Download "Генетика. Двойное сцепление с полом и летальный ген | ЕГЭ 2022"
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252Fa-8yHa56q7M%252Fmaxresdefault.jpg&w=1139&q=75)
Please wait. We're preparing links for easy ad-free video watching and downloading.
Similar videos from our catalog
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252FimSXd5rPtH8%252Fmaxresdefault.jpg&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252Fum-ArxlP_OE%252Fmaxresdefault.jpg&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi_webp%252F1ljDx0IoYO8%252Fmaxresdefault.webp&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi_webp%252F2IiN3eAZUFY%252Fmaxresdefault.webp&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi_webp%252F973EH4LioD0%252Fmaxresdefault.webp&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi_webp%252FDwlxIXzj2fE%252Fmaxresdefault.webp&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi_webp%252FDzlT0j7zUUs%252Fmaxresdefault.webp&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi_webp%252FP4K_tPZCgKA%252Fmaxresdefault.webp&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi_webp%252FQeCLLTO31gE%252Fmaxresdefault.webp&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi_webp%252F_3q-aYjG3DU%252Fmaxresdefault.webp&w=1139&q=75)
Video tags
Subtitles
Russian
00:00:01
Hello everyone guys today we have our00:00:04
last lesson recording everything else but00:00:06
online and I really miss you00:00:09
tomorrow we will meet as usual00:00:12
well now let’s practice solving00:00:15
problems the first task let’s look at it00:00:17
and try to understand what00:00:19
type of00:00:21
person it is between the alleles of the gene absence of00:00:24
sweat glands and hemophilia type A, crossing over00:00:26
occurs,00:00:29
hemophilia is linked to the x chromosome,00:00:33
since with Dima, between the gene where the00:00:36
cartoon is and the absence of sweat glands,00:00:38
crossing over occurs,00:00:40
this tells us that get the00:00:42
absence of sweat glands is also located00:00:45
on the x chromosome, and then it’s also beautiful it’s00:00:47
true that it won’t happen, too, let’s not forget about him,00:00:49
let’s not forget about the00:00:53
disease not being specified, a00:00:57
woman, a woman who does not have the specified00:01:00
diseases,00:01:02
whose father had hemophilia, whose mother had an00:01:06
annual diploma, the absence of sweat00:01:08
glands00:01:10
means the mother was de homozygous00:01:17
de homozygous and she had the absence of00:01:20
sweat glands, which means she was and the00:01:22
little one is small and homozygous, but00:01:26
she didn’t have hemophilia, so she was already00:01:28
big, the woman was00:01:31
healthy in both respects, the father of this00:01:35
woman, well, look, Matilda is00:01:37
so much like that x why will this woman be a00:01:40
little h big, the father of this woman00:01:43
suffered from hemophilia then yes, he was x and00:01:47
nothing much is said that he00:01:50
had atrophy, lack of sweat glands,00:01:53
so of00:01:56
course, he gives his daughter exactly the00:01:59
x chromosome, otherwise it’s not a daughter anymore,00:02:02
but a big one, and00:02:06
so it00:02:09
turns out this woman00:02:11
marries a man not having these00:02:13
diseases x and the00:02:16
big one is already big y let's see what they00:02:20
will have what kind of gametes x the small one is00:02:23
big x and the big one is even a little bit will00:02:27
crossing over happen here00:02:30
before crossing over will happen in00:02:33
any case but will there be changes yes00:02:36
if you change even the big one00:02:38
small in some places, the gametes will00:02:40
fundamentally change and the00:02:44
children will become like this, the gametes that00:02:46
were formed as a result of a beautiful lake,00:02:48
but these two gametes were formed as a00:02:51
result of the sci initial linkage of genes00:02:55
born in this marriage, well, let's00:02:57
find the shooting range of their offspring f100:03:01
we will get00:03:04
forgiveness for more gametes already gametes the father was not00:03:07
found because of the big big y now00:03:11
let's find their offspring I should have00:03:13
written here but I don't have enough space00:03:16
so I write here a little higher and the00:03:19
braid is small h big x big as much as big x as00:03:23
small as big as big y00:03:28
x and big as much as small x a00:03:32
big-big and00:03:33
x and big is even small y these children00:03:38
were the result of linked00:03:40
inheritance, the00:03:41
remaining children00:03:44
