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A woman with cystic fibrosis marries her first cousin. What is the risk that their first child will have CF?

a. 1/2
b. 1/4
c. 1/8
d. 1/16
e. 1/32

plz explain if possible using illustration ?

im really having hard time with these type of questions


First to get the probabilty You must multiply her probability of moving the disease to her child in 1/8 of inhereitance through her cousin

so it would be 1/4 * 1/8 = 1/32

SO I think The answer would be E


If Iam wrong please let me know.. any body??


cystic fibrosis is an autosomal recessive disease

so the affected mother will have two recessive allels for the disease "aa"

while her cousin will be probably a homozygous normal "AA" or heterozygous carrier "Aa"

and because the question assume the child have a risk of having CF so we should assume that the cousin is heterozygous carrier "Aa" (because if he is homozygous normal one of the children will have the disease, remember that CF is autosomal recessive and requires two mutant allels to become manifest)

so the results of the mating of "aa" with "Aa" will be:

1/2 affected
1/2 carriers

so I think the answer to the question is : a. 1/2


what's the right ans?? shocked


I WAS thinking that i was the only one with so much confusions in genetics . well every body else having the same prob,ems with genetics .

i also go with A . as its autosomal recessive disease



probability of passing cf from mother=1(as she has disease)
probability that father will pass cf to child=1/4(as father is first cousin,so 1/4)

now probabilty mother and father will have cf child =1/2
so 1*1/4*1/2= 1/8
so answer is c! for sure


wishikhan no doubt its autosomal reccessive disease but you have to take into account the probability too


c is right nod

the question is what is the chance to have a gene allele from your grandfather ?
1: your father has 50% chance to get it from your grandfather
2: you have another 50% to get it from your father
so you have 25% of have an allele from your grand father

back to question
because the woman has cf so she must has two mutant alleles one from father and one from mother
and her parents got that allel from her grand parents
because she and her cousin share grand parents so he may also has this mutant allele
so the risk of her cousin to get that mutant allele from their common grand parents is 25%

last point what is the risk that this cousin pass the mutant allele to his child is 50%
so over all risk is 0.25 x 0.5
or 1/4 x 1/2 = 1/8

look at pic

1/2 % risk to pass mutant allele from genration I to genration II
1/2 % risk to pass mutant allele from genration II to genration III
1/2 % risk to pass mutant allele from genration III to genration IV " as the mother is homozygoes diseased she only can pass the mutant allele so if her cousin pass the mutant allele the child will get the disease

so over all risk is 1/2 x 1/2 x 1/2 = 1/8
as if u have multiple risk u multiply them

Attached Files:
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Edited by madoo on Sep 16, 2009 - 7:34 PM


Passerby wrote:
If Iam wrong please let me know.. any body??

i think u got this from kaplan when they mention the " coefficient of relationship"
as first cousin share 1/8 of their genes and second cousins share 1/32 of their genes

i don't claim to understand this issue but i don't think this is the same concept here


agree with madoo


thanks sumeet . i will definitely buy that book genetics seems easy when you do it from kaplan notes well its to o difficult when i am solving problems.poor concepts


sumeet that book consists of questions or it has some notes in it


its questions.. biochemistry pretest...if u want... send me ur email..i will send u


i hope u will be able to download 4rm here..if not then i will send u...
after downloading go thru genetics tests and explanation...they have very basic questions and numerics that helps in making concepts...


thanks alot i have downloaded it . i t is really a great help as i know my concepts of genetics it is easy to do from book but to difficult when i do questions .thanks again


Agree with C . Nice explanation, Madoo


Sumeet, thanx for the pretests grin


simply,,any child get 1/2 genes from each parents(general role)
chance of women to get disease gene is 1/2x1/2 (general principle)
chance of 1st cousin get disease gene is 1/2x 1/2(same principle)
so chance of both=1/2x1/2x1/2x1/2=1/16
cystic fibrosis inheritence(autosomal recessive)=women homozygus(affected)and cousin is( carrier -consangous) so inheritance=1/2
1/2x1/16=1/32 risk for child to be affected ,,,,, E,,,


I thing the correct answer is D 1/8
first : the chances of the cousin is CF carrier is 1/4
why ??
cuse 1/2 (from his grand parent) * 1/2 (from his father) = 1/4 (assuming grandparent are carrier)

second : the probability of getting a child having CF (homozygos) from affected mom and carrier father is 1/2 (punett square)

so the answer is 1/4 * 1/2 = 1/8

plz correct me if Iam wrong

Edited by dr_toffee on Jan 02, 2015 - 6:36 PM

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