BSCI 410 
Molecular Genetics 
Fall 2008  

 

Instructor:
Steve Mount 


 

TA:
Zenas Chang 


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Questions about Exam 1 (pre-Exam review).

The following are questions and answers from students regarding Exam 1 (2008) prior to the exam.

Many questions refer to the study sheet. For material that was covered in lecture, I refer you back to the text or to online sources (NCBI books, especially Molecular Biology of the Cell). Use the index of the text to find relevant sections for specific terms.
From a student:

Hi Professor
I have a couple of questions while reviewing for the exam, and I would greatly appreciate if you could help clear up the confusion.
1. On Exam 1, 2006, #1-4, the question concerns complements and the results of a dihybrid cross.  Can you explain how to do the cross to obtain the answers because I am a little confused about the wording of the question. one number?

You cross mice homozygous for cp1 with mice homozygous for cp2. 
That produces F1 mice that are heterozygous at both loci (cp1/+: cp2/+)
When you mate these mice together you get the F2.  The answers are posted
http://www.chemlife.umd.edu/ classroom/BSCI410/HomExam.html
Dihybrid crosses are explained in chapters 2 and 3 of the text; in particular the complementation test is explained on page 62.  

From a student:

Many of the amino acid answers concern conservative changes. What exactly do you mean by this?

When an amino acid is replaced by one with similar chemical properties that is a conservative change.  Conservative changes have higher values on blosum matrices such as blosum62.

From the review sheet:

A woman finds a particular trait of her father's to be extremely annoying. She (of course) marries a man who does not have this trait. She learns by reading an article in the newspaper about genetic research that this common trait is monogenic and is controlled by autosomal recessive alleles at a single locus. The overall allele frequency of the causitive alles in the population is 1 in 3, so about 1 person in 9 shows the trait. What is probability that her child will have this trait?

Assume that she is not homozygous (does not show the trait). First, we need to calculate the probability that her husband is a carrier. From the Hardy-Weinberg distribution, p = 1/2, q = 2/3 and 2pg = 4/9. He is not affected. We can use conditional probability. If A is the event that he is a carrier, and B is the event that he is not affected, AB is the intersection (that he is a carrier and not affected). Since carriers are not affected P(AB) = P(A). So, P(A|B) = P(AB)/P(B) = (4/9)/(8/9) = 1/2. 8/9 is the sum of the carrier and unaffected homozygotes. The probability that he is not a carrier is therefore 1/2 and the desired probability is (1/4)(1/2) where 1/4 is just the Mendelian ratio. So, the probability that her child will have the trait is 1/8.


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