were the result of beautiful faith00:03:46
will be numerically less than00:03:59
a night, about numerically they don’t ask here, we will00:04:02
definitely indicate for all children00:04:07
the tasks where the coupling with gender occurs, we will00:04:10
definitely indicate gender and00:04:13
sign the phenotype00:04:16
let’s write for ourselves a long time ago it’s already given00:04:20
x and big is00:04:23
without the00:04:25
absence of sweat glands x small00:04:29
is the absence of sweat glands xh large00:04:32
without hemophilia x h small this is00:04:36
hemophilia then here will be00:04:39
without the absence of sweat glands and without00:04:43
hemophilia00:04:45
here will be the absence of sweat glands and00:04:48
there will be no Dima fillet00:04:51
here00:04:53
without the absence of sweat glands and without00:04:57
hemophilia without the absence of sweat glands00:05:00
but with hemophilia00:05:02
healthy in all respects00:05:04
[music]00:05:07
from the absence of sweat glands from00:05:10
hemophilia unlucky does not sweat and the00:05:13
blood does not clot00:05:16
[music]00:05:18
here without the absence of sweat glands and without00:05:21
hemophilia, this is a healthy boy,00:05:24
completely00:05:28
draw up a scheme for solving the problem,00:05:31
indicate the genotypes and phenotypes of the parents,00:05:35
otherwise this one needs to write a healthy phenotype for the Russian Federation,00:05:38
wait even further further,00:05:40
continuation of those born in this marriage,00:05:43
Monna is homozygous, I am a healthy daughter, still00:05:45
here, monokuma, a second healthy daughter,00:05:48
right here this is, for example, a healthy daughter, and for us00:05:52
it is homozygous for one trait and00:05:55
hydra zygote in another way, where else is it like this,00:05:59
this00:06:02
is heterozygotes to die homozygote, everything00:06:06
fits these two,00:06:09
she comes out healthy men, they have a00:06:11
child, Dima helix, so that00:06:14
they have a child, hemophilia she must00:06:16
be a carrier of the hemophilia gene, so out of00:06:19
these two daughters we will choose which one we00:06:22
will choose the second because she has the00:06:24
hemophilia gene, which means we take this girl and00:06:35
marry her to a healthy man00:06:55
will this girl have a00:06:58
singular correction, yes it will happen but00:07:01
gametes will not fundamentally change if00:07:04
you change as much as a big one as a small one in some00:07:07
places you will still have a big one00:07:09
and a big one as much as a small one and a00:07:13
big-big one00:07:15
I'm a big one as much as a small one 2 identical two00:07:18
times identical gametes we do not write00:07:20
so here we do00:07:22
not indicate any crossover additional comets we immediately00:07:25
write the descendants we get x00:07:29
and the big one is already small x and the00:07:33
big one is even00:07:38
big x and the big one is even small y00:07:59
without hemophilia without00:08:03
absence of hemophilia healthy in00:08:06
all respects and00:08:08
here, too, is it00:08:11
possible in the first marriage to give birth to a child suffering from00:08:14
this00:08:17
disease, that00:08:20
is, two in the first marriage we00:08:23
have a boy who had the absence of00:08:26
sweat glands and hemophilia, both at the same time,00:08:31
why explain the answer here they want to00:08:34
tell you they want you to tell them that00:08:38
ghetto like 2 genes are linked and00:08:41
crossing over occurs,00:08:42
crossing over leads to the appearance of00:08:45
such gametes and when this gamete00:08:47
is combined with a game we get an00:08:50
organism we get a boy with this00:08:53
genotype like this further in00:08:56
humans alleles of chicken genes blindness,00:08:59
aka night acid, oh number and red-00:09:04
green tanto nismo on one chromosome,00:09:06
so we have color blindness, night00:09:09
blindness,00:09:10
they are on the same chromosome, we00:09:13
know that color blindness is linked to the00:09:15
x-chromosome, ara blindness, and just didn’t00:09:17
know one chromosome of that blindness is also on the same00:09:20
one chromosome and there will be blindness00:09:23
without the specified diseases, a00:09:26
woman00:09:30
whose mother had night blindness and00:09:33
color blindness, that is, her mother had both00:09:37
night blindness and00:09:40
color blindness,00:09:42
was to write and00:09:43
[music] let’s00:09:46
write immediately before for00:09:49
color blindness is00:09:53
always a recessive trait and00:09:58
night blindness should be a recessive00:10:01
