How to solve pedigree probability problems

[@GeneticsLessons]

Here is a link to my Playlist "Pedigree probability problems": kzbin.info/aero/PLzBiAPKi8vOPHfZwgAiq0skc8CFZ7aa5L

[@memeal6328]

Great explanation. I just had to get used a bit to the accent :) Thank you so much that really helped a lot.

[@casualcasual1234]

Hello thanks a lot for the tutorial :) For this question, can I think like this: III-1, III-2 both are heterozygous since IV-1 is homozygous recessive; IV-4 has dominant phenotype, and to obtain child with trait, IV-4 must be heterozygous, hence 2/3 IV-5 is heterozygous since III-5 is homozygous recessive but IV-5 has dominant phenotype, hence 1/1 IV-4 x IV-5 give 1/4 of trait so 2/3 * 1/1 * 1/4 = 1/6?

[@nainasinghbanyal3598]

It is this way only 🙂

[@GeneticsLessons]

Watch my video about mnemonic techniques: How to memorize abstract numbers kzbin.info/www/bejne/eH26fGSDpapmnrs

[@ButterflySimmer]

Very helpful video. These are the questions that you just need to break it down and to complete punnet square.

[@GeneticsLessons]

Glad it was helpful!

[@popyenihaikali5489]

beautifully explained! thank u so much Nikolay

[@GeneticsLessons]

You're welcome! Thanks for watching.

[@satinderjit4]

Thank you for this. I was struggling and it finally clicked!

[@GeneticsLessons]

Glad it helped!

[@andykilker8960]

Thanks for the video! Very informative and easy to follow along.

[@GeneticsLessons]

You are welcome! Thanks for watching.

[@memecartoontv1328]

In generation {5-6} will be heterozygous/homozygous dominant Aa/AA .So take it with both.

[@GeneticsLessons]

In which generation? III? 5 is obligate aa and 6 is AA.

[@tonyd7829]

Lol, that arnold motivational reference tho 👌

[@Mike-gf1bl]

The kids grand parents are cousins lol

[@GeneticsLessons]

In Iran such consanguineous marriages are norm - in order to keep the land within a family. And this is a norm in top reach families of America to keep the money in a family - lol.

[@temporarilyvanishing796]

@@GeneticsLessons Not only in Iran, but in Turkey, Iraq, Kuwait,...etc most of the Islamic world. It's not really to keep things within a family, but most people grow up around their cousins, usually the culture isolates them from other sexual interests, and so it's easy for them to end up that way since there's no Islamic law against marrying your cousin.

[@temporarilyvanishing796]

It's still advised against though ofc.

[@GeneticsLessons]

​@@temporarilyvanishing796 It is a good practice to have a blood test before marriage. If one suspects some hereditary disease or any other problem then he/she should seek the advice of a medical expert in this field.

[@temporarilyvanishing796]

@@GeneticsLessons Indeed. Thank you for the great explanations by the way! You've been a lifesaver for me.

[@brcarter1111]

Since III.1 is not a family member, you would normally assume they are AA, but since they have an affected child, you know they are heterozygous. Wouldn't you have to assume and calculate for if III.6 was AA OR Aa?

[@GeneticsLessons]

Nothing indicates that he is Aa, and this is rare genetic disorder, so we have to consider him to be AA.

[@GeneticsLessons]

@Richard Please watch the video to the end - I give the answers.

[@SanSimeonTheCar]

How would we solve if it was not a rare gentic disorder and that the dad could be either Aa or AA.

[@odekeFrancis-i3n]

Ilove u so much.I am at the university in Uganda.

[@niyatisharma7162]

very good explanation..

[@lastchance5444]

I was doing my calculations and my conclusion was that the father of the unknown child was an orphan because (by my calculations) he didn't recieved any feature from his adoptive parents 😂😂. But luckily all my calculations were wrong 😁😁

[@bouchser000]

Why didn’t you multiple 1/2 for the last step?

