’t

00:52:16

have to write the direction there 53

00:52:18

end of 3 4 end of 220 they carried out such

00:52:21

updates in Unified State Examination that now you need to

00:52:24

write each chain with a direction like

00:52:28

this, I came across problem

00:52:32

28 of the line, it’s already yes now like

00:52:35

this if the table had not

00:52:39

only the Central Administrative District at the beginning also Olesya, we’ll

00:52:42

analyze this type today, see

00:52:48

50/50

00:52:53

and what do

00:53:02

you give me to do? of course That is, yes, I can

00:53:05

decide

00:53:09

or I have to Although I can’t vote

00:53:14

Okay, let’s take a break for about

00:53:18

5 minutes

00:53:21

here

00:53:23

Well, that is, until 18:00 it’s already even 4 and Let’s go

00:53:27

on to rest,

00:53:33

look, I can disappear on the webinar

00:57:38

like this

00:57:41

All I’m back

00:57:45

too, rested or not rested

00:57:52

[music]

00:57:54

or just to go have fun, yes it

00:57:56

turns out

00:57:57

5 I think more

00:58:02

so everyone moved on to the

00:58:07

next problem to solve

00:58:10

problem number eight

00:58:13

here you will read what you will notice

00:58:16

Touch it

00:58:21

is known that the complementary chains

00:58:23

of nucleic acids are antiparallel Yes,

00:58:25

the synthesis of amino acids begins 53 ends

00:58:29

Or it itself moves along markets direction

00:58:31

The gene has coding coding regions,

00:58:35

here we are given a DNA molecule,

00:58:38

again it is not indicated which chain Which

00:58:40

determine the sequence of

00:58:42

amino acids the beginning of the path

00:58:44

no explain the sequence so

00:58:47

on

00:58:48

in this problem

00:58:50

no yes and it is not said that we can say

00:58:54

that this problem is similar to the seventh type

00:58:56

here it’s simple there is no clarification of how many

00:58:59

amino acids there should be in a polypeptide, on

00:59:03

the other hand, this means even simpler than

00:59:04

the past

00:59:08

Well, let’s solve it, look

00:59:13

Let’s assume

00:59:16

Let’s assume that the lower chain is

00:59:18

transcribed let’s

00:59:24

assume the

00:59:26

lower chain is

00:59:28

transcribed

00:59:35

for this

00:59:41

I won’t let you rewrite it and how

00:59:43

would it be Imagine what’s here a chain is a

00:59:46

chain

00:59:49

and at the bottom we’ll write find and RNA go RNA

00:59:53

let’s find GT it turns out like this

00:59:58

53 GT it will be with and then Transform it

01:00:04

turns out

01:00:11

the main thing now again Very small

01:00:14

written like this

01:00:16

from a turned

01:00:22

um m g further

01:00:33

further from And we Transform into y a y

01:00:40

here also at

01:00:43

tct we Convert a g a g a

01:00:56

So wait Oh yeah

01:01:02

Yeah and it turns out the last TG and the stroke so

01:01:14

up to 10 there’s no type I’ll add it Then I’ll add it

01:01:18

because but I decided to sort it out today I

01:01:20

already wanted to sort it out in a separate lesson,

01:01:21

I finally decided let’s give today it’s better to

01:01:24

finish everything to the end of the type and

01:01:26

of course everything is still on the palindrome and the problems

01:01:28

that mine appeared later we’ll analyze it

01:01:30

because it won’t be at the game

01:01:33

we’ll analyze it when at the additional

01:01:35

webinars on Sunday

01:01:39

it’s generally easy

01:01:41

everyone has now found the amino acid of methane that

01:01:48

we can just take this phrase from the table

01:01:51

Gene code

01:01:53

amino acid no

01:01:56

corresponds to codons aug

01:02:00

Let's find

01:02:04

that's

01:02:09

all and we've already found it,

01:02:14

then we need to find that after amino acids

01:02:16

there is no C Hz we will have that

01:02:20

cgc so c

01:02:24

this is ark

01:02:27

ark then g.gu

01:02:37

shares

01:02:39

we get this or

01:02:46

a y a y a is this too or

01:02:57

so

01:03:01

So let's

01:03:07

g a g it's stupid

01:03:14

here we all guessed right the first time

01:03:18

So where did I go wrong

01:03:20

Tell me