trait00:10:07
because that this is painful, it is00:10:10
not very common, recessive00:10:12
traits are less common than00:10:14
dominant, so we chose00:10:16
recessive,00:10:18
look, this woman’s mother00:10:21
was like that, so she00:10:23
only gets one type of gametes and she00:10:26
can only give her daughter a00:10:27
small d small and00:10:30
which one her father was a father who had the00:10:32
indicated diseases, that is, he was like00:10:35
this,00:10:36
she married a man who was colorblind, it is not00:10:40
said that he had night00:10:43
blindness, so we will assume that he does00:10:45
not have it, neither he is colorblind,00:10:48
the man has gametes and to the kaida itself small00:10:53
x big d big xa big you are00:10:56
small and y in the offspring we get x00:11:00
and small where small x and big and00:11:04
small00:11:05
x and small you are small y00:11:09
xa highest d big and x a big de00:11:13
small not last xa big for00:11:16
big y be sure to indicate gender gender00:11:19
should always be obtained 50 to 50 5000:11:22
percent girls 50 percent boys00:11:25
and this girl will be00:11:29
healthy due to night blindness but00:11:34
with color blindness00:11:36
this girl is healthy due to blindness00:11:40
and00:11:41
from I apologize he has night blindness00:11:43
and00:11:44
color blindness00:11:47
this girl00:11:49
it was a boy this girl is healthy and00:11:53
healthy and a boy healthy healthy00:11:58
born here was crossing over00:12:03
born in this marriage daughter colorblind00:12:06
married a man00:12:08
theorist look theoretically here it would be possible to00:12:11
carry out crossing over but00:12:14
do we need it00:12:20
here they don’t say anything about it00:12:23
here too we don’t have this information00:12:26
look further born in this marriage00:12:28
definitely colorblind married a man who does00:12:31
not have these diseases,00:12:34
we have only one colorblind daughter so far, let’s00:12:37
assume that it’s her00:12:39
now,00:12:41
you know how we’ll find out whether we’ll paint a00:12:43
guyver here or not, we’ll see further if00:12:46
they give birth to some kind of00:12:48
child, which is impossible00:12:50
get from this girl here means we will00:12:53
need to get00:12:54
more crossover gametes and then we will get00:12:58
four more children here and we will00:12:59
take another girl but let's00:13:01
try with this one00:13:03
x a small00:13:07
small d small x a big you00:13:11
little00:13:16
mouse married a healthy man by00:13:19
all indications,00:13:27
a man is00:13:28
big porridge y and00:13:32
in the offspring we get x and big and00:13:40
small you are small x and00:13:44
big00:13:46
like I have here in your why I had to do00:13:49
things I’m here00:13:57
x and big and small x and big00:14:00
yes big00:14:05
x small you are small y x big for00:14:12
small y half cash register00:14:19
Now let's see in this family a00:14:21
child was born with night blindness and color blindness, now00:14:30
we have a child with night blindness00:14:33
and color blindness,00:14:38
so draw up a scheme for solving the problem,00:14:44
indicate the genotypes and phenotypes of the parents, indicate the00:14:46
genotype, we will also indicate the phenotype,00:14:49
healthy woman,00:14:51
healthy man, according to the first sign,00:14:54
suffering from color blindness,00:14:57
woman here is a girl healthy butt the00:15:01
first sign suffering from color blindness00:15:04
man00:15:07
[music]00:15:10
healthy healthy00:15:14
genotypes and phenotypes sex of possible00:15:16
offspring here everything here we will also indicate00:15:18
the phenotypes healthy boy color blind00:15:22
healthy healthy healthy healthy all00:15:26
this was indicated is it00:15:28
possible in a first marriage to have a00:15:30
child with night blindness and00:15:33
no color blindness00:15:41
here is such a moment, if there00:15:44
was no beautiful lake, then we say that no, it’s00:15:49
impossible, the second option, we solve the00:15:53
second option, if there was a crossing over,00:15:55
because we didn’t have any clues about the00:15:58
presence of a crossing over,00:16:01
here we do a00:16:03
small d large and x and a00:16:06
large d small and then we have there will be00:16:09
four more descendants and a braid small d00:16:13
big x a big and small xa00:16:16
small d big y00:16:21
xa big for little x big and00:16:24
small and x a big00:16:29
you small y00:16:35
indicate gender00:16:39
phenotypes this is healthy and healthy with00:16:43
chicken slip boy with night blindness00:16:46