[@GeneticsLessons]

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[@angelicalara1776]

very helpful thank you! didn't realize it was that easy

[@GeneticsLessons]

Glad it helped!

[@adityapatel8156]

My all time favourite teacher in exploring genetics World

[@yourvideoclipsandvlogs1356]

I love this video so much, thank you for you effort

[@GeneticsLessons]

You're so welcome!

[@fatimajaweria5874]

Thanks you so much for sharing these videos... These are really helpful ...kindly share more videos on risk calculation

[@GeneticsLessons]

Sure I will

[@albazepeda221]

Your video helped me so much. Thank You.

[@GeneticsLessons]

You're welcome! Thanks for watching.

[@th3shadow]

thank you for your videos! :)

[@ebubeiseh2780]

Thanks. However the question asks about the probability of having a child with the trait not an affected(diseased) child. In this case wouldn"t the last multiplication factor be 2/4? since both parents 4 and 5 are carriers?

[@GeneticsLessons]

I understand that trait means effected - if the question would be about probability of heterozygocity the question would be about probability of the child being a carrier of the trait (not having a trait). But if you understand it as being a carrier, then we have to multiply by 1/2.

[@Geggie1000]

Ebube Iseh I was thinking he same thing.

[@Studywithme-live]

thank you so much ,do you have videos about operons genetics?

[@GeneticsLessons]

No I do not have videos about operons, may be later I will post.

[@subhojitpaul6377]

Would u post any video on the probability of sex linked disease related pedigree ??

[@GeneticsLessons]

I have a ton of such videos - just search my channel for videos with pedigrees.

[@bismel786]

Nikolay's Genetics Lessons

[@DreaKuntjoro]

so SO helpful!!! thank you!!!!

[@GeneticsLessons]

Glad it was helpful!

[@Ashley-qz5nv]

How can I work out which parent is carrier when no information has been given ?

[@GeneticsLessons]

You can figure it out from the pedigree pattern.

[@farhanhuszaimimpajar3713]

Thank you, this video is helpful.

[@GeneticsLessons]

You are welcome! Thanks for watching.

[@alpanarajput7766]

pedigree is of inheritance of a common phenotype controlled by an autosomal recessive allele. in pedigree I. female(I-1) is normal mate with a diseased male (I -2). They have 3 children . A normal female( II -1), a diseased male (II-2) and normal female(II-3). This third normal female (II-3) is mated with a normal male coming out of pedigree (II-4). Now the question is what is the probability that a child from parents II-3 and II-4 will show the phenotype. THE probability of carriers in the population is 1/3. options are; a. 1/16 ,b. 1/18 ,c. 1/36, d. 3/16 ans s given as b. 1/18. but i am unable to understand how. pls reply asap

[@GeneticsLessons]

Could please post a link to the pedigree picture?

[@beewinwood]

Im struggling with genetics currently and I need help! My university has a shortage of genetics tutors :( if I email you some problems could you post videos?

[@GeneticsLessons]

The best way to post them under any of my the videos and provide links to the pictures if any.

[@GeneticsLessons]

Yes.

[@mohammadsultan8894]

but in a multiple choice exam we don't have time to do all these things!

[@GeneticsLessons]

Exams is not always multiple choice - in genetics some exams often given to solve at home or there are exams with only 10-12 questions + it took me long to explain but if I would solve this question on exam this should not take me longer then couple minutes.

[@mohammadsultan8894]

yeah but unfortunately i have this subject in a multiple choice exam :/

[@GeneticsLessons]

With a practice this will not take more then couple minutes..

[@benudharapati464]

This type question ll not take more than 1-2 min... thanks sir

[@jrath2001]

Definitely such a long question will not come, but the concept involved here will be used in other questions!

[@summerfatima2525]

Thank you 🙏🏼

[@GeneticsLessons]

You’re welcome 😊

[@beewinwood]

Thank you!! Very helpful

[@GeneticsLessons]

You are welcome! Thanks for watching.

