[Catalist] Genetics question for Bio/ Human Bio Teachers

[CLARK Julie [SIDE - Sch of Isol & Dist Edu]]

Hi Greg,

The probability of the next child being an affected daughter is ¼ or 25%. Students can look at their punnet square to get this answer. Alternatively, it is probability of one event after another where you can multiply the probabilities ie. Probability of a daughter = ½ and the probability of a daughter having the condition = ½ , so ½  X ½ = 1/4

If the parents know that the next child is a daughter, then you only look at the possible genotypes of the daughters. So probability is 50%.

Regards,
Julie Clark
Science teacher
Mon - Thurs

School of Isolated and Distance Education
Secondary School
Building A
164 -194 Oxford St.
Leederville 6007

Phone:  (08) 9242 6330



From: Catalist [mailto:catalist-bounces at lists.stawa.net] On Behalf Of Greg Munyard
Sent: Thursday, 26 October 2017 9:52 AM
To: 'catalist at lists.stawa.net'
Subject: [Catalist] Genetics question for Bio/ Human Bio Teachers

Good morning folks

My colleagues and I are having some discussion around this question and I wondered if there is a consensus in the Bio/ Human Bio community on this:

"If II-8 (XnY) and II-9 (XNXn) have more children, what is the probability that they will produce an affected daughter?" (Genotypes not in the original question - had to be ascertained from a pedigree.)

One colleague's approach is to say that there is a 50% chance for producing a daughter and 50% chance that a daughter will be affected so the combination of these is a probability of 25%.

My approach is to say that there is zero chance for sons to be daughters so that they represent a null sample. Considering only daughters then the chance of an affected daughter is 50%.

Any guidance out there in the ether?

Regards

Greg Munyard
Senior Teacher - Science
[kennedy_email-footer-t4]

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