and00:16:47
healthy on the basis of color blindness00:16:50
the boy is healthy on the basis of night00:16:53
blindness but with color blindness00:16:58
healthy with color blindness is00:16:59
also not a girl and a boy is a healthy00:17:02
organism, but in this case there is a00:17:04
possibility that a child will appear who00:17:07
suffers only from blindness but he00:17:11
will not have color blindness only on the condition that00:17:13
there was a Kassin of the year so this and you need to write down,00:17:16
I recommend doing this task in the following00:17:19
way: write the first00:17:21
option without painting Hoover and write the00:17:24
first option if there was no crossing over, the00:17:26
second option if00:17:29
there was crossing over and00:17:30
then solve the second crossing scheme and00:17:33
then in the answer indicate that before the00:17:35
possible birth of a child in00:17:39
case there was a color ambition00:17:48
further in a person, they or the genes for00:17:52
optic nerve atrophy and the00:17:54
red-green water of the body are00:17:56
on the same chromosome, but the same thing begins again, the00:17:58
large x is atrophy of the00:18:02
photos, I ask for the reason, on the contrary, it’s great00:18:05
for this sign, and the small x is00:18:08
atrophy00:18:09
xb large, this is color blindness00:18:16
xb small00:18:20
color blindness a00:18:21
x would be big big it is healthy according to00:18:26
this sign a00:18:30
woman a00:18:34
woman whose mother had00:18:37
optic atrophy and color blindness that is,00:18:39
her mother had a small00:18:44
small d small and a small d00:18:47
small00:18:59
accordingly this woman gives this00:19:02
girl only this such gametes the00:19:06
father did not have the indicated diseases, that00:19:09
is, he was a big d big and scream00:19:12
means the girls he gave big to big00:19:15
she married a colorblind man a00:19:18
big d small y let's write the gametes x a00:19:23
small you are small x a big d00:19:26
big x big and small00:19:30
y00:19:33
offspring x can it happen here00:19:36
we paint coffee can it be written is this a task00:19:39
again no so let’s do it this way let’s00:19:42
formalize it directly this is the first crossing00:19:47
without a beautiful faith or y so if00:19:50
crossing over doesn’t happen00:19:55
doesn’t00:19:56
happen00:20:01
we get x and small00:20:06
yes small x big and small00:20:10
x00:20:12
small d small y00:20:16
x a large yes I am large x large and00:20:20
small x a large d large y we00:20:29
all indicate the gender and00:20:35
sign the phenotypes00:20:39
healthy with color blindness00:20:42
girl boy00:20:45
with atrophy and with color blindness00:20:52
completely healthy girl and00:20:56
completely healthy boy00:20:59
born in this marriage daughter color blind00:21:02
here she married a man the00:21:06
second crossing did00:21:23
not have the specified diseases the man00:21:26
was healthy00:21:31
we indicate the gametes00:21:37
we find the offspring00:21:42
I swapped them here but it’s00:21:45
not important you can’t change places like00:21:48
this you could leave it00:21:58
here for sure crossing over it will00:22:01
happen but it won’t be visible here00:22:04
what they write in this family, a00:22:06
child was born with optic nerve atrophy and00:22:09
before a nismo the00:22:11
girl is00:22:13
healthy healthy the00:22:16
boy with atrophy and with color blindness00:22:21
the girl is healthy and healthy and the boy is00:22:26
healthy with color blindness00:22:30
so indeed there is one such00:22:33
child was born everything coincides make a00:22:37
diagram for solving the problem00:22:39
indicate their genotypes and phenotypes of the parents00:22:42
phenotypes must be signed before signing healthy00:22:45
healthy healthy colorblind00:22:49
girl healthy in atrophy but with00:22:53
colorblindness boy man more precisely healthy healthy00:22:56
in children everyone indicated whether it is possible in the first00:23:00
marriage to have a child with00:23:02
optic nerve atrophy and no00:23:04
colorblindness and here we will write00:23:07
the second crossing not 2 crossings the00:23:10
second option00:23:11
here is the first we write like this the first crossing00:23:16
if00:23:18
crossing over occurs00:23:25
if crossing over occurs and we indicate00:23:29
the woman x and small yes small x00:23:34
big d more in general what are we00:23:36
writing here we are rewriting everything the same only00:23:39
in addition we are writing two more types of gametes00:23:42
this is how it was in the previous problem00:23:44