[@kankanabanerjee7541]

Very helpful.. Thank you.

[@GeneticsLessons]

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[@siyashrivastav9900]

thank u sooo oooo much sir really helpful

[@biologyp]

Awesome explanation 👏👏👏

[@niayesh1990]

why did u not say its 1/2 that 4 will be a carrier... why dont u consider the aa?

[@GeneticsLessons]

Because aa phenotipically different.

[@niayesh1990]

if we were saying: what is the probability that the man and girl have 3 children and one of them is affected... how would we solve that? thank you

[@GeneticsLessons]

Not enough information.

[@dutley26]

Anybody mind explaining why III6 is AA rather than Aa? I know it's not aa, but I don't understand why it is AA for certain since it could also be Aa. All their children are dominant so they could still get the dominant 'A' from their father but get the recessive gene 'a' from their mother. Correct? Thanks in advance!

[@GeneticsLessons]

Because mutations are rare and we have to assume that all people that are not direct descendants of effected people are not carriers.

[@nadiashareef4500]

Clear

[@emilianwaojei1331]

Thank you so much i got it immediately

[@GeneticsLessons]

You're welcome! Thanks for watching.

[@kankanabanerjee7541]

Very very helpful!

[@tayebe3457]

Hi Why didn't you calculate the probability to transfer the allele from generation 1to3?

[@GeneticsLessons]

No needed - I mentioned that.

[@tayebe3457]

Thank you

[@marina-sq9fn]

thanks for the vid:) since ure so good in biology can u plese help me w/ this problem??? An individual has a blood type genotype of AB and is homozygous recessive for the Rh factor. Which blood types can this individual receive in a blood transfusion? Blood types are O-, O+, A-, A+, B-, B+, AB-, AB+ PLZ HELP!!

[@GeneticsLessons]

O-, AB-, A-, B-

[@GeneticsLessons]

Here is a solution of your problem kzbin.info/www/bejne/rpPOpWV8n8uMa7M

[@marina-sq9fn]

Thank You!!! спасибо:) that helped a ton

[@presiannas]

The risk of the child being affected in the end is 1/6 but there is also consanguinity because the parents are related so this increases the risk right?

[@GeneticsLessons]

Inbreeding increase the risk of the inheritable genetic disorders.

[@udghoshikabhattacharya6428]

This video is very helpful for me sir.... but I have a problem on another que...will u give me a solution for this question..my que is ..=given that husband and wife are both heterozygous for a recessive allele for albinism.if they have dizygotic twins, the probability that both the twins will have the same phenotype for pigmentation will be.

[@GeneticsLessons]

Ok, will make a video today and send you a link.

[@GeneticsLessons]

Here is solution: kzbin.info/www/bejne/nnzRc6Bmq8uprqM

[@katherinele2207]

Thanks so much for this video :)

[@GeneticsLessons]

You are so welcome!

[@dhruvabarua5265]

What if you got a problem when it doesn’t say whether it is autosomal dominant or recessive? (I know for a fact that it’s not sex-linked) What would I do?

[@GeneticsLessons]

You have to build a pedigree of this family and analyse inheritance patern.

[@thejac4098]

3rd generation 5&6 how did they make 100% carrier in 4th generation's 5th person..

[@GeneticsLessons]

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[@bvkapadnis3816]

Good job sir. Thanks

[@aul9745]

What a voice you got

[@GeneticsLessons]

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[@soniyasherpa1999]

Thanks for this explanation! Had to replay some of it because of the accent hahaha. But good Job!

[@GeneticsLessons]

Glad it helped!

[@ghazal3340]

Thanks so much!

[@simontong8330]

thanks, great job!

[@kaarthikgauthum5328]

Greetings sir, it will be really helpfull if you could help me to understand the answer for the following question Peter Smith is affected with MCAD disease but is now asymptomatic. He has just been diagnosed through family studies. His partner Sarah is concerned regarding their 2 children, Ian (2yrs) and Sophie (1yr). What is the a priori risk of Peter and Sarah’s children of being affected?