01:03:29

like this GT

01:03:37

Let's better tell me

01:03:47

after Aue there's an extra letter

01:03:54

like that Let's check there

01:03:58

is such a moment like this gtt this

01:04:03

cyaniva has a bad influence

01:04:07

Saa further

01:04:10

y and

01:04:15

y cg

01:04:28

so I understood everything

01:04:31

here the extra letter

01:04:34

should be correct It turns out We are

01:04:37

wrong up to amino acids

01:04:40

now it’s normal

01:05:00

so no further

01:05:10

so Allah

01:05:13

okay then you didn’t screw up the Unified State Exam before

01:05:17

gts

01:05:19

so gts G then g u and

01:05:23

this is the shaft

01:05:26

shaft

01:05:29

then

01:05:32

gua this

01:05:36

gcg also took it or something

01:05:45

and this is from us Val

01:05:51

u and this

01:06:01

works out for us

01:06:32

So let's make this chain shorter again,

01:06:34

so you'll bring it as

01:06:37

close as possible so as not to get confused

01:06:40

5th stroke like this C Yes we have the beginning

01:06:48

then what we go with

01:06:52

the tact it will be

01:06:55

and

01:06:58

so that is, we don’t take any more

01:07:01

further from And we won’t take this from us

01:07:08

By the way, a life hack so as not to

01:07:11

get confused, put such dashes,

01:07:15

too, we remove

01:07:18

further

01:07:25

then it goes from us,

01:07:31

we remove further goes it

01:07:38

turns out we will have

01:07:40

y and y

01:07:43

and

01:07:48

then we also have y

01:07:56

then go shopping center it turns out that a

01:08:01

g

01:08:04

a and the last letter T should now be

01:08:09

all right

01:08:12

now we remove these things so that they

01:08:14

don’t interfere with us finding aug

01:08:19

So A already turns out here

01:08:27

further on the table Gen code we find no

01:08:32

no

01:08:34

further

01:08:39

we found it with you further

01:08:43

this is what we get

01:08:47

ggu

01:08:49

this gly

01:08:52

and

01:08:55

this or and

01:08:57

this or and

01:09:02

this too or

01:09:06

and g a g this is

01:09:11

stupid

01:09:27

I mean it was right and they didn’t

01:09:30

prank or something

01:09:32

damn it you give guys

01:09:37

Well, it’s your fault that it

01:09:47

turned out the same way Well, it turns out

01:09:50

We solved this problem correctly But

01:09:54

no,

01:09:58

we solved everything correctly and it wasn’t even worth

01:10:02

assuming that the Upper chain is

01:10:03

transcribed in fact, in order to

01:10:06

show the complete solution to this

01:10:08

problem you need to assume that they are the

01:10:10

Upper the chain is transcribed And you

01:10:13

have to write the second paragraph, suppose the

01:10:17

top chain is

01:10:20

transcribed you then What do you do

01:10:24

then, as if you drink it, write it out already in the

01:10:27

right direction, say that

01:10:30

since the semantic chain How so As the

01:10:34

transcribed chain at all, in fact, this should not be

01:10:37

done Wait

01:10:39

Let's we won’t do this because

01:10:41

in fact, that year I

01:10:44

said what needs to be done, but then our

01:10:46

Student went This type of task, he did not

01:10:49

imagine that the Upper goal was

01:10:51

transcribed wrote such a phrase,

01:10:52

look,

01:10:54

he wrote like this

01:10:58

[music]

01:11:01

in the end we got

01:11:05

we got

01:11:08

4 amino acids

01:11:12

in a polypeptide

01:11:19

synthesis begins with an amino acid no

01:11:25

amino acid no

01:11:28

hence the

01:11:33

Bottom chain is the

01:11:38

Bottom chain is transcribed

01:11:43

and the Upper chain is semantic

01:11:46

Why in fact you can do this and

01:11:49

write the correct option right away because in the

01:11:52

Keys the correct option is immediately written down

01:11:54

there is not there at all assumptions when you

01:11:58

write an assumption it simply shows

01:12:00

the expert to really think about the problem

01:12:03

and what the conclusion is, you understand the logic of this

01:12:07

problem. This is how we generally understand now whether

01:12:10

we write an assumption or not write it.