we additionally write two types of gametes: a00:23:47
small d large and x a large for00:23:51
small they will be beautifully correct and a00:23:53
whole series of descendants that are formed as00:23:56
a result of crossing over and there you00:23:58
will get a child with optic00:24:00
nerve atrophy with the absence of color blindness, you00:24:03
answer like this and write down that if00:24:04
crossing over occurs, then it is00:24:07
likely that there is a possibility of the birth of00:24:10
such a child with trophies but without00:24:12
color blindness, the00:24:13
crossing scheme will be similar to how00:24:16
we indicated here,00:24:19
so00:24:22
on in humans, the alleles of the genes for night00:24:25
blindness and hemophilia include hemophilia00:24:29
type A and are located on the same chromosome,00:24:31
well here again begins sex large is00:24:34
healthy for night blindness x00:24:36
small is blindness00:24:39
xh large is healthy xh small is00:24:43
hemophilia monk homozygous and a woman00:24:47
does not have the indicated diseases, that is, a00:24:49
woman00:24:51
she is healthy mono homozygous for one00:24:57
trait she was00:25:00
homozygous for another heterozygous00:25:03
mother whose mother had night blindness00:25:08
so the mother was starting to be blind00:25:14
about her00:25:17
hemophilia, nothing was said, we’ll do it like00:25:20
this, and the00:25:21
father did not have the indicated diseases,00:25:24
but he was, and the big one is big, which means00:25:29
the father gave it to her, and the big one, even the00:25:31
small one, the mother suffered from night blindness,00:25:35
she will give it to her exactly, and the small one, and what kind of one00:25:38
she is will give the second one as much as a big second00:25:41
gene as much as a big only a small one right there00:25:43
can be as much as a small one here maybe a00:25:45
very big one is needed for this woman00:25:47
to turn out homozygous she is already00:25:51
heterozygous by the first sign then00:25:54
by the second she should be homozygous00:25:57
now now we have one mass here but00:25:59
if we put a small one here,00:26:02
then she would be heterozygous, which does not00:26:04
coincide with our conditions, her00:26:07
mother married a man with00:26:11
hemophilia,00:26:12
that is, everything is fine here after rubbing her00:26:14
legs mafia trouble,00:26:17
we’ll find comets here and a small h00:26:21
big-big-big00:26:23
you’ll be here we paint the year, maybe00:26:26
it will be, but in any case nothing00:26:27
will change00:26:31
descendants x and the small one is already big x and the00:26:37
big one is already small here x is small00:26:40
h big y here x and the big one is already big00:26:44
x and the big one is even small and00:26:48
like that,00:26:57
who can say it’s great there thank you00:27:00
here we will indicate the gender we will00:27:03
sign the genotypes healthy00:27:06
healthy with blindness but healthy00:27:10
healthy healthy healthy healthy00:27:14
here good industrial Dima chemists no super00:27:18
blindness with chicken blindness is starting blindness you00:27:21
can still live in peace those00:27:24
born in this marriage a healthy daughter00:27:27
married to a man who does not have these00:27:29
diseases here there are00:27:31
2 healthy daughters, you need to choose one in00:27:35
this family, a child was born with night00:27:38
blindness and hemophilia, the00:27:40
girl needs to be a carrier of the gene for both00:27:43
night blindness and hemophilia, this one has00:27:47
only hemophilia, and this one has the gene for both00:27:50
blindness and hemophilia, so00:27:52
this girl is suitable we choose this girl by00:27:55
you the first is you crossing00:27:58
this is the second crossing00:28:03
we write here x and the little one attributed00:28:07
why the genotype of this girl00:28:09
she marries a healthy man00:28:16
gametes00:28:45
and indicate the gender00:28:50
girl boy girl boy00:28:54
indicate phenotypes00:28:57
healthy healthy with blindness but great00:29:02
great I’m healthy and a boy00:29:10
so the boy turns out00:29:14
healthy with hemophilia, but we needed a00:29:16
boy with night blindness, Kim and Philly, this is00:29:19
where we can make a krasin grief00:29:22
and the gametes will really change here00:29:25
then there will be a small00:29:31
small h small 1 crossover on I00:29:34
gamete and a big one even big one00:29:38
crossover on I gamete these crossover00:29:41
gametes now let's use them we get00:29:43
x a small h small x a00:29:47
big-big00:29:49
x small h small y00:29:52
