Prep for USMLE 
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Author  27 Posts  
GinaRMD 
Answer is C; 3/4 For each pregnancy, the probability that the child will be affected is onehalf. Therefore, the probability that all three children will be affected is the product of three independent events (1/2 Ã— 1/2 Ã— 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.  
abdalla
Forum Senior Topics: 0 Posts: 101 
hg  
abdalla
Forum Senior Topics: 0 Posts: 101 
;j  
spinal shock
Forum Newbie Topics: 23 Posts: 241 
E, each reproductive process is statistically independent from all other event in single gene mutation diseasea  
mohammem
Forum Newbie Topics: 2 Posts: 198 
EACH ONE OF THESE KIDS HAS 50% CHANCE TO GET THE DISEASE,,,,USE BUNNET SQUARE ,,,,THANKS,,,,,E,,,E,,,EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE  
Varuag
Forum Newbie Topics: 0 Posts: 11 
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 midfifties, 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 (onehalf ) 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.  
Varuag
Forum Newbie Topics: 0 Posts: 11 
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 onehalf. 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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