Fall 2004
Animal Science 213 Animal Genetics
EXAM 3
125 Points
Exam must be completed in INK!
Name:__KEY________________________
Multiple Choice (4 points each)
1. If an individual plant homozygous dominant for the single gene tall (DD) is mated to an
individual homozygous recessive for the dwarf phenotype (dd), what proportion of the offspring
will be tall?
a. 0%
b. 25%
c. 50%
d. 75%
e. 100%
2. In the cross mentioned above, how many of the offspring will be heterozygous?
a. 0%
b. 25%
c. 50%
d. 75%
e. 100%
3. Flower color, is controlled by alleles at the W locus (W red, w white). The results of a WW x
ww cross are pink F1 flowers. When the F1 plants with pink flowers are self-crossed, the
proportions of the resulting F2 generation are 25% red, 50% pink, 25% white. What is the
relationship between the W and w alleles?
a. W is dominant to w
b. w is dominant to W
c. W and w are partially dominant
d. W and w are codominant
4. If heterozygous tall plants with pink flowers (DdWw) are self-crossed, what proportion of the
offspring will be tall with pink flowers?
a. 25%
b. 37.5%
c. 50%
d. 75%
e. 100%
5. In Labrador retrievers, two of the loci controlling coat color (black, chocolate and yellow) are
the E locus and the B locus. At the B locus, black (B) is dominant to chocolate (b). At the E
locus, homozygous recessive alleles at the E locus (ee) results in yellow pups regardless of
genotype at the B locus.
a. E is epistatic to B
b. B is epistatic to E
6. If a chocolate lab is mated to a yellow lab, is it possible to have black pups?
a. Yes (if you went with the information from class; there is a third allele)
b. No
c. Not enough information (if you went strictly off the information from question 5.
7. Research at an institution in the US has isolated a mutation in mice that results in the absence
of a tail in homozygous recessive individuals. At the same time, researchers in the Netherlands
also identify a recessive tailless mutation. The researchers exchange breeding stock and cross US
tailless mice with Netherlands tailless mice. The resulting offspring all have tails. Do the
mutations complement each other?
a. Yes
b. No
c. Not enough information can be generated from one mating.
8. Which of the following is an example of a temperature-sensitive effect?
a. phenylketonuria
b. Tay-Sach’s disease
c. Point (snout, ear and paw) color in Himalayan rabbits
d. None of the above
9. Which of the following is an example of a nutritionally sensitive effect?
a. phenylketonuria
b. Tay-Sach’s disease
c. Point (snout, ear and paw) color in Himalayan rabbits
d. None of the above
10. When a gene that is normally expressed is “turned off” when it is moved to a portion of the
chromosome that is largely heterochromatin, this is an example of a
a. tumor suppressor
b. permanent mutation
c. heterochromatin translocation
d. positional effect
11. The (2n + 1) rule allows an individual to calculate the number of
a. alleles controlling a trait
b. the number of genes controlling a trait
c. the predicted outcome of a genetic cross
d. none of the above
12. Traits exhibiting continuous variation are often
a. polygenic
b. quantifiable
c. influenced by additive alleles
d. a and b
e. a and c
f. all of the above
13. Erythroblastosis fetalis is a condition caused by incompatibility of
a. secretor status
b. parental status
c. ABO status
d. Rh factor status
14. Lethal alleles can be which of the following?
a. dominant
b. recessive
c. conditional
d. all of above
15. From a dihybrid cross, it was concluded that the 2 genes were linked with a recombination
frequency of 20 percent. How many of the resulting offspring were nonparental (recombinant)?
a.
b.
c.
d.
e.
40%
80%
20%
10%
60%
16. Which of the following is NOT a characteristic of Mendelian genetics?
a.
b.
c.
d.
independent segregation
simple dominance/recessiveness
linkage
genes occur in pairs
True/False (1 point each)
___F__ 17. Phenotype always reflects genotype.
___T__ 18. It is possible to have more than two alleles for one locus.
___T__ 19. Mitochondria and chloroplasts replicate their own DNA
___T__ 20. Traits that segregate independently are not linked.
___T__ 21. Mendel’s “unit factors” are now known to be genes.
___T__ 22. The alleles for the genes controlling presence of the A/B/0 antigens are codominant
___T__ 23. Substance H is involved in blood type
___T__ 24. An individual may have no more than two alleles for a locus.
___T__ 25. The greater the distance between 2 linked genes, the greater the recombination
frequency.
___F__ 26. If two genes are following Mendelian patterns of inheritance, the genes must be
linked.
Problems (points as assigned)
27. (4 pts) In Angus cattle, two coat colors are possible, red and black, controlled by a single
locus with two alleles (B black, b red). Over several breeding seasons, a black bull, A286, was
mated to 200 red cows, who gave birth to 104 F1 black calves, 96 F1 red calves. For each color,
half were heifer calves. When the 100 heifer calves grew to maturity, they were all mated to a
red bull. The 52 black F1 cows gave birth to 25 black F2 calves and 27 red F2 calves. The 48 F1
red cows gave birth to 48 F2 red calves.
a. Based on the results of the above matings, what is the relationship between B and b?
B is dominant to b.
b. What is the genotype of the original black bull, A286?
