Challenge question: practicing probability Homework •...

Challenge question: practicing probability Homework • Unanswered Two parents want to know if their child will be affected by a rare recessive disorder that runs in both their families. Both parents are unaffected but have a 50% chance of being a carrier. What is the chance their child will be unaffected, but still be a carrier of the recessive allele? This is challenging because each parent has a 1/2 chance of being homozygous for the unaffected allele so they can only pass on an unaffected allele...OR each parent has a 1/2 chance of being heterozygous AND thus have a 1/2 chance of passing on the dominant allele. Because a parent cannot be both DD and Dd (these are mutually exclusive possibilities), we'll need to use the sum rule for part of this calculation to take both possibilities into account. A 1/8 B 1/16 C 3/8 D 3/16 E 1/512 Human pedigree pattern identification: Example 1 Homework • Unanswered The allele that causes the affected phenotype is _____. (Can you explain why?) A definitely autosomal dominant B definitely autosomal recessive C could be either dominant or a very common autosomal recessive allele

Transcript

00:01 In order to answer this question, let's talk about inheritance.

00:02 It says two parents want to know if their child will be affected by a very recessive disorder that grants in both of their families.

00:09 Both parents are unaffected but have a 50 % chance of being a caguer.

00:13 It means that you have here the mother, well the father, i'm sorry, here you have the mother, okay? and according to this information, both parents have 50 % chances of being a carrier.

00:23 It means both of them have one -half chances for being heterocycle because 50 % is the same as one -half, okay? and it says, what is a chance that their child will be unaffected, but still be a caliber of the excessive allele? in this case, they are telling us, or this is an auto -somer excessive condition.

00:41 In this case, the homocycles dominant, and the heterocybinocytes are going to be normal, and the homo -saclysis is going to be for disease.

00:50 Now, they want a child, and according to this question, you want to know the probabilities that the child is healthy but heterocylus.

00:59 It means a calmer, okay? so let's make this course.

01:02 Heterocygote with another heterocygose, you have here homocygous dominant, heterocygote, heterocygous and homozygose.

01:13 Now, what are the chances of being heterozygous? you have one and two out of four.

01:20 It means two quarters that is the same as one half.

01:22 So you have one half chances for this child to be heterosegios.

01:26 But in order for this to happen, first these two things need to happen.

01:30 So in this case you have one half, multiplied by one half, multiplied by one half, and this is equal to.

01:36 This is the probability for these parents to have a child with the heterocygary.

01:43 This is the answer, but remember that this is only one possibility, because in order to get heterocygose, both parents can be heterocygote.

01:51 But the other recommendation is that one parent is heterocygous.

02:02 Well, we also said that the second parent has one has chances for being heterocygous.

02:06 But it says that both parents are healthy.

02:09 It means the remaining half is going to be for the homo -sigo -dominal and the remaining half is going to be for heterocygous.

02:16 So in this case, we already performed this possibility here.

02:21 So what is another possibility that can give you a child with a heterocygoyne? well, it can be that one parent is homo -sigalamic, and the second parent is heterocycles.

02:32 Okay? you have one -half chances for being homo -sigital and one -half chances for being heterocygous, and when you perform the cross, they're going to have homozygous domino, heterocygous, gyrosovo, and heterocygos.

02:47 So again, two out of four, it means one half is the probability for getting heterocygous.

02:53 In this case, you also have one half, multiple by one half, and multiple by one half is equal to one eight.

03:00 Now, you have one eight chances for this possibility between if both parents are heterocygios, one half chances if the father is homozygous dominole and the mother is heterocygose.

03:08 And another possibility is that the model is the homozygous dominant and the father is the heterocybos.

03:17 So in this case, you're going to have also, this is the same cross as you get here...

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