01:12:13

When you solve such problems, you

01:12:15

begin to assume that the lower chains are

01:12:17

always transcribed and we

01:12:19

start the lower chain and then if The lower chain

01:12:22

does not fit,

01:12:23

go to the upper one if you have The lower

01:12:27

chain immediately gets that it is

01:12:28

transcribed for you Everything matches,

01:12:30

write this and write the conclusion and if

01:12:35

not, then you write down the assumption

01:12:37

starting from the lower chain, it

01:12:40

doesn’t work there, write it all down and write it down

01:12:41

and then go to the upper one, there you

01:12:43

can translate it in a different

01:12:47

direction. As a result, you get the correct

01:12:49

answer like this, that is, your solution

01:12:52

should according to the template, it starts with

01:12:53

an assumption about the lower chain, so this is

01:12:56

the maximum. Logically,

01:13:00

this whole problem is solved

01:13:06

Regina I think there is no need to write it down,

01:13:10

so you write, you write immediately the result That is,

01:13:14

when you assume that the Lower one

01:13:18

turns out to be real, so you don’t need to

01:13:20

write everything about the upper chain and write it down, it

01:13:23

just deduces that if we found it, it’s

01:13:25

no longer possible for DNA to have 2

01:13:27

transcribed ones. This is what logically

01:13:29

happens,

01:13:38

so with this task we understand more

01:13:40

here they checked the correctness of the solution

01:13:42

in general they could have solved this problem in three minutes

01:13:44

if it weren’t for your pranks in the chat

01:13:51

Why 4 amino acids But because it

01:13:54

happened that way

01:13:55

oh

01:13:57

they are 4 6

01:14:03

I looked here at the wrong

01:14:06

solution to problems

01:14:17

That’s the truth Here I also got to

01:14:20

move here and in the end we get 6 to 6

01:14:28

so OK Let's move on now let

01:14:32

's move on to more adult

01:14:34

problems this is 9 Type and 10

01:14:37

in general these problems are similar

01:14:40

in that

01:14:43

the topic is not similar Let's see it is

01:14:45

known that the complementary chains of

01:14:48

nucleic acids are antiparallel to the

01:14:49

acids the 5th stroke of the end begins or the

01:14:51

soma moves from 5 to 3 The gene has a

01:14:53

coding non-coding region a

01:14:55

coding region where it is called an

01:14:57

open reading frame see

01:14:59

here a new term therefore a new type of

01:15:01

task see Gene has a coding it

01:15:04

turns out the Gene is a section of a DNA molecule

01:15:06

everyone I hope they remember and it has a coding

01:15:08

and a non-coding letter H yes regions

01:15:12

so here the coding region is called the

01:15:16

open reading frame,

01:15:19

this is what you should know about the end fragment of the Gene

01:15:22

has the following

01:15:23

nucleotide sequence The lower strand is template Here

01:15:26

we are told that the lower strand is template

01:15:28

here we don’t need to assume anything

01:15:31

further with the worldwide reading frame that

01:15:33

is, here is the coding region and find the

01:15:35

amino acid sequence in the

01:15:37

end fragment of the

01:15:39

field the end of the peptide chain is important, it

01:15:43

is known that the final part of the polyptide

01:15:45

encoded by this gene has a length of more than

01:15:47

4 amino acids.

01:15:49

Explain the sequence of solving

01:15:51

the problem, and so on. Now let’s

01:15:54

remember about this important moment of

01:15:56

the end of the polyp and look at problem number 10,

01:16:00

here the beginning is the same, we are given DNA

01:16:04

The lower strand is transcribed to us writes,

01:16:06

but look here Determine the correct

01:16:09

frame, the correct open reading frame, and

01:16:13

find the sequence in the fragment, the

01:16:15

beginning of a

01:16:18

more peptide Chain, that is, this

01:16:21

task differs in that from

01:16:23

that task you see, as it were, the end of the chain.

01:16:26

You must find it. And in this task, you are

01:16:28

looking for the beginning of a polymorphous chain. Maybe

01:16:31

just it seems that what a difference there

01:16:34

really is a difference. There is a big difference.