Bb
28. (4 pts)The presence of scurs, which are small, moveable pseudohorns is another trait
controlled by two alleles at a single locus in cattle (Sn normal, Sc scurred). A286 possessed scurs,
but none of the 200 P1 cows did. Of the F1 calves, all of the bull calves were scurred, but none of
the 100 heifer calves were scurred. The red bull mated to the F1 heifers was also scurred. When
the F2 calves were born, there were 50 each bulls and heifers. All of the bull calves were scurred,
but only 25 of the heifer calves were scurred.
a. Is the scurs trait sex-limited? No
b. Is the scurs trait sex-influenced? Yes
c. Is scurs dominant in males? Yes
d. Is scurs dominant in females? No
29. (6 pts) The developmental mutation, bicoid (bcd-) is recessive in mice. Normal bicoid protein
must be present for embryos to form normally; if bicoid is not produced, the embryos will die.
However, development can occur in homozygous recessives. Consider a cross between a female
heterozygote (bcd+/bcd-) and a male homozygote bcd-/bcd-).
a. How can males homozygous for the mutation survive?
There is a maternal effect. Male embryos contain normal bicoid mRNA from the egg, which
is then translated into normal bicoid protein.
b. Predict the outcome (normal vs. failed [lethal]) embryogenesis in the F1 offspring of the
cross above.
All of the embryos will develop normally because of the maternal contribution of bicoid
mRNA then translated into bicoid protein.
c. If the F1 offspring are crossed, predict the outcome (normal vs. failed [lethal])
embryogenesis in the F2. (Describe the genetic conditions that would allow survival,
those that would not. No more than two short sentences are sufficient to answer this)
Any embryo born to a bcd-/bcd- mother would fail (no normal bicoid mRNA to contribute).
Any embryo born to a mother with at least one copy of bcd+ would develop normally.
30. (4 pts) Describe an example of infectious heredity in flies.
There were two examples given in class. The first was a sensitivity to CO2 anesthesia in flies
infected with the sigma virus. Infected flies do not recover from anesthesia while uninfected flies
do. The sensitivity and the virus are passed on to subsequent generations.
The second example is a temperature-sensitive alteration in sex ratios of fruit flies born to
females infected with a protozoan. When reared at room temperature, there is a predominance of
females born. The condition is passed only to females.
31. (4 pts) Below is a table showing inheritance of leaf pigmentation in a particular variety of
four o’clock plants.
Ovule Source
Pollen Source White Green Variegated
White
White Green W, G, var
Green
White Green W, G, var
Variegated
White Green W, G, var
a. What source (ovule or pollen determines the color of leaves in this variety of four
o'clocks? Ovule
b. What is the source of DNA controlling leaf pigmentation?
Chloroplast (I will accept organelle)
32. (8 pts) The diagram on the next page is highlighting the key steps of meiosis, focusing on a
single chromosome and 2 genes (A and B) that are linked on that chromosome. Using what
you’ve learned about linkage and the process of meiosis, complete the following:
(2 pts) In the blanks next to each primary germ cell, write the genotype.
These answers are written in on the separate Powerpoint document
(1 pt) Identify (draw an arrow to) the tetrad structure.
These answers are written in on the separate Powerpoint document
(.5 pt each) Fill in the gametic genotypes, and identify what percentage of each would result
(recombination frequency = 20%).
These answers are written in on the separate Powerpoint document
(1 pt) The mating illustrated in this scenario, where one individual is mated with an all-recessive
individual (recessive phenotype for all traits of interest) is known as what type of cross?
Test cross
33. (10 pts) The genes for body type and coat color are 15 map units apart. Dwarf (dw) body
type is recessive to long (wild-type) body; likewise, albino (a) coat is recessive to brown (wildtype) coat. A normal individual (long body, brown coat) is mated to an albino, dwarf individual,
and all of the F1 offspring are wild-type.
Note: Use the horizontal line(s) format to show genotypes, for example: _dw_______a_
(you can use “DW” and “A” instead of “+” if you prefer)
+
+
(2 pts) What is the genotype of the normal individual in this mating?
(2 pts) What is the genotype of the F1 offspring?
+
+
+
+
+
+
dw
a
(2 pts) Explain why recombination does not have an effect on the outcome of this mating.
Both parents are homozygous at both loci.
Next, an F1 female from the mating above is crossed with a recessive (dwarf, albino) male to
generate the F2 offspring.
(2 pts) What percentage of the F2 offspring will be dwarfed with brown coat?
7.5%
(2 pts) What are the recombinant genotypes? +
a
and
dw
+
dw
a
dw
a
34. (12 pts) 3-point mapping problem. Three recessive traits in fruit flies are being studied, short
wings (sw), black body (b), and yellow eyes (y). The wildtype (dominant) traits for each gene
are long wings, grey body, and red eyes. In order to set up an informative test cross, an initial
mating was conducted between a short-winged female (wildtype for other 2 traits, and
homozygous at each locus) and a black-bodied, yellow-eyed male (wildtype for wing type,
homozygous). All of the resulting F1 offspring were wild type, as expected. Using horizontal
line format to indicate genotypes (as in previous problem), diagram this cross and the resulting
F1 (4 pts).
sw
+
+
sw
+
+
x
+
b
y
+
b
y
NOTE: The order of genes does NOT matter at this point.
F1:
sw
+
+
+
b
y
Subsequently, an F1 female was mated to a male recessive for all 3 traits. Out of 1000 offspring
counted, the following numbers were observed:
F2 phenotype
Black body, short wings
Black body
Short wings
Yellow eyes
Short wings, black body, yellow eyes
Black body, yellow eyes
Short wings, yellow eyes
Wildtype
(4 pts) What is the correct gene sequence?
28
146
310
32
3
322
154
5
b sw y OR y sw b (sw is in the middle)
(4 pts) What is the map unit distance between the body type (b) gene and the wing type (sw)
gene?
6.8 map units