Could someone please answer this : Why isn't the probability of being a carrier is 1/2 but 2/ 3? 1/ 4 = exhibits trait/ All 2/3 = carriers / the ones who do not exhibit trait. Why is the denominator of second ratio shouldn't be 4? ( Which means out of all of them) If impossibilities are removed when calculating the first ratio it would be 1/1 which is wrong. Why do u remove impossibilities only for the second ratio??
The probability of being a carrier is 2/3 because there is a 2/3 possibility that the recessive allele (Rr) was inherited from their heterozygous parents. Looking at the punnet sqaure, we see that the possible genotypes if parents A and B were to mate are: 1/4 rr (carries the disease), 2/4 Rr (carries the recessive allele) and RR (no recessive allele, no trait expressed) From the pedigree, we know that both parents are unaffected, leaving only 3/4 genotypes that are possible. RR and Rr. Looking at the punnet square, we can see that there is a 1/3 chance of parent A inheriting the RR genotype and a 2/3 chance of parent A inheriting the Rr genotype. The same goes for parent B. Since We know that it it is an autosomal recessive disease (meaning the kid's genotype must be rr) we know that neither parent can have the RR genotype because the child would end up not expressing the trait and there is no way the child could inherit the rare autosomal disease. So that leaves us with the 2/3 chance that the parents are Rr. We then multiply that 2/3 by the chance of the heterozygous parents having a child with the genotype which is 1/4. So youre going to get (2/3 * 2/3 * 1/4)
Thank you so much ma'am..u are really just awesome..u clear all my confusion..plz make make more videos on probability problems...i wish u can tech me from my college days...
No, because you are sure the parents don't have the rr genotype, as they don't have the disease, so you can just leave that out of your calculations :D
Good job Dr!!!I wish I could have seen this video while I was studying Genetics for Step 1.Thanks for sharing!!!I will share it with some of my IMGs friends.
thank you so much. i was very confused. Wish i had found this video before my first midterm
wow, this is actually the best video I watched about the pedigree calculating probabilities.
Really maam !!!! I needed this for my step 1 exam ❤❤❤
Thank you so much for posting this video! I was so confused when it came to this topic and you cleared this subject up for me!
Why is it that this we didn't ignore the RR but added it to the probability making it 1/4 and rather not 1/3 for AA
YOU ARE THE BESTTT
Thank you ma'am
Fantastic! Do you private tutor? Be glad to do it on Skype or Zoom! I need help! Thanks!
Thank you... I did well in my exam..
very resourceful
Thank you
Why A B are Rr heterozygous ? Why not RR and Rr
Thanks a lot..you are the best
Could someone please answer this :
Why isn't the probability of being a carrier is 1/2 but 2/ 3?
1/ 4 = exhibits trait/ All
2/3 = carriers / the ones who do not exhibit trait.
Why is the denominator of second ratio shouldn't be 4? ( Which means out of all of them)
If impossibilities are removed when calculating the first ratio it would be 1/1 which is wrong. Why do u remove impossibilities only for the second ratio??
The probability of being a carrier is 2/3 because there is a 2/3 possibility that the recessive allele (Rr) was inherited from their heterozygous parents. Looking at the punnet sqaure, we see that the possible genotypes if parents A and B were to mate are: 1/4 rr (carries the disease), 2/4 Rr (carries the recessive allele) and RR (no recessive allele, no trait expressed)
From the pedigree, we know that both parents are unaffected, leaving only 3/4 genotypes that are possible. RR and Rr. Looking at the punnet square, we can see that there is a 1/3 chance of parent A inheriting the RR genotype and a 2/3 chance of parent A inheriting the Rr genotype. The same goes for parent B.
Since We know that it it is an autosomal recessive disease (meaning the kid's genotype must be rr) we know that neither parent can have the RR genotype because the child would end up not expressing the trait and there is no way the child could inherit the rare autosomal disease. So that leaves us with the 2/3 chance that the parents are Rr. We then multiply that 2/3 by the chance of the heterozygous parents having a child with the genotype which is 1/4. So youre going to get (2/3 * 2/3 * 1/4)
@@twerkformemes Thanks a lot!!!
great video! :-)
Genetics grade = saved
Thanks
Thank you so much ma'am..u are really just awesome..u clear all my confusion..plz make make more videos on probability problems...i wish u can tech me from my college days...
Im confused, isnt the probability of the parents being carriers is 1/2?
No, because you are sure the parents don't have the rr genotype, as they don't have the disease, so you can just leave that out of your calculations :D
is nobody going to say that those car are fucking orange
Plzz explain pedigree
👍👍💓
Superbly done...thank u Dr Crowder frm South Africa
Still confused as fuck tho
dr crowder is one of the best teachers I have ever had. so glad I took her for human heredity. really wish I could've had her for genetics.
i dont understnad hwy we are just calculating probably that the parent is hetrogenous
Thnx a lot for this useful vedio
Good job Dr!!!I wish I could have seen this video while I was studying Genetics for Step 1.Thanks for sharing!!!I will share it with some of my IMGs friends.
Carly Steeve Augustin hey how are you bro? I'm also a med student. What med school are you attending?
thank you, this video helped a lot!