01:16:37

Why is this so in this problem

01:16:42

now let's

01:16:44

decide what we have to do here

01:16:47

first of all

01:16:55

we have to find by the template chain of

01:16:58

this

01:17:00

RNA the first point I have to write

01:17:06

we find and RNA is the

01:17:17

open reading frame guys this is the

01:17:20

coding part of the Gene consists of

01:17:23

coding regions and non-coding

01:17:25

conditions it is written that the

01:17:27

coding part is that which

01:17:29

forms proteins at the end through it then

01:17:32

RNA is formed and then the protein is by and RNA and

01:17:36

not coding but this is the one that leaves

01:17:38

Then as a result of spicing in the RNA

01:17:41

molecule

01:17:42

and that’s why the coding part

01:17:44

in another way they call an open reading frame,

01:17:48

so we find RNA, why because we are

01:17:51

asked to As a result, find amino acids How

01:17:54

can we find amino acids RNA so here I

01:17:58

will write all this myself so that they don’t

01:18:00

prank me again, so we write like this here

01:18:04

it means a and then C Hz u a u a g

01:18:14

so

01:18:17

Let's

01:18:33

3 3

01:18:37

1 2 3

01:18:49

share Hz

01:18:51

a u a

01:18:58

y it turns out let's a g.g

01:19:02

cgts u a y A.G cia u a g a

01:19:09

gsu

01:19:14

Vera I hope Let's just count like this

01:19:17

one two three

01:19:20

6 9 12 15 18 21

01:19:24

24

01:19:30

9 12 15

01:19:34

should also be 25

01:19:44

maximum we double-check ourselves one

01:19:47

two three 4 5 6 7 8 8 by 324 + 1 25

01:19:56

Absolutely true The main thing is that the letters

01:19:59

match

01:20:02

so

01:20:08

everything is correct Absolutely true probably the class

01:20:13

was the most difficult moment in this the

01:20:15

task of finding and RNA, then what should we do

01:20:21

since this is the end of the polepian Chain, then

01:20:24

look,

01:20:25

that is, for example, we have a chain

01:20:28

of amino acids, then we have

01:20:30

an amino acid, once there, then

01:20:33

amino acid 2 And in general, you should

01:20:36

remember here how protein synthesis generally ends,

01:20:39

it stops us, why

01:20:42

amino acids

01:20:46

they don’t join further What is

01:20:50

the end of the translation reaction

01:20:54

Absolutely true in order to understand

01:20:57

where it ends

01:21:04

here somewhere in RNA

01:21:08

and we look for this stop codon in the

01:21:10

Gene Code table,

01:21:12

look, remember guys that stop codons

01:21:15

in the Gene Code table can be found, these are

01:21:17

characters That is, and this is a stop cadon here and

01:21:23

then AG also has

01:21:27

a stop cadon

01:21:30

and y or rather g a also has a stop cadon

01:21:36

But you have a stop cadon we have,

01:21:39

we write the

01:21:42

second point and RNA we are looking for stop cadons

01:21:48

this is y a g this is y a

01:21:53

this

01:21:59

Let's find these stop-kadons

01:22:01

here like this Let's

01:22:06

find it Let's oge Maybe there is somewhere,

01:22:11

but here's

01:22:14

a stop-kadon further on,

01:22:22

here's another stop-kadon And that there are two of them,

01:22:25

these Olesya asked me, now

01:22:27

we'll understand what to do here, I do

01:22:32

n't see any more Let's see but

01:22:34

there is, but look

01:22:42

here we can find as many as three stacks of

01:22:45

houses

01:22:46

and maybe even more more I don’t see

01:22:49

here

01:22:52

it’s interesting

01:22:57

to do in this case

01:22:59

here as many as three stacks are found we

01:23:02

will choose in order to choose the

01:23:04

right stacks We need to look

01:23:06

at the conditions it is known that the final part of

01:23:09

the polypeptide encoded by these genes

01:23:11

slowly more than 4 amino acids

01:23:14

look if we take Yes, let's

01:23:18

assume that this is a stopcadone,

01:23:22

then it is clear that

01:23:24

the reading goes in this direction

01:23:26

from here

01:23:28

and after y And it is clear that the synthesis

01:23:31

will stop and there will be no more amino acids, which

01:23:34

means synthesis will be involved

01:23:38

somewhere here involved precisely with these

01:23:41

canons,

01:23:42

or we will succeed, for example,

01:23:44

it will give its own amino acid and cdc will give more and

01:23:48

more amino acids already here after And

01:23:52

no more amino acids will be added. That is,

01:23:55

if we take and here, then we will have

01:23:57

only two amino acids

01:23:59

and According to the condition, there must be more than four

01:24:03

And as a stop- Kadon doesn’t fit, we

01:24:06

don’t take it further, it

01:24:11

also doesn’t fit Why look

01:24:15

Firstly, because there is another stop Kadon in front of it,

01:24:17

and even if this

01:24:21

happens, 2 amino acids also don’t

01:24:25

fit our conditions,

01:24:27

but AG, which is here,

01:24:31

fits the conditions,

01:24:32

therefore

01:24:34

Why is it suitable? Let's count

01:24:38

the amino acid,

01:24:49

it's not a whole chain. Maybe there's a

01:24:51

continuation somewhere, it

01:24:53

already gets how many times two three 4

01:24:55

5 amino acids and fits our

01:24:57

conditions, so it's AG that fits

01:25:01

our conditions like a stack of houses and we have to

01:25:13

we write it in as much detail as possible to describe the conditions of the third paragraph,

01:25:15

since the

01:25:17

stop code

01:25:19

at the end of the Gene is the open

01:25:23

reading frame, that is, the coding part

01:25:27

that gives the amino acids

01:25:29

will be located

01:25:33

do located

01:25:35

from the stop cadon to the left of the stop cadon

01:25:43

here on the left this means on the left here

01:25:51

[music]