[@GeneticsLessons]

Close to 0 cause this is is autosomal recessive disorder MCADD is most prevalent in individuals of Northern European Caucasian descent. The incidence in Northern Germany is 1:4000, currently the highest in the world. Northern Europe is also the origin of the common mutation in MCADD. For populations without origins in Northern Europe, the incidence is significantly lower, 1:51,000 in Japan and 1:700,000 in Taiwan. The common mutation has not been identified in MCADD cases identified in Asian populations.

[@kaarthikgauthum5328]

sir but i was given that the carrier frequency is 1:50.. Still the answer remains the same?

[@GeneticsLessons]

Then why are you posting a problem and hiding substantial information needed for calculations?

[@kaarthikgauthum5328]

I was asked to comment on the method of diagonse for the condition so i thought the carrier frequency was stated for that. I am extremely sorry sir.

[@GeneticsLessons]

Please post your problem text exactly as given and I will show you a solution.

[@adavydov7636]

Thank you so much for this video!

[@hanngo8162]

That helps a lot!

[@gracechong727]

thank you!!!

[@GeneticsLessons]

You are welcome! Thanks for watching.

[@manishasaboo9445]

WHICH SOFTWARE DO YOU USE ????

[@GeneticsLessons]

Ipad and Explaineverything app.

[@sonikx38]

thnx bud

[@castro4832]

so the Dominant always first?

[@GeneticsLessons]

You can write aA or Aa - both would be correct because in the nucleus there is no such thing as "right" and "left" - but in the textbooks heterozygous genotypes noteted as Aa.

[@hemeleh8683]

amazing

[@GeneticsLessons]

Thanks

[@crushhurts]

why did u multiply by 1/4

[@GeneticsLessons]

This is probability of two heterozygous parents to have effected child.

[@crushhurts]

Oh! Makes much sense now. Thanks!

[@puterielaika9209]

Nikolay's Genetics Lessons so you mean that, if i wanna get the probability of their child, i need to multipy the parents first then the child ?

[@estherlarbi8514]

agyeiwaa123

[@Saint_Ale]

how do i find out the probability of, 2 out of 4 daughters having such a trait?

[@GeneticsLessons]

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[@dwaipayandattaroy9801]

Malik are you teaching me for gre biochemistry personally, if yes, I will come to your city, do a job somewhere and learn from you , you are awesome, else I will not be able to avail you and pay learning fee and how I will survive

[@GeneticsLessons]

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[@rahmahdalhat8484]

I dont understand how 1/4 is added...why

[@GeneticsLessons]

Aa x Aa will result in 1/4 AA, 2/4 Aa and 1/4 aa

[@rahmahdalhat8484]

@@GeneticsLessons so it is compulsory to add the 1st generation... probability to have a homozygous recessive ?

[@GeneticsLessons]

All we need to know a probability for 4-4 and 4-5 to be heterozygous, next we have to multiply this probabilities and multiply by probability of getting "aa" in a progeny.

[@rahmahdalhat8484]

@@GeneticsLessons Thanks so much..so if it was dominant how will the final calculation be?

[@alpanarajput7766]

hi sir i hav a question

[@GeneticsLessons]

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[@Dontuseyourrealname134]

Brother and sister....

[@dwaipayandattaroy9801]

III (1) is carrier not affected female

[@nikolayk.8237]

I give explanation at 8th minute.

