Biology Genetics: Colorblindness and Blood Type

AI Thread Summary
A colorblind man with blood type AB and a heterozygous woman with normal vision and blood type A are analyzed for the probability of having a daughter who is both colorblind and blood type A. The genetic crosses indicate that 1/4 of the offspring will be colorblind females and 1/2 will be blood type A. By multiplying these probabilities, the final odds of having a daughter that is colorblind and has blood type A are calculated to be 1/8. The discussion emphasizes the importance of treating the traits as independent for accurate calculations. This genetic scenario illustrates the application of basic principles of inheritance in predicting offspring traits.
shawonna23
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1. Homework Statement

A colorblind man with blood type AB marries a woman with normal vision and blood type A who is heterozygous for both traits. What are the odds of having a daughter that is colorblind and has blood type A?

2. Homework Equations

Xc Y IA IB x XC Xc IA i


3. The Attempt at a Solution

Dont know if it is 2/8=1/4 or 2/16=1/8
 
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Since these traits are not linked.. it might be easier to visualize these crosses as separate then bring the information together at the end. If we cross the Xc Y x XC Xc, then (1/4) of the offspring to be colorblind females. If we cross the blood types IA IB x IA i, then we get (1/2) of the offspring to be blood type A.

Since the target phenotype was a female that was color blind AND blood type A... we multiply the answers.

(1/4)(1/2)= (1/8) will be colorblind type A females <----- Final Answer
 
thanks for the explanation...i understand it even better now!
 
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