01:25:55

we will write further

01:25:58

fourth point

01:26:01

stop kadons we must now

01:26:04

describe all the stop kadons why

01:26:07

the stop kadons in a and A.G are not suitable, that is, these ones

01:26:11

that were here that gave only

01:26:14

two amino acids on the left

01:26:17

which begins

01:26:20

which begins ugan we must

01:26:23

write which one is in AG because that there are

01:26:25

two of them here so one two three 4 5 6 7

01:26:30

8 9 10

01:26:34

nucleotides here it is

01:26:37

10 nucleotides

01:26:43

and RNA

01:26:46

do not correspond

01:26:49

these two codons do not correspond to the conditions of the

01:26:52

problem

01:26:55

since you

01:26:56

cannot get

01:26:59

more than 4

01:27:02

amino acids of

01:27:05

amino acids

01:27:07

in the field

01:27:11

and the stop codon is

01:27:15

in AG which begins

01:27:20

with C what nucleotide, let's see one

01:27:24

two three 4 5 6 7 8 9 10 11 12 13

01:27:29

nucleotide

01:27:34

18 nucleotide

01:27:39

it

01:27:42

corresponds to

01:27:46

the conditions of the problem

01:27:56

and now we can do what

01:27:59

we can find use the table Gene

01:28:01

code 5 point

01:28:04

using the encode table

01:28:08

we will find the

01:28:09

sequence of amino acids

01:28:14

polypeptides

01:28:18

everything

01:28:21

for this we must now carefully

01:28:23

from A y g find these triplets

01:28:27

from him a triplet is once a triplet 23 plaid

01:28:30

33 years

01:28:32

4 and 5

01:28:36

5 so but I will look at the table of the Gene

01:28:39

Code and immediately write it down

01:28:44

so it turns out gcg what is this for us G cg

01:28:49

this we have Alla next C y and it will be

01:28:54

laziness

01:28:58

and this or

01:29:03

so

01:29:06

gca is Alla

01:29:10

and

01:29:14

we found all this with it turns out

01:29:20

and we also need to write the

01:29:23

correct open reading frame for

01:29:26

this you can make such a trick,

01:29:28

look right in the picture, more precisely on the chain

01:29:32

you can mark it just take it like this,

01:29:34

right on the Unified State Examination Society,

01:29:38

put square brackets like this

01:29:42

just without crossing out the letters. This is how it

01:29:45

will turn out with a different color,

01:29:48

take it like this

01:29:54

and up to here, that is, without including the stop code

01:29:59

and write

01:30:01

the correct open reading frame or

01:30:04

open reading frame

01:30:06

open frame absorption

01:30:16

and here, too, in AG you can circle it with a

01:30:20

pen and write with a

01:30:23

stop code

01:30:30

in parentheses write the place where

01:30:33

protein synthesis ends

01:30:37

Well, or a polypeptide

01:30:40

here This is the

01:30:47

picture we get

01:30:52

Why don’t we take the first two I didn’t listen to

01:30:55

because we don’t take them because there’s not

01:30:57

enough the third nucleotide for the triplet,

01:30:59

so we don’t take the

01:31:02

open reading frame, so

01:31:05

we were told in the end fragment the field of the

01:31:08

peptide chain, the end of the polyclinicity

01:31:09

begins here and further to the left we don’t

01:31:12

know what’s going on,

01:31:31

the problem is solved,

01:31:34

how is this problem on I think it seemed

