Name __________________________________
Bio 3070 Genetics F06
Examination 1
Dr. E. M. Golenberg
October 9, 2006
Write your name on every page of this exam. You have 14 problems on this test.
Please read the questions carefully and understand what is asked of you. Organize your
thoughts and your work, so that the graders can clearly see what you are doing. Answer
all question on these pages. If you need more room, write on the back of the same page.
Put your name on all of the pages. If you come to a question and you don’t know how to
attack it immediately, don’t panic. Skip the question and go on. When you are done with
the rest of the test come back to it. You can get partial credit. Any evidence of cheating
will result in your failing this exam. If you need it for the exam the following are
theoretical chi-squared values. (20.05, df=1= 3.84; 20.05, df=2= 5.99; 20.05, df=3= 7.82; 20.05,
df=4= 9.49)
The first 9 problems are short answer problems and are worth 5 pts. or 10 pts. each. Each
can be answered by a single sentence and or calculation.
1. Define pleiotropy. (5 pts.)
Pleiotropy is the condition where a single gene directly influences the phenotype of
several traits or characters in an individual.
2. The following pedigree is for myopia in humans. Determine the most probable
genotype for each individual based on your prediction of the pattern of inheritance
(dominant/recessive, autosomal/X-linked).(10 pts.)
Autosomal recessive. aa A-(probably AA)
Aa Aa
Name __________________________________
Aa
Aa
Aa
Aa
A- aa A-
aa
A- A-
3. Using the binomial distribution equation, determine the probability of a family of
five offspring having two children afflicted with sickle-cell anemia (an autosomal
recessive disease) when both parents are heterozygous for the sickle-cell allele..
(5 pts.)
5!/2!3! (.25)^2 (.75)^3=0.264
4. You cross a true-breeding colorless individual with a true-breeding red individual.
All F1 offspring are red. When you cross two F1 individuals, you get 70 reds and
58 colorless. The reciprocal cross gives similar results. Test whether this
distribution fits a 3:1 ratio. (10 pts.)
X2= (96-70)2/96 + (32-58)2/32=7.04+21.13=28.2 df=1. Significantly differs from
expected. Reject 3:1 ratio.
5. You set up the following cross: AA Bb Cc Dd Ee Ff X Aa Bb cc Dd Ee ff. All
genes are autosomal and unlinked. BB is lethal. D and d are codominant. What
proportion of the living offspring are expected to have a dominant A, dominant B,
dominant C, codominant D (heterozygous) , recessive E, and recessive F
phenotype? (5 pts.)
Name __________________________________
1 X 2/3 X 1/2 X 1/2 X 1/4 X1/2=1/48
6. Assume that A and C are linked and are separated by 10 cM. Answer problem 5
again. (10 pts.)
This would mean that from the first parent, we would get gametes AC or Ac 90 % of the
time. However, the recombinants would also be Ac or AC. The same goes for the other
parent in terms of the effect of recombination. The ratio remains the same.
7. At what stage are chiasmata first seen? (5 pts.)
Diplonema
8. Define bivalent. (5 pts.)
A bivalent is a structure that is a pair of homologous chromosomes found in meiosis I.
9. In a cross between a dihybrid individual and a tester, the offspring were of the
following phenotypic classes (A means dominant A trait, a means recessive; B
means dominant B trait, and b means recessive). 117 A B. 109 a b, 120 A b, 108
a B. Are the loci linked and if so what is there map distance? (5 pts.)
The loci are unlinked.
Name __________________________________
Name __________________________________
10. In Drosophila, a heterozygous female for the X-linked recessive traits a, b, and c was
crossed to a male that phenotypically expressed a, b, and c. The offspring occurred in the
following phenotypic ratios:
+ b c
460
a + +
450
a b c
32
+ + +
38
a + c
11
+ b +
9
No other phenotypes were observed.
a. What is the genotypic arrangement of the alleles of these genes on the X
chromosome of the female? (5 pts.)
b. Determine the correct sequence and construct a map of these genes on the X
chromosome. (15 pts.)
c. What progeny phenotypes are missing? Why? (10 pts.)
a. + b c and a + +
b. The gene order is a b c. The distance between a and b is 7 mu. The distance
between b and c is 2 mu.
c. The double crossing over class ++c and ab + are missing. The expected number is
1.4. There may be interference or the sample size is too small.
Name __________________________________
11. Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three
pairs of autosomes, referred to as chromosomes 2, 3, and 4. A genetics student
discovered a male fly with very short legs. Using this male, the student was able to
establish a pure breeding stock of this mutant and found that it was recessive. She then
incorporated the mutant into a stock containing the recessive gene black (body color
located on chromosome 2) and the recessive gene pink (eye color located on
chromosome 3). A female from the homozygous black pink short stock was then mated
to a wild-type male. The F1 males of this cross were all wild type and were then
backcrossed to the homozygous b p sh females. The F2 results appeared as shown in the
table that follows. No other phenotypes were observed.
