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 Waardenburg  




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Author27 Posts
  #21

Answer is C; 3/4

For each pregnancy, the probability that the child will be affected is one-half. Therefore, the probability that all three children will be affected is the product of three independent events (1/2 × 1/2 × 1/2 = 1/8).

The probability that all three children will be unaffected is the same. When evaluating the prob- ability that one of the three children will be affected, there are three of eight possible birth orders that have one affected child. For two of three children to be affected, there are also three of eight possible birth orders.

grin





  #22

hggrinnod


  #23

raised eyebrow;smiling faceconfusedmadgrinsmiling facewinkj


  #24

E, each reproductive process is statistically independent from all other event in single gene mutation diseasea


  #25

EACH ONE OF THESE KIDS HAS 50% CHANCE TO GET THE DISEASE,,,,USE BUNNET SQUARE ,,,,THANKS,,,,,E,,,E,,,EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE


  #26

Huntington disease (143100) is an autosomal dominant, progressive
neurodegenerative disease that causes uncontrolled physical movements
and mental deterioration. A husband and wife have three children, two
boys and one girl. The husband was diagnosed with Huntington disease in
his mid-fifties, as was his father. The wife is normal, as were both her parents.
What is the probability that all three of his children will eventually
develop Huntington disease?
a. 1/8
b. 1/4
c. 1/2
d. 3/4
e. All three will develop Huntington disease

ANSWER IS "A"
Since Huntington disease is autosomal dominant, both
homozygotes and heterozygotes for the Huntington allele will develop the
disease. The husband is a heterozygote, as his father had Huntington but
his mother did not. The wife is most likely homozygous recessive, as she
has not developed the disease. Thus the children each have a 50% chance
(one-half ) probability of having inherited the dominant Huntington allele
from their father and a 100% chance of inheriting a recessive allele from
their mother. Thus, the individual probability for each child developing
Huntington disease is 1/2 and the probability that all three children will
develop the disease is 1/2 × 1/2 × 1/2 = 1/8.


  #27

I FOUND THIS QUESTION :

Waardenburg syndrome (193500) is an autosomal dominant condition
that accounts for 1.4% of cases of congenital deafness. In addition to
deafness, patients with this condition have an atypical facial appearance,
including lateral displacement of the inner canthi (eye corners), hypertelorism
(widely spaced eyes), poliosis (white forelock), and white patches
of skin on the ventral midline (partial albinism or piebaldism). A mother
has Waardenburg syndrome, her husband is unaffected, and they plan to
have a family with three children. What is the probability that one of the
three children will be affected?
a. 1/8
b. 1/4
c. 3/4
d. 1/3
e. ½

The answer is c
For each pregnancy, the probability that the child will be affected is one-half. Thereforethe probability that all three children will be affected is the product of the three independent events—that is, 1/2 × 1/2 × 1/2 = 1/8. The probability that all three children will be unaffected is the same. When evaluating the Probability that one of the three children will be affected, it must be noted that there are three of eight possible birth orders that have one affected child (Www, wWw, wwW). For two of three children to be affected, there are also three of eight possible birth orders (WWw, WwW, wWW).






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