In: Biology
Swannamotosis is a variant of Neurofibromatosis-type 2. It is an autosomal dominant condition that shows both incomplete penetrance and variable expressivity. Sixty percent of individuals with at least one mutant allele will show the condition in the phenotype. Of those showing the phenotype, 20% have a severe version, 50% have a moderate version, and 30% have a mild version. If two heterozygous parents have a child, what is the chance that is will show the most severe form of the disorder? To calculate this we'll have to multiply _______ X ________ X ________ X _______. If a child of two heterozygous parents, does not have the condition, what is their chance that they do NOT have the allele? ________
Blank 1 Options: 1/2, 1/4, 2/3, 3/4, 1
Blank 2 Options: 0.4, 0.6, I dont need either of these values
Blank 3 Options: 0.2, 0.3, 0.5, I dont need any of these values
Blank 4 Options: 1/4, 1/2, 2/3, 3/4, I dont need any of these values
Blank 5 Options: 1/4 x 0.4 x 0.2, 2/3 x 0.4, 1/4 x 0.4(3.4), (1/4)/(0.4 x (3/4)), (1/4)/((1/4) + 0.4(3.4))
It is given that the Swannamotosis is an autosomal dominant condition, so presence of even one dominant allele will cause the condition.
Parents are heterozygous ………….(For parents being heterozygous their parents will be either AA X aa or Aa X aa)
Aa X Aa
AA, aa, Aa, Aa
So 25% will be severe with AA, 50% moderate with Aa, nd 25% mild with aa.
So for a child without this condition the probability will be
Blank A: 1/2 (probability of parents with this condition which will be crossed)
Blank 2: no need of values
Blank 3: no need of values
Blank4: ¼ ( Probability of child without this condition)
Their chance that they do NOT have the allele is
Blank 5: 2/3 x 0.4, 1/4 x 0.4(3.4),
As ¼ is the probability of not being affected, 0.4 is the percentage of the least affected, and the 2/3 is the heterozygous population among the non dominant or non severely affected individual that may or may not have allele