Biology 321 Answers to Problem Set 3

BIOLOGY 321 SPRING 2013 ANSWERS TO ASSIGNMENT SET #3
Answers to text questions:
Chapter 2
http://fire.biol.wwu.edu/trent/trent/IGA_10e_SM_Chapter_02.pdf
Answers to Edition 9 text questions:
http://fire.biol.wwu.edu/trent/trent/answers3.9.pdf
Basic Pedigree Problems
Problem 1
a. (3/4) 3
b. (1/3) 3
c. All possibilities except all dominant and all recessive=
1- [(3/4) 3 + (1/4) 3 ] = 64/64 – 28/64 = 36/64
Problem 2
In general, guidelines indicating progeny ratios are not particularly useful because
of the small sample size most pedigrees represent.
autosomal recessive: first part of guideline #2 must be met
autosomal dominant: guidelines #1 and #2 must be met. With respect to #3, the
trait, of course, need not appear in the offspring of an affected individual.
X-linked recessive: guidelines #1 and #4 must be met; #3 will usually be observed
X-linked dominant: guideline #1 must be met
Problem 3
a. No b. Dominant trait are expressed even in heterozygotes
c. Unrelated individuals are unlikely to both be carriers of a rare allele
d. Recessive traits may appear dominant if the trait is common in the population
and there are many heterozygotes in the pedigree. In other words, the trait may
appear in every generation. If a dominant trait exhibits incomplete penetrance or
variable expressivity, it may appear to be inherited as a recessive trait. These
terms will be defined in class next week.
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Problem 4
a--excluded: X-linked dominant
if rare: autosomal dominant is best answer
b--excluded: nothing is absolutely excluded, but if the trait is rare, autosomal
dominant or X-linked dominant are the best answers
c--excluded: X-linked dominant; X-linked or autosomal recessive
only possibility: autosomal dominant
d--excluded: X-linked or autosomal dominant and X-linked recessive
only possibility: autosomal recessive
e--excluded: X-linked dominant or autosomal dominant
X-linked recessive is best answer, but autosomal recessive is possible
f--excluded: autosomal dominant and X-linked dominant
if trait is very rare, X-linked recessive is best answer
Problem 5 (i) a. (ii) b
Problem 6 (i) c. (ii) d
Problem 7
__C___ Autosomal recessive inheritance of the trait; recessive allele is common in the
population ARC
__E__ Autosomal Recessive inheritance (recessive allele is very rare) ARR
__C__ X-linked recessive inheritance (recessive allele is common) XRC
__E__ X-linked dominant inheritance XD
__E__ Autosomal Dominant AD
__E__ X-linked dominant; only males show the trait; dominant allele is very rare XDM
__C__ Autosomal Dominant and only males show the trait ADM
_E__ Y-linked -- the gene specifying the trait is on the Y chromosome Y
ARR excluded because too many unrelated members of the pedigree would have to be het to
make this pedigree work
XD affected individuals have unaffected parents
AD ditto
XDM excluded because of male to male transmission --unrelated female in second generation
(mother of two affected sons) would have to carry the allele
Y -linked excluded because affected males can have unaffected sons and trait is passed via
mother in some generations
Problem 8
.__C _ ARR (note inbreeding)
.__C _ ARC
E XRC
E XRD
__C __ AD
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. The counselor’s was assuming that the trait was autosomal dominant. Since neither parent
exhibits the blue trait, they would then be homozygous wildtype and could not produce a blue
child.
c. This trait could be autosomal recessive. IN this case, there is a possibility that both parents
are heterozygous since each is in direct descent from an affected individual.
Problem 9
E = this mode of inheritance is excluded by the data
C = this mode of inheritance is consistent with the data
__C__Autosomal recessive inheritance of the trait; recessive allele is common in the population
ARC
_E___ Autosomal Recessive inheritance (recessive allele is very rare) ARR
_E___ X-linked recessive inheritance (recessive allele is common) XRC
_E___ X-linked recessive inheritance (recessive allele is very rare) XRR
_E___ X-linked dominant inheritance XD
_C___ Autosomal dominant inheritance AD
Problem 10
• NOTE: the easiest way to work this problem is to figure out all possibilities except 2:2
and then subtract the sum of all these other possibilities from 1. BUT, the weakness in
this approach is that if you have made a mistake in one of the other calculations, your
answer for 2:2 will be wrong as well.
• Alternatively, for 2:2, figure out all possible sibships (birth orders) of 2 affected and 2
unaffected. You can figure this out by trial and error OR (OPTIONAL) learn to use the
appropriate mathematical tool (the binomial) at this site:
http://www.stat.yale.edu/Courses/1997-98/101/binom.htm
dominant recessive probability
4 0 (3/4) 4 81/256
3 1 4 (3/4) 3 (1/4) 108/256 = 42%
2 2 6 (3/4) 2 (1/4) 2 54/256
1 3 4 (3/4) 1 (1/4) 3 12/256
0 4 (1/4) 4 1/256
Problem 11 See class discussion
Problem 12
a. probability that groom is het = 1 probability bride is het = 2/3 X1/2 = 1/3
both het = 1 X 1/3 = 1/3
b. probability that neither are het = 0 since groom has to be het
probability that one is het and the other homozygous = 1-(1/3) = 2/3
or another way to think about it: probability that groom is het = 1
probability bride is not het = 1 -(1/3) = 2/3
c. 1 (probability groom is het) X 1 X 10 -3 X ¼ = 1 x 1/1000 X ¼ = 1 in 4000
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