xa big as big x big it's00:29:56
big00:29:59
so and finally we got a child00:30:03
suffering from blindness and hemophilia, let's show00:30:07
them all gender,00:30:14
sign the phenotypes everywhere and00:30:20
voila, everything is ready, indicate the genotypes and00:30:24
phenotypes of the parents for the offspring,00:30:27
explain the birth of a child suffering from these two00:30:29
diseases in a family of healthy00:30:31
parents, the fact is that a00:30:35
beautiful number occurred, this led to the00:30:38
appearance of such a gamete00:30:40
x small h small and as a00:30:44
result when these are gametes combined with the00:30:47
y-chromosome,00:30:48
the result was a boy from Egypt with such a00:30:51
genotype and with such a phenotype, so the00:30:54
next task in ducks is the crest trait is00:30:58
tasty and the perennial quality is a dream and and are00:31:01
not linked, well, like everything is fine in a00:31:05
homozygous dominant state, and the00:31:07
big, big crest gene tasty causes00:31:11
death of embryos in this crest gene is tasty00:31:14
then big and small is something like whether there00:31:17
is a crest or not, it is not yet00:31:21
entirely clear which of them is dominant00:31:23
which is recessive plumage b-large and00:31:27
would be small that we have00:31:31
crossbreeding grab them00:31:35
with normal plumage ducks and crested00:31:40
drakes00:31:44
with normal00:31:46
plumage, you know the date, such00:31:49
drakes are vodka males, some of the offspring00:31:53
turned out00:31:55
without a tuft00:31:57
and00:31:58
with silky plumage of00:32:02
silky pyramids, look at this00:32:05
sign does not appear in the parents, but00:32:09
appears in the offspring, since this00:32:11
trait was suppressed here, then00:32:14
the holidays will be red sin and that means there is a00:32:16
small one without a crest, and a small one is00:32:19
silky plumage, then and a big one00:32:23
is the presence of a crest, but only in a00:32:25
heterozygous state,00:32:27
because in a homozygous state, a00:32:30
crest may develop in them but does not00:32:32
die, both are large, this is normal and00:32:35
the plumage now00:32:39
we can write them the genotypes are crested,00:32:42
that’s for sure, but the big one is small, why00:32:45
can’t we choose the big one, and the big one,00:32:47
because they are not dead, normal, big ones,00:32:51
since we have silky ones in our offspring, it00:32:53
would be big, it would be00:32:55
small, by the way, with negligence, the same thing is the00:32:57
second reason why we are writing00:32:59
here the small one this is because she00:33:02
is born without crests and the little one is00:33:05
small, that is, she was the carrier of the00:33:07
gene without a crest tasty here is the same thing and00:33:11
oh no we get the Punnett lattice00:33:26
thalac00:33:31
here we need the Punnett lattice we will write00:33:36
it00:33:44
more accurately we will draw it00:34:19
ok but we guess everything that should00:34:22
turn out the ratio day 33 k1 but there00:34:26
will be death and you may already00:34:29
remember in the last lesson we said00:34:31
when such a genotype causes00:34:34
death00:34:36
what should be the ratio what but00:34:40
usually it is typical there the number is so00:34:42
specific00:35:15
and now cross out all of these00:35:19
because we are not00:35:22
signing the dead phenotypes this species crested00:35:26
normal crested normally00:35:31
crested normal crested silky00:35:38
without crest00:35:41
normal00:35:47
approach and normal00:35:51
crested00:35:53
silky00:35:55
without crest00:35:57
normal and without crest choco eats and so00:36:09
now00:36:11
by crossing the resulting first00:36:14
generations of crested ducks with normal00:36:18
plumage and00:36:20
[music]00:36:23
drakes with the same genotype it turned out00:36:26
2 phenotypic groups so when00:36:28
crossing in the first generation00:36:31
crested normal00:36:33
homozygous00:36:35
paths we take 2 crossings brem crested crested00:36:40
then only large small00:36:43
normax normal plumage homozygotes00:36:47
they are here directly task of writing00:36:49
a duck and00:36:51
we cross them and00:36:54
series them with the same genotype of00:36:58
gametes a large large00:37:02
small large a big b big00:37:05
small big in the offspring we get a00:37:09
big a big big big a big small b big00:37:13
-big a big small big00:37:17
big I00:37:21
small small big big let's00:37:23
sign the phenotype for00:37:27
him, this dead one, don't sign anything00:37:30
better, the phenotypes for the one who00:37:33