01:31:36

more difficult to you because or not it’s also

01:31:40

like this simple

01:31:57

I think this year, too, many

01:32:00

will come across it because few people have coped with this

01:32:02

task. I

01:32:03

wait literally 5-6 seconds and we

01:32:07

move on to the final tasks, the

01:32:20

tenth type here turns out The difference is

01:32:24

what is written in the fragment of the beginning of a

01:32:27

more efficient chain, we do it in the beginning

01:32:30

in the beginning Just take and we are looking for RNA

01:32:33

let's do it like this

01:32:39

and RNA again now we look

01:32:43

and RNA as

01:32:50

carefully as possible like this 5 4 end

01:32:53

and at

01:32:55

[music]

01:32:56

ggc and at g a a at AC

01:33:04

gc g

01:33:11

and

01:33:14

we need to double-check Yes, here we

01:33:16

get 1 2 3 4 5 6 7

01:33:22

21 22 here

01:33:31

and it doesn’t match

01:33:34

cou

01:33:36

ggc aug

01:33:41

a y a y AC

01:33:45

bcg

01:33:50

5 c A and d forgot at the end

01:33:58

now it’s correct,

01:34:02

great, the

01:34:12

main thing is that the extra letter is not here, you can’t remove it

01:34:15

So, damn, I removed it again,

01:34:19

so once

01:34:22

two,

01:34:26

everything should be right now

01:34:45

So we found RNA with you. Next, what

01:34:48

do we do next? They talk about Met.

01:34:51

It turns out.

01:34:53

Let’s see what to do. Look,

01:34:55

since this is a fact, there is no beginning of the semi-leptile

01:34:57

Chain, that is, we are guided by what we are

01:34:59

guided by here, not by the stop

01:35:00

Don, but by the start codon, because this is the

01:35:02

beginning this should be noticed

01:35:05

there was an end, so we were guided

01:35:07

by the stack here already by a meter,

01:35:11

so

01:35:14

we write that according to the Gene Code table,

01:35:18

according to the encode table, we find

01:35:22

meteonine, that is, aug

01:35:30

We must find it in RNA,

01:35:36

but I

01:35:43

’ll take a smaller font

01:35:50

here, look here, it’s the other way around if

01:35:54

when we said about stopkado we

01:35:55

looked here to the left here we must

01:35:57

look to the right because the synthesis

01:35:59

will begin from here to go further that's why we

01:36:07

look We must have more than

01:36:10

4 amino acids Let's count For

01:36:12

example This time an amino acid is two

01:36:15

amino acids 3 amino acid 4 5

01:36:19

converges perfectly

01:36:24

but we still have to check What if there is

01:36:26

some kind of stop-kadon here? We need a

01:36:28

gene code table and we need to build a

01:36:30

protein.

01:36:36

Let's do this,

01:36:43

wait, maybe there is another auga,

01:36:47

but wait,

01:36:55

now what else should we do? We have two augas here,

01:36:57

which one will we take exactly? of these

01:37:01

Or that one

01:37:20

has been killing you for a month for pranks,

01:37:26

now I pranked you

01:37:28

Yes, there are two augs

01:37:31

So now we have to choose which

01:37:33

one of them

01:37:35

Well, let's say Let's say that we

01:37:38

take the first aug Let

01:37:40

's That is, aug it will be no

01:37:45

no further

01:37:48

after the aug comes GTA GTA

01:37:54

then after Hz

01:38:02

[music]