Wild Pink Black Short Black Pink Short
Females 63
58
55
69
Males
59
65
51
60
a. Based on these results, the student was able to assign short to a linkage group (a
chromosome). Which one was it? Include a step-by-step reasoning. (15 pts.)
b. The student repeated the experiment, making the reciprocal cross, F1 females
backcrossed to homozygous b p sh males. She observed 85 percent of the
offspring fell into the given classes (above), but 15% or the offspring were
equally divided among b + p , b + +, + sh p and+ sh + phenotypic males
and females. How can these results be explained and what information can be
derived from the data? (15 pts.)
a. Black and short always appear together or wildtype with wildtype. In terms of pink,
you have 123 pink –wildtype, 106 wildtype- short, 122 wildtype-wildtype, and 129 pink
short. This is close to independent assortment. There is no sex-specific distribution.
Males don’t recombine. Altogether, short must be on the same chromosome as black,
chromosome 2.
b. The distance between black and short must be 15 map units. (There is recombination
in the females.)
Name __________________________________
12. In birds, the male contains two like sex chromosomes and is designated ZZ. The
female contains one Z as well as a nearly blank chromosome designed W (females are
ZA). In parakeets, two genes control feathe color. The presence of the dominant Y allele
at the first gene results in the production of a yellow pigment. The dominant B allele at
the second gene controls melanin production. When both genes are active, a green
pigment results. If only the Y gene is active, the feathers are yellow. If only the B gene is
active, a blue color is exhibited. If neither gene is active, the birds are albinos.
Therefore, with our conventional designation, phenotypes are produced as follows:
Y- Bgreen
Y- bb
yellow
yy Bblue
yy bb
albino
A series of crosses established that one of the genes is autosomal and one is Zlinked. Based on the results of the following cross shown here, where both parents are
true-breeding, determine which gene is Z-linked.
P: green male X albino female
F1:
1/2 green males, 1/2 green females
F2:
6/16 green males
2/16 yellow males
3/16 green females
1/16 yellow females
3/16 blue females
1/16 albino females
Support your answer by establishing the genotypes of the P parents and working the cross
through the F2 generation. (30 pts.)
You can tell right away from the F2 males that Y is on Z. All males get the Y from the
mother, but not always the B. This implies that the mother is heterozygous for Bb which
means it must be autosomal. The crosses and genotypes are as follows:
P: green male YY BB X albino female yW bb
F1:
1/2 green males Yy Bb, 1/2 green females YW Bb
F2:
6/16 green males Y- B 2/16 yellow males Y - bb
3/16 green females YW B1/16 yellow females YW bb
3/16 blue females yW B1/16 albino females yW bb
Name __________________________________
13. Pigeons may exhibit a checkered or plain pattern. In a series of controlled matings,
the following data were obtained:
F1 Progeny
P Cross
Checkered
Plain
a) checkered X checkered
36
0
b) checkered X plain
38
0
c) plain X plain
0
35
Then, F1 offspring were selectively mated with the following results (the P cross giving
rise to each F1 pigeon is indicated in parentheses):
Progeny
F1 X F1 Crosses
Checkered
Plain
checkered (a) X plain (c)
34
0
checkered (b) X plain (c)
17
14
checkered (b) X checkered (b)
28
9
checkered (a) X checkered (b)
39
0
How are the checkered and plain patterns inherited? Select and define symbols for the
genes involved, and determine the genotypes of the parents and offspring in each cross.
(25 pts.)
The trait is a simple one gene two allele system with Checkered (Ch) dominant over plain
(ch ch ).
a. Ch Ch X Ch Ch
b. Ch Ch X ch ch
c. ch ch X ch ch
d. Ch Ch X ch ch
e. Ch ch X ch ch
f. Ch ch X Ch ch
g. Ch Ch X Ch ch
Name __________________________________
14. During oogenesis in an animal species with a haploid number of 6, one dyad
undergoes nondisjunction during meiosis II. Following the second meiotic division, the
involved dyad ends up intact in the ovum. How many chromosomes are present in (a) the
mature ovum, (b) the second polar body, and (c) the first polar body? (d) Following
fertilization by a normal sperm, how many chromosomes are present? (e) What
chromosomal condition is created? (25 pts.)
a.
b.
c.
d.
e.
7
5
6
13
trisomy.