kalsam
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A man and a women r both affected by an autosomal dominant disorder that has 80% penetrance in all affected individuals. They r both heterozygotes for the disease causing mustation.
What is the probability that they will prduce phenotypically normal offspring?????
a. 20%
b. 25%
c. 40%
d. 60%
e. 80%
Answer with explanation pleeese 
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| piter
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c. 40%
25% - normal genotypically and phenotypically +
20% normal phenotypically of the 75% affected = 25%+15%=40%
hope this helpes
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| kalsam
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I am sorry still I did not understand...
How did u get that 25%...normal geneotypically and phenotypically..
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| piter
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if 2 parents are dom. & heterozygote - 1 of 4 children will be normal geneotypically and phenotypically. Maybe you confused with "80% penetrance" - it is only related to affected children - who had that gene, and 20% of them are phenotypically normal.
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| kalsam
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thanks!!!!!
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| mjl1717
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kalsam thx for bring up the ghost of H/W because Ineed it fresh in my mind in order to do it. Ill have to take my time and sequentially explain it in different posts. [but I wont dance around it](for now Ill ignore what H/W means Verbatum and plod thru the mathematical work of those 2 geniuses) :|
p2 +2pq +q2 =1 / also p+q=1
p2-homozygous normal
2pq-heterozygotes
q2-homozygotes with disease
**ok megs post was redone on the other "p2+2pq+q2 =1" post(Im saved)
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| usmleasr
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ans is 40%...becos...AD disease...both parents r heterozygos ...means...75% of children will have disease causing gene...
but only 80%penetrance...means 75% x 0.8 = 60% will express disease phenotypically. so 40 % normal phenotypically..hope this helps
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| kalsam
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Thanks usmleasr..!!!
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| elsa
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I still dont understand it. Please help me. How do you get 75% please explain further :oops:
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elsa
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| intern_doc
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Ok Elsa,
let me make it simple....just follow an example
dad = >Aa
mom=>Aa
kids = > AA, Aa,Aa, aa (so this mean 75% have the gene causing disease. Only 25% is NORMAL)
Now take that 75% x 80(penetrance) => 60%
So what this mean is 60% will express phenotypically and subtract from 100 => 40 ( NORMAL)
Hope this help :lol:
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| sar
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thanks, intern_doc you have made it very easy for all of us.
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| santaclara
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thanx to all to explain in such agreat way.
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| elsa
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Thanks! In your example the Capital A is the affected gene and the small aa is the normal? Can you explain to me by examples the meaning of Autosomal dominant? Heterozygous and the other stuff please thanks a lot! Intern_doc I am right?
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elsa
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| intern_doc
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Ok Elsa,
Autosomal Dominant Inheritance
This mean that it take ONLY 1 gene to show the disease. Therefore BOTH of the parents have to have the gene .
example
mom - Aa
dad - Aa
So the outcome possibility
Aa x Aa = > AA, Aa, Aa, aa
This mean=> 25% affected (AA)
50% carrier (Aa)
25% NORMAL (aa)
NOTES: there is NO skipped generation in Autosomal Dominant b/c 2 unaffected parents cannot transmit the disease to their offsprings
homozygote => alleles at a locus are the same (homo = same)
heterozygote=> alleles at a locus are different
So homozygote is referred to aa (same)
heterozygote is referred to Aa (different)
REMEMBER ANOTHER CONCEPTS
if the questions ask what percentage of the offspring is affected in heterozygote mating with heterozygote ?
Aa x Aa => AA, Aa, Aa, aa (so 75% will be affected)
or heterozygote mating with homozygote ?
Aa x aa => Aa, Aa, aa, aa ( so 50% will be affected)
I hope this help,
Good luck
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| sar
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The parents of a girl with Tay-Sachs disease decide to pursue bone marrow transplantation in an attempt to provide a source for the missing lysosomal enzyme. Preliminary testing of the girl's normal siblings is performed to assess their carrier status and their human leukocyte antigen (HLA) locus compatibility with their affected sister. What is the chance that one of the three siblings is homozygous normal (i.e., has a good supply of enzyme) and HLA-compatible.
choice
A. 1/2
B. 1/3
C. 1/4
D. 1/6
E. 1/12
The answer is: C
The probability that any one sibling is homozygous normal is 1/3. The human leukocyte antigen (HLA) cluster on chromosome 6 consists of several loci that are each highly polymorphic. Because the loci are clustered together, their polymorphic products form haplotypes (i.e., A1-B8-DR2 on one chromosome and A9-B5-DR3 on another chromosome). Since recombination among HLA loci is unlikely, the chances of two siblings being HLA-identical are essentially those of inheriting the same parental chromosomes, that is, 1/4. The chance for a sibling to be both homozygous normal for Tay-Sachs disease and HLA-compatible is 1/3 × 1/4 = 1/12. Since there are three siblings, the total chance is 1/12 × 3, or 1/4.
Can anyone explain, it to me that.... Why is the probability of one sibling to be homozygous normal is 1/3??
Why the chances of two siblings to be HLA identical is 1/4???
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| mash
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tay sachs is AR
if the girl has tay sachs it means both parents r carriers(Aa and Aa)
now 4 possibilities wud be AA, Aa, Aa, aa
girl's normal siblings wud either be carriers or homozygous normal( i. e they r not homozygous fr mutant allele)
so, we r left with 3 options Either Aa or Aa or AA
out of these 3, AA is homozygous normal => chances of her sibling being homozygous normal r 1/3
now coming to 2nd part of the q
2 sibs wud be HLA compatible if they inherit same chromosomes
if the pt got allele with A1-B8-DR2 frm the father , the chances of other sibling getting the same ch frm father are 50%(1/2)
AND if pt has got ch with A9-B5-DR3 frm mother then the chance of dat sib getting same ch frm mother is 50%
the chances of inheriting same parental ch r 1/2*1/2 = 1/4
so, the chances of being homozygous normal and HLA compatible r 1/3*1/4 = 1/12
chances of one of the 3 siblings being homozygous normal and HLA compatible r 1/12 +1/12+1/12 =1/4(X or Y or Z)
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| elsa
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Thanks intern_doc
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elsa
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| sar
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Thanks Mash. I am so impressed how well you all are in genetics. How many days did you give for genetics, and how many readings?
I now understand the explanation
I have some more question.
1-What does (x, y or z) means?
2- In 2nd part of the question,
2 sib to get a gene from a father, it means that there are 2 choices, either having that gene from father or not, in this way you got 1/2 chance (50%).
3-what is Haplotype?
4- There would always be 4 combinations AA, Aa, Aa, or aa, whether the disease is recessive or dominant.
Thanks for giving a good explanation.
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| mash
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well sar, i ve used x/y/z variables fr 3 sibs , just to make it clear dat ques is asking abt one out of 3 being MHC compatible and homozygous normal..
2. u r right , there is 50% probability of getting it frm father or not
3.its been a while dat i opened up my genetics book so, i really dont remember the def of haplotype
it is genotype of different loci close together on a chromosome for eg.A1-B8-DR2 on one chromosome
these haplotypes do not separate on recombination( this is wat i remember)
4 yeah there wud always be 4 combinations but in AD ds Aa and AA will ve the ds
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| sar
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Thanks mash. sorry for another dumb q
You said, in AD, Aa and AA will have disease. so in AR, aa and Aa will have disease?
What is the diffeence b/w homologous and homozygous?
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| intern_doc
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In AR, ONLY AA will have the disease
aa = normal and
Aa = carrier
b/c it take 2 genes to get affected
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| sar
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Thank you so much intern_doc!
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