01:38:03

and this we will have a dash

01:38:09

dash here it stands, that is, but we

01:38:13

can’t take the corners

01:38:14

because here Although this is

01:38:19

an amino acid, there is a

01:38:21

stopkadon here that

01:38:24

completes protein synthesis Therefore, you won’t

01:38:27

get more than 4 amino acids, this

01:38:30

moment disappears therefore we should take the

01:38:33

second one

01:38:35

which is here Let's check it too

01:38:37

What if in general the problem has a solution there

01:38:39

will also be a stack here

01:38:43

so Let's not It will be clear

01:38:47

first of all

01:38:48

it's us now in the draft if we

01:38:50

're filling out anything you should do this draft

01:38:52

further and y and y and this is ours will it be

01:38:57

or

01:39:01

then at AS at AS it is

01:39:14

gcg this is

01:39:18

Alla,

01:39:22

we took it

01:39:34

about it

01:39:38

and we have one letter left at the end,

01:39:42

we don’t seem to grab it, it

01:39:45

turns out so we converge, we

01:39:48

get 5 amino acids in the condition it

01:39:51

says length more than 4 everything converges

01:39:54

it turns out

01:39:57

He is the real start codon,

01:40:04

so the first one is like that, that is, it does

01:40:07

n’t fit, so we have to write the

01:40:10

following phrase

01:40:14

amino acid

01:40:18

no corresponds to

01:40:22

codon

01:40:24

5 aug 3 stroke of such canons 2

01:40:31

but the synthesis

01:40:32

begins with the second of them,

01:40:37

that is, with the one that begins the 7th

01:40:40

nucleotide

01:40:43

I’m here I already calculated it myself 7 nucleotide

01:40:46

367

01:40:49

and already the first letter from the first letter

01:40:51

we look at the 7 nucleotide the

01:40:57

first five stroke begins the

01:41:02

three stroke that starts

01:41:05

from the second nucleotide

01:41:10

does not fit

01:41:13

because the policy breaks off

01:41:19

in the reading frame

01:41:22

there is a stop codon

01:41:25

this should not be so, that’s

01:41:30

basically all that is

01:41:33

needed we should mark the open reading frame the

01:41:37

open reading frame will

01:41:40

now be marked

01:41:49

and the last one Gni grab

01:41:57

Yes, damn it, I don’t get it,

01:42:04

this is the open reading frame

01:42:18

the problem has been solved

01:42:24

here, the main thing is to explain Yes, that’s why

01:42:26

we take exactly this one, as if this is the meaning of the

01:42:30

tasks,

01:42:34

that is, you are now really if

01:42:36

so divide these tasks into levels Yes, here’s the

01:42:39

first, second, third, and in these

01:42:41

two lessons we’ve reached the entire 10th level. That

01:42:43

is, you’re already like if you’re

01:42:46

still solving all your homework, it will be solved as

01:42:49

correctly as possible the first time, it

01:42:52

’ll be a blast, it’s clear that

01:42:57

nothing ever

01:42:58

works out perfectly the first time, so you have to

01:43:00

repeat these tasks, don’t forget about them, you

01:43:03

can even open the Rokhlav collection

01:43:07

and start all these tasks there, you can

01:43:09

take all 28 lines there to solve it because

01:43:13

it can already be done. It’s

01:43:15

clear in

01:43:17

some variant, but

01:43:20

If you understand all these types of

01:43:23

problems here, you really understand

01:43:24

the propalindrome, is there even a way to figure it out? Well, of

01:43:27

course, we’ll look at this in a separate

01:43:29

lesson,

01:43:31

so in principle, this is the

01:43:38

whole

01:43:42

question about homework and it’s better to

01:43:45

solve it yourself, or you can follow the example from the notes

01:43:49

if it’s really difficult for you to solve these

01:43:52

problems physically, you can look

01:43:54

somewhere, try to decide for yourself,

01:43:57

let it be wrong Well, just so

01:43:59

that you can decide for yourself. So you will

01:44:01

understand better. It’s really possible to make a real

01:44:04

gap,

01:44:06

but not of course not a list;

01:44:10

all the

01:44:12

stickers are already needed;

01:44:21

let’s give some of you stickers as a gift;

01:44:29

write whatever you want. in the chat, I’ll give you a

01:44:33

gift,

01:44:41

people immediately appeared who didn’t

01:44:44

write anything so I woke up

01:44:50

I can really prank you just

01:44:52

click end the broadcast of stickers no

01:44:56

okay not tough and not such a tough

01:44:59

person you are at the webinar today so

01:45:04

that’s it

01:45:07

Let’s hop

01:45:09

Yulia Kastyrskaya wins Yulia

01:45:12

Kostylskaya may not write in a personal message

01:45:15

and within 24 hours I will give you

01:45:20

everything

01:45:21

Yuliya Kostylskaya is waiting for you

01:45:24

everyone else bye bye

01:45:27

See you

01:45:29

on Friday because the next two Webs

01:45:32

put this on the prom too

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