[@dwaipayandattaroy9801]

Am I correct or incorrect

[@GeneticsLessons]

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[@biologist8540]

In generation 4th of 5(parent) why u take 1/1

[@GeneticsLessons]

Because we know he is 100% carrier. 1/1 is 100%

[@biologist8540]

Ok sir thank you

[@biologist8540]

If all generation are carrier what will be 1/1

[@dwaipayandattaroy9801]

4 female is carrier , male troubled, affected, the trait is X chromosome based and is dominant in expression of the carrier chromosome , y chromosome is non infected , only X chromosome is the flawed or rewarded one , 100% 4 is carrier, bit white , 100% 5 is black male 😁

[@GeneticsLessons]

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[@himanshuranjan9276]

Love from india

[@dwaipayandattaroy9801]

Correct me if wrong please

[@GeneticsLessons]

Watch my new video: Did you know about The Jewish Autonomous Region, situated in the Russian Far East? kzbin.infoQ3WC9wjBemU

[@richaparashar9788]

спасибо большое ! :)

[@GeneticsLessons]

Пожалуйста.

[@saintofchelseathomascarlyl5713]

i dont think we can assume III1 is AA because she has a sick offspring with aa so she must be Aa edit: nvm i understand now thank you very much =]

[@nikolayk.8237]

You are welcome, thanks for watching!

[@GeneticsLessons]

Watch my new video: Did you know about The Jewish Autonomous Region, situated in the Russian Far East? kzbin.infoQ3WC9wjBemU

[@butx1555]

I got 1/6

[@GeneticsLessons]

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[@jacobmellick6207]

probability problem probability

[@GeneticsLessons]

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[@dwaipayandattaroy9801]

Here, the male will be victim simple reason the anomali is X dependent 5 affected father, 4 it's biologically impossible to say which X 4 will carry, but that will be carrier at Max , as the father 2 is normal, so off spring all affected , I know daughters receive x from father and mother , son received x only from mother and according to the problem one cannot be smart dhope to give 3 off spring not four to eliminate the chance of finding the gender of last offspring , then I would have said if the father was carrier recessive or dominant, seriously , it works that way too, just depending on round black aa affected mother , which is a guarantee that the son will be affected , so both son affected and both daughter 6/7 carrier, I said both son cause any son would be a dhope if mum is aff cted,

[@GeneticsLessons]

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[@shamimshahed7365]

What the hell is your explanation?

[@GeneticsLessons]

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[@GeneticsLessons]

New practice problems for you: How to calculate phenotype frequency: kzbin.info/www/bejne/n3vZlWZ3h7Fpa5I How to solve probability problems in genetics: kzbin.info/www/bejne/jWqlkHiIhbODrpY How to calculate probability of the specific phenotype: kzbin.info/www/bejne/rYS1h2OKn7Z8p5I How to calculate probability of inheritance of recessive genetic disorder: kzbin.info/www/bejne/oHPEdHuKaZV5mdk Simple probability problem and solution: kzbin.info/www/bejne/o6OpZnipqcaMZ9k How to Calculate Species Extinction Probability: kzbin.info/www/bejne/q4aXlqlmjZacis0 How to use forked-line method to solve probability problems in Genetics: kzbin.info/www/bejne/poa6oKhmnqeJbNk How to solve probability problems in Genetics: kzbin.info/www/bejne/l4mUoXWVf6h-rNk How to solve DNA sequence probability problems: kzbin.info/www/bejne/nHOtfaesj8-qkKM When we have to sum probabilities and when we have to multiply them: kzbin.info/www/bejne/aoe4l2uHisZppM0 DNA sequence combinatorics problem: kzbin.info/www/bejne/j2Ksk5J3Zr9njqs ABO blood group allele frequency brain teasing problem: kzbin.info/www/bejne/momxp2mBpaeKZ7c Interesting probability problem (Genetics): kzbin.info/www/bejne/r2STemSOi8mmqZo How to Find a Variance of the Population: kzbin.info/www/bejne/n3jVZoSXetBgj8U ABO Blood groups, Rh+- Factor and Forked Line Probability Calculations Explained: kzbin.info/www/bejne/rGSaqqBnoLxniJY How to solve simple probability problems: kzbin.info/www/bejne/Z3_TpWaYl9CCqpI

