Genetics 320 (EEB/MCB)  TEST ONE SOLUTIONS


(1)    [10 points]   Two parents who are both carriers (heterozygotes) for a 
recessive disease have a one in four chance that their child has the disease.  
Unfortunately, genetic counselors often hear such parents say this is not a 
problem as they only plan to have three children.   

What is the probability that at least one of their three children show the 
disease?


=1-Prob(none of three have it) 
= 1 - (3/4)(3/4)(3/4) = 1-27/64 = 37/64 = 0.58
 

Suppose their first two child were disease-free.  What is the chance that the
third has the disease?    1/4  

(2)     [10 points]     In class it was mentioned that Mendel likely "cheated" as 
he only considered seven traits, exactly the number of pea chromosomes.  If 
more traits were considered, they would not have shown independent 
assortment.   What is the probability  in peas  of picking seven traits at 
random and having them all fall on different chromosomes (assume that all 
pea chromosomes have an equal number of genes)?
 

= 1* (6/7)*(5/7)*(4/7)*(3/7)*(2/7)*(1/7) 
= 6!/7^6  = 720/117649 = 0.006


#3)  [25 Points]    2000 Gametes from an AbC / aBc individual are scored, with 
the following numbers seen:
ABC	162	aBC	 22		
ABc	165	aBc	661
AbC	657	abC	155
Abc	  20	abc	158

Compute (1) the gene order, (2) the recombination frequency between all sets 
of loci, (3) the coefficient of coincidence, and (4) the interference.  

  
A-B pair:  Recom gametes are AB, ab = 162+165+155+158 = 0.32
A-C pair:  Recom gametes are Ac, aC = 165+20+22+155 = 0.181
B-C pair:  Recom gametes are BC, bc = 162+20+22+158 = 0.181
Suggested order:  ACb/acB.  This predicts the double recombs
as Acb and aCB, which indeed are the rarest classes.
COC = Obser / expected =  (20+22)/(2000*0.181*0.181) = 0.64
interference = 1-COC = 0.36


#4)  [15 Points]     The A and B loci are linked, with recombination frequency 
1/4.  A cross is made with  Ab/aB  as the father,  ab/aB as the mother.

(a)   What are the gamete frequencies for the father?

 
     Ab      (1/2)(1-1/4)   = 3/8           AB  = (1/2)(1/4)  =  1/8
     aB      (1/2)(1-1/4)    = 3/8           ab  = (1/2)(1/4)  =  1/8
 

(b)  What are the gamete frequences for the mother?

      ab     1/2         aB    1/2


(c)  What is the chance of getting an Aabb   offspring?

    Prob(Aabb) = 
   Prob(Ab from dad, ab from mom) +  Prob(ab from dad, Ab from mom)
          = (3/8)(1/2) + (1/8)(0) = 3/16 


(d)  What is the chance of getting a AaBb   offspring?

    Prob(AaBb) = 
      Prob(AB from dad, ab from mom) +   Prob(ab from dad, AB  from mom) +
     Prob(Ab from dad, aB from mon) +   Prob(aB from dad, Ab from mon)
               =      (1/8)(1/2) + 0 + (3/8)(1/2) + 0 = 4/16 = 0.25


#5)    [5 points]    Define both  Penetrance  and expressivity.    How are these 
different?   

Penetrance is the number of individuals with a genotype that
express that particular phenotype

Expressivity is the extent to which that phenotype is seen in 
individuals  displaying the trait.


#6)   [10 points]   A child has blood type O.    There are two possible fathers, Joe 
Nosering (a mega-million dollar rock star) and Joe Sixpack (an unemployed 
truck driver).  The mother (blood  type B) claims Nosering (blood type A) is 
the father.  

a.  Based on this data, can Nosering be excluded?  Why?

No, Nosering could be Ai, mother Bi, so could have an ii offspring


b.  Joe Sixpack had an AB child from a previous marrage.  Can Sixpack be 
excluded?  Why?

No, Sixpack could be  either Ai or Bi.



#7)   [10 points]     Fungal mapping

(a)  Consider the following 1000 linear octads and their frequencies
	1	AAAAaaaa		350
	2	AAaaAAaa		150
	3	aaAAaaAA		150
	4	aaaaAAAA		350

Which octads are first division segregrants?    1, 4  

Which are second division segregrants? 2, 3 

What is the recombination frequency between the marker and centromere?

  = 1/2 Freq(2nd div seg) = (1/2)(150+150)/(1000) = 0.15 

(b)     Suppose the observed frequency of second division segregrants is 2/3. 
What can we say about the recombination frequency?

 Since freq (2nd div seg) cannot exceed 2/3, all we can say is that the
recombination frequency is between 1/2 (unlinked) and 1/3.



#8)     [5 points]    Suppose a parent is a carrier for a dominant autosmal 
disease gene, where D = disease allele, d= normal.  D is closely linked 
(recombination frequency c) to a marker locus with alleles M and m.  If the 
parent has phase DM/md, what is the probability that an M-bearing gamete 
from this parent also carries the D allele?   (Be sure and show how you got 
your result!)


P(DM | M) = P(DM) / P(M) = (1/2)(1-c)/(1/2) = 1-c




#9)   [10 points]    Consider chromosomes 1 and 2 below and the resulting 
translocation between them.	

Draw the chromosome pairing at metaphase in a translocation heterozygote.  
Use arrows to show how an example of how the centromeres could segregate 
to give rise to a set of normal euploid gametes. 

  Euploid gametes are produced if the centromeres N1 and N2
 go to one pole, T1 and T2 to the other.