Download The lactase gene is involved in the breakdown of lactose in the

I) In a population of 100 individuals 33% are of the BB for eye colour. What
percentage is expected to be BG assuming Hardy-Weinberg equilibrium?
A. 9%
B. 21%
C. 49%
D. 51%
E. insufficient information to answer question
Answer: C
q^2 = 33% or .33,
so q = .574 and p=.426.
The genotype BG is heterozygous so the frequency = 2pq = 2(.426)(.574) =
48.9%
II) The lactase gene is involved in the breakdown of lactose in the human, allele +
(p=0.8) functions typically, allele – (in a recessive pattern) causes the development of
lactose intolerance at young adulthood. Assume Hardy-Weinberg equilibrium has been
reached.
1. What is the frequency of all possible genotypes?
In the year 2115, the dairy industry has infiltrated every government in the world (all 2 of
them) and has bribed enough politicians to manipulate the food industry into including
some dairy derivative in all food production.
This all of a sudden reduced the relative fitness of lactose intolerant people to half that of
their pizza-eating peers (not because of any real deleterious effects but rather because it is
nearly impossible to find a date when afflicted by unchecked flatulence).
2. Within 2 generations, what is the allele frequency of lactase – ?
Answers:
#1
p = 0.8
q = 1-p = 0.2
G0 f(+/+) = p^2 = 0.64
G0 f(+/-) = 2pq = 0.32
G0 f(-/-) = q^2 = 0.04
G0 = generation 0 (parental)
#2
w+/+ = 1
w+/- = 1
w-/- = 0.5
G0 Waverage = f(+/+)*w+/+
+
f(+/-)*w+/-
+
f(-/-)*w-/- = 0.98
After 1 generation:
G1 f(+/+) = w+/+ * G0 f(+/+) / Waverage = 0.65
G1 f(+/-) = w+/- * G0 f(+/-) / Waverage = 0.33
G1 f(-/-) = w-/- * G0 f(-/-) / Waverage = 0.02
At G1, q = 0.02 + 0.33/2 = 0.185
G1 Waverage = G1 f(+/+)*w+/+
+ G1 f(+/-)*w+/-
+ G1 f(-/-)*w-/- = 0.99
After 2 generations:
G2 f(+/+) = w+/+ * G1 f(+/+) / Waverage = 0.66
G2 f(+/-) = w+/- * G1 f(+/-) / Waverage = 0.33
G2 f(-/-) = w-/- * G1 f(-/-) / Waverage = 0.01
At G2, q = 0.01 + 0.33/2 = 0.175
Since q is getting smaller with each generation, there seems to be an erosion of genetic
diversity regarding the lactase gene.
III) Individuals who are A/A are considered wild type and their red blood cells remain in
circulation for 120 days. Individuals that are A/a are heterozygous for the sickle cell trait
and their red blood cells function normally, however have a reduced lifespan providing
them resistance against the parasite Plasmodium falciparum whose development requires
an incubation period longer than that provided by the heterozygotes. Individuals with a/a
are affected with sickle cell anemia and if left untreated will die before reaching
reproductive age.
An outbred population within a large village in Botswana is quite frisky and mating
randomly. 1234 individuals genotyped and the following results were obtained 334
individuals were A/A, 764 were A/a, 136 were a/a.
Medical treatment there is weak and only 1/20 a/a individuals will survive to
reproductive age, 10/20 that are A/A will die of malaria before reproducing and all
heterozygotes are healthy.
a) What will be the A and a allele frequencies after the following generation?
b) What will be the allele frequencies at equilibrium?
Answers:
a) In the present generation (G0): f(A)=p={334 +764(0.5)}/1234=0.58
f(a)=q=1-0.58=0.42
f(A/A) = 334/1234= 0.27
f(A/a) = 764/1234= 0.62
f(a/a) = 136/1234= 0.11
W A/A=0.5
W A/a= 1
W a/a= 0.05
G0 Waverage = 0.5*0.27 + 1*0.62 + 0.05*0.11 = 0.76
After one round of selection:
f(A/A) = 0.27*0.5/0.76 = 0.178
f(A/a) = 0.62*1/0.76 = 0.816
f(a/a) = 0.11*0.05/0.76 = 0.007
p= 0.178 + 0.816/2 = 0.585
q= 1-0.585= 0.415
b) At equilibrium p= s/(s+t)
s = selection co-efficient against the a/a genotype (w= 1 - s)
t = selection co-efficient against the A/A genotype (w= 1 - t)
p =0.95/(0.95+0.50) = 0.65
q=1 - 0.65 = 0.35 or (t/s+t)= 0.50/(0.95+0.50)= 0.35
Reference p.629 in textbook
IV) You have two markers linked to QTL on separate chromosomes that contribute to
plant height in cat-tail (Typha latifolia). The RF value for marker 1 and its locus is 0.2;
that for marker 2 and its locus is 0.05. Each marker has two alleles characterized by the
following phenotypic effects (a unit refers to the allele’s contribution to the phenotypic
differences between the pure-breeding lines mentioned below)
M1 = 8 units
m1 = 1 unit
M2 = 4 units
m2 = 0 units
You have two pure-breeding lines, one short, one tall and cross them to generate an F1
which is then allowed to self to produce an F2 segregating for the marker alleles. The
genotype of the tall line is M1/M1;m2/m2 and that of the short is m1/m1;M2/M2
If your aim is to increase plant height which markers should you select for in the F2?
How many units of progress can you make by selecting these markers?
Answers:
You would select progeny that are M1/M1;M2/M2.
The average phenotypic effects of the allele classes are as follows
M1 – 8(0.8)+1(0.2) = 6.6
m1 – 8(0.2)+1(0.8) = 2.4
M2 – 4(0.95)+0(0.05) = 3.8
m2 – 4(0.05)+0(0.95) = 0.2
The phenotypic effect of gametes carrying:
[M1 ; M2] = 6.6+3.8 = 10.4
[m1, m2] = 2.4+0.2 = 2.6
M1/M2 homozygotes = 10.4 + 10.4 = 20.8 height units
m1/m2 homozygotes = 2.6 + 2.6 = 5.2 height units
The difference (20.8-5.2 = 15.6 units) reflects maximum height units of improvement
toward by selectively breeding plants containing M1 and M2 markers.
V) Consider the following pedigree:
I
II
III
IV
?
The effected individuals have a rare (fictional) disorder that involves amnesia, dyslexia,
color blindness, and an extremely acute sense of hearing, called Heinsbergen syndrome.
The wild type allele (H+) is present in the population at a frequency of 0.995. Assuming
that the gene responsible for Heinsbergen is in Hardy-Weinberg equilibrium, what is the
probability that child IV-1 will be affected?
Answer:
If H+ is 0.995 frequent, we can write p=0.995, and q=1-p or 0.005 is the
frequency of the H- allele. From this we calculate the probability that individual III-1 is a
carrier (P(III-1)) as P(H+/H-)=2pq.
I
H+/H-
H+/H-
II
P (H+/H-) = 2pq
=2*(0.995)(0.005)
=0.00995
III
P (H+/H-) = 2/3
IV
?
P (H-/H-) = P (H-/H- | III-1, III-2)P (III-1)P (III-2)
=(1/4)(0.0095)(2/3)
= 0.00166
Probability that the father is a carrier:
Probability that the mother is a carrier:
2/3 (using pedigree information)
0.00995 (using population estimates)
Probability that each will pass on H-:
½*2/3*1/2*0.00995 = 0.00166 (probability of affected child)