dies in the first place are empty waste of00:37:35
time secondly, he died, we don’t even00:37:37
take it into account in the splitting,00:37:39
this will be crested normal crested00:37:44
normal00:37:46
without crest00:37:49
normal and00:37:52
it was said that it was said that00:37:56
we got 2 Finns typical groups00:37:58
we really get crested normal00:37:59
and without crests normal00:38:01
draw up a scheme for solving problems determine the00:38:04
genotypes of the parent individuals00:38:06
genotypes and here it would also be worth signing00:38:09
the phenotypes it was worth the00:38:12
genotype parental genotypes and00:38:14
phenotypes of the resulting offspring here from the00:38:17
first crossing we have a00:38:18
Punnett grid which has both genotypes and00:38:20
phenotypes again do not duplicate this in the00:38:23
second squeak with the second crossing this is00:38:25
all indicated determine and explain the00:38:28
phenotypic split in the first and the00:38:30
second crossing, you say, since a00:38:33
big, big one causes death, then in the00:38:35
first crossing, wasps without00:38:37
such a genotype die with such with such and with00:38:39
such about the author and a split of 6 is obtained,00:38:43
three to two to one; in the second00:38:47
crossing, the organism with00:38:50
such a genotype dies and we get00:38:52
a splitting of two to one, such a00:38:55
problem is interesting00:38:57
and the last problem is the repetition00:39:01
in Drosophila of dominant genes00:39:04
that control the gray color of the body and00:39:07
the development of bristles, that is, they immediately00:39:11
tell you that gray paint is a00:39:14
dominant gray color and00:39:18
bristles are00:39:20
also a dominant feature,00:39:23
then what kind of body is small? black00:39:26
body00:39:28
b there are no small bristles and00:39:31
it is said that these two genes are located on the00:39:35
same chromosome,00:39:37
what does this tell us that these two00:39:40
genes are inherited linked, the00:39:43
recessive alleles of these genes that00:39:46
cause the00:39:48
black color of the body and the absence of bristles00:39:51
are located on the other to the malaga chn and00:39:53
chromosome00:39:55
what genotype and phenotype of00:39:58
the offspring and in what00:40:00
percentage can be expected from a cross00:40:03
where heterozygous sulfur females have00:40:07
developed bristles sulfur00:40:10
bristles00:40:12
with a black male without bristles00:40:18
black I00:40:20
ask you to forgive them they should be00:40:24
without bristles00:40:26
without bristles, provided that the female has 5000:40:31
percent of gametes were a crossover, well,00:40:34
let's look here at four types of00:40:36
gametes,00:40:38
how linkage will be indicated here, you00:40:42
need to carefully read00:40:44
the first and second sentences and you00:40:47
will understand everything, now I will indicate all possible00:40:50
types of gametes and we will understand what they are00:40:52
linked, re-read again in the morning00:40:55
production dominant genes00:40:58
controlling the gray color of the development of the00:41:00
bristles are localized in one chromosome,00:41:02
that is, initially it is the dominant genes00:41:05
in one chromosome, that is, they00:41:07
are linked; the00:41:09
recessive alleles of these genes00:41:11
causing the black color are00:41:13
absent are located in another00:41:16
chromosome,00:41:18
here is another chromosome, and therefore we can00:41:22
conclude that the linked ones are a00:41:25
large big I'm small would be small00:41:28
and these two00:41:30
were formed as a result of Hoover's beauty00:41:34
here and we will be fifty percent00:41:37
here 1 meter and small would be small in00:41:41
f1 we get first we will find the individuals00:41:44
that were the result of00:41:45
linked inheritance and big00:41:48
small big would be small and this is00:41:50
a small small boo small would be00:41:53
a little now let's find those that00:41:55
resulted from Krasin Goda Rao00:41:57
big small boo small would be00:41:59
small I would be small small would be00:42:02
big would be a little let00:42:04
's sign their phenotypes sulfur with bristles00:42:10
black without bristles now with crossovers00:42:14
children00:42:15
gray without bristles and00:42:19
black with00:42:21
bristles it is said that 50 percent the gametes00:42:25
were crossover, that is, in fact, it00:42:27
turned out to be exactly equal, half will be00:42:30
linked half crossover, this is00:42:32
very rare, but it happens this way,00:42:35