[@GeneticsLessons]

How to analyze pedigree chart: kzbin.info/www/bejne/eYGZco2taN6bnM0 Pedigree vs. Arrow diagram (path analysis): kzbin.info/www/bejne/annIfYCmhteUbbs How to identify the mode of inheritance: kzbin.info/www/bejne/h6Cspaicq8yaj7s Pedigree analysis explained: kzbin.info/www/bejne/jai8oHx_mdKXoqc Pedigree and disease probability calculations: kzbin.info/www/bejne/j4Tdg2uwp76Iq5Y Unusual genetic disorder inheritance explained: kzbin.info/www/bejne/rYeWaIRofrWomMU How to analyze Pedigree (for medical students): kzbin.info/www/bejne/f4isc6SXedZ-pq8 Pedigree probability problem explained: kzbin.info/www/bejne/qHecgHdnocuhpbs Pedigree chart and probability calculations: kzbin.info/www/bejne/paalqaR6q7qnl68 Interesting inheritance pattern explained: kzbin.info/www/bejne/sJ_WhWqri7V9Zrc Pedigree probability practice problem explained: kzbin.info/www/bejne/aKaUmneXrteMaqc How to analyse pedigrees and solve problems: kzbin.info/www/bejne/bYKrdI16ntaWb7s Pedigree probability practice problem: kzbin.info/www/bejne/gHmcf6Buq8l4fKc How to calculate probability of genetic disease: kzbin.info/www/bejne/p2mcmYuXbdJgmpY

[@GeneticsLessons]

More probability problems for you: Three Methods how to Solve Probability Problems: kzbin.info/www/bejne/bZfal4GCqKyqj9E How to Solve Probability Problems in Genetics: kzbin.info/www/bejne/l6Swk3iudpqSm6s How to solve simple probability problems in genetics: kzbin.info/www/bejne/npOwe2CDmrissNE How to solve Probability Problems Using Pascal's triangle: kzbin.info/www/bejne/emqnd6KYh9dnptk

[@GeneticsLessons]

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[@GeneticsLessons]

How to solve genetics problems using Pascal's triangle: kzbin.info/www/bejne/hJ2nhXhunqhnrZY What is a probability a family of 3 children, has at least 1 girl, given that at least 1 is a boy? kzbin.info/www/bejne/hYvFoaOJlN2nodU How X-linked recessive genetic disorder inherited kzbin.info/www/bejne/mWPIqnmYip6kmpY A family has 4 children, 3 of them are girls, what is the probability that the last one is a boy? kzbin.info/www/bejne/mofToaR4lphkiKc How to Solve Probability Problems Using Pascal Triangle kzbin.info/www/bejne/rZ2ynYShm6xgrMU How to Solve Population Genetics Problems kzbin.info/www/bejne/eV7Mqpd-dslogLM How to solve pedigree problems kzbin.info/www/bejne/o3bOeKBtbcqXqtE How to Solve Pedigree Problems kzbin.info/www/bejne/sJLChqFnq9t2fs0 How to solve simple probability problems kzbin.info/www/bejne/nmbPkpSkgdeNlZI How to solve simple probability problems in genetics kzbin.info/www/bejne/Zqa7k2CjgpdngqM Simple probability problem kzbin.info/www/bejne/q3zIgIqHoa-tlcU How to Find Probability of Genotype or Phenotype in a Progeny kzbin.info/www/bejne/hKu5lap_obCci8U How to find all possible genotypes kzbin.info/www/bejne/iaTOiYN-br5roZY

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[@GeneticsLessons]

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[@abrahamtibebu7341]

thank you!

[@GeneticsLessons]

You are welcome! Thanks for watching.

[@zacharyharris469]

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[@Mehta0034]

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[@justinoettle5658]

Thank you!

[@GeneticsLessons]

You're welcome!

[@TristanWeekss]

thank you!

[@GeneticsLessons]

Welcome!