these will be fifty percent of00:42:38
these, although they are not crossover,00:42:40
although usually we have less cursor ones,00:42:42
but what do they want? by00:42:45
genotype and phenotype,00:42:48
what percentage of their offspring00:42:49
can be expected from00:42:52
such a crossing,00:42:54
we will get bristles like this00:43:02
with bristles00:43:06
with gray bristles with bristles, there will be a quarter of these00:43:11
because these will also remain 5000:43:13
percent for reading they will be 2500:43:15
percent like this00:43:19
black without bristles there are00:43:22
also 25 percent of them, the remaining00:43:27
crossover of00:43:31
25 percent,00:43:37
also 25 percent, be sure to indicate00:43:41
that these two were formed as a00:43:44
result of linkage,00:43:50
linkage of genes, and big for big and00:43:55
small would be small, but these00:44:00
two genotypes and phenotypes were formed00:44:04
as a result of a00:44:07
beautiful lake,00:44:13
this is the problem I hope Everything was00:44:17
clear to you, if you have any questions,00:44:18
write in the general group, I will00:44:21
definitely be in touch to answer your00:44:23
questions about the problem, that’s all for now00:44:27
until tomorrowDescription:
► Вход в личный кабинет - https://biosch.ru ► Инстаграм: https://www.facebook.com/unsupportedbrowser ► Группа ВК: https://vk.com/biologyalena ► Подпишись на канал: https://www.youtube.com/c/%D0%90%D0%BB%D0%B5%D0%BD%D0%B0%D0%9A%D0%BB%D0%B8%D0%BC%D0%B5%D0%BD%D0%BA%D0%BE?sub_confirmation=1
Preparing download options
* — If the video is playing in a new tab, go to it, then right-click on the video and select "Save video as..."
** — Link intended for online playback in specialized players
Questions about downloading video
How can I download "Генетика. Двойное сцепление с полом и летальный ген | ЕГЭ 2022" video?
http://unidownloader.com/ website is the best way to download a video or a separate audio track if you want to do without installing programs and extensions.
The UDL Helper extension is a convenient button that is seamlessly integrated into YouTube, Instagram and OK.ru sites for fast content download.
UDL Client program (for Windows) is the most powerful solution that supports more than 900 websites, social networks and video hosting sites, as well as any video quality that is available in the source.
UDL Lite is a really convenient way to access a website from your mobile device. With its help, you can easily download videos directly to your smartphone.
Which format of "Генетика. Двойное сцепление с полом и летальный ген | ЕГЭ 2022" video should I choose?
The best quality formats are FullHD (1080p), 2K (1440p), 4K (2160p) and 8K (4320p). The higher the resolution of your screen, the higher the video quality should be. However, there are other factors to consider: download speed, amount of free space, and device performance during playback.
Why does my computer freeze when loading a "Генетика. Двойное сцепление с полом и летальный ген | ЕГЭ 2022" video?
The browser/computer should not freeze completely! If this happens, please report it with a link to the video. Sometimes videos cannot be downloaded directly in a suitable format, so we have added the ability to convert the file to the desired format. In some cases, this process may actively use computer resources.
How can I download "Генетика. Двойное сцепление с полом и летальный ген | ЕГЭ 2022" video to my phone?
You can download a video to your smartphone using the website or the PWA application UDL Lite. It is also possible to send a download link via QR code using the UDL Helper extension.
How can I download an audio track (music) to MP3 "Генетика. Двойное сцепление с полом и летальный ген | ЕГЭ 2022"?
The most convenient way is to use the UDL Client program, which supports converting video to MP3 format. In some cases, MP3 can also be downloaded through the UDL Helper extension.
How can I save a frame from a video "Генетика. Двойное сцепление с полом и летальный ген | ЕГЭ 2022"?
This feature is available in the UDL Helper extension. Make sure that "Show the video snapshot button" is checked in the settings. A camera icon should appear in the lower right corner of the player to the left of the "Settings" icon. When you click on it, the current frame from the video will be saved to your computer in JPEG format.
What's the price of all this stuff?
It costs nothing. Our services are absolutely free for all users. There are no PRO subscriptions, no restrictions on the number or maximum length of downloaded videos.