BIO240 Exam #1 Answer key
Professor Jill Carroll
I stated that some of the end of chapter questions or review questions would be seen on the exam – either the
exact same question, or a similar one. For those of you that did not believe me:
Question #38a – on review sheet
Question #39 – similar to question 3.29
Question #40 – similar to question 3.33 (genotypes altered)
Question #41 – on review sheet (observed numbers were changed)
Question #42 – similar to question 4.33
Question #44 – similar concept to question 5.22 from review sheet
Question #45 – question 5.35
Other end of chapter questions or review questions/concepts of them were also seen elsewhere on the exam!
37. Contrast meiosis I and meiosis II.
The same events occur in both stages – Prophase, Metaphase, Anaphase, and Telophase. In Prophase I,
homologous chromosomes pair, and crossing over occurs. In Metaphase I, homologous pairs align at
equatorial plate, and pairs separate during Anaphase I. In Meiosis II, there are no homologous pairs
since the cells at that point are haploid. Chromosomes line up individuals during Metaphase II, and
Anaphase II separates sister chromatids into individual chromosomes (events are Meiosis II are similar
to those of mitosis)
38. Contrast each of the following pairs of concepts:
a. Incomplete penetrance vs. variable expressivity
Incomplete penetrance indicates than less than 100% of individuals with the same genotype will
express the expected phenotype. For example: even though polydactyly is a dominant trait, some
heterozygous individuals will have the normal number of digits. Their dominant allele, however,
can be expressed in their offspring.
Variable expressivity indicates that a single genotype can yield a range of phenotypes, from mild
to severe. For example, if CB = black, and CW = white, CBCW individuals can exhibit colors from
pale gray to almost black.
b. Sex-linked trait vs. sex-influenced trait
Sex-linked traits are genes that are located on a sex chromosome, which often is different
between the sexes. For example, white-eye color in fruit flies is an X-linked recessive trait.A
female must have two white-eyed alleles in order to have white eyes; males only have a single X
chromosome – and expressed whichever ONE allele they have (in the XX-XY determination
system)
Sex-influenced traits are autosomal genes – both males and females have two copies of then
gene. The EXPRESSION of the heterozygote condition varies from male to female. In one sex
(usually the male), the trait is dominant, but it is recessive in the female. For example, horns is
dominat in males, but recessive in females. A Hh male would display horns, but a female with the
same genotype would be hornless.
39. In the jimsonweed, purple flower (P) is dominant to white (p) and spiny pods (S) are dominant to smooth
(s). In a cross between a homozygote for white flowers and spiny pods and a homozygote for purple
flowers and smooth pods, determine the phenotype of: (show your work)
a. The F1 generation
Since the P generation is ppSS x PPss, 100% of the F1 will be purple, spiny (PpSs)
b. The F2 generation
The F2 generation will have an expected yield of 9 purple spiny: 3 purple smooth: 3 white spiny:
1 white smooth
Gametes
PS
Ps
pS
ps
PS
Ps
pS
Ps
PPSS
Purple spiny
PPSs
Purple spiny
PpSS
Purple spiny
PpSs
Purple spiny
PPSs
Purple spiny
PPss
Purple smooth
PpSs
Purple spiny
Ppss
Purple smooth
PpSS
Purple spiny
PpSs
Purple spiny
ppSS
white spiny
ppSs
white spiny
PpSs
Purple spiny
Ppss
Purple smooth
ppSs
white spiny
ppss
white smooth
c. Progeny of a cross of the F1 back to the white, spiny parent
PpSs x ppSS
Expected ratio is 1 purple spiny: 1 white spiny
Gametes
pS
PS
Ps
pS
ps
PpSS
Purple spiny
PpSs
Purple spiny
ppSS
white spiny
ppSS
white spiny
40. In cucumbers, dull fruit (D) is dominant over glossy fruit (d); orange fruit (R) is dominant over cream fruit
(r); bitter cotyledons (B) are dominant over non-bitter cotyledons. The three characters are encoded by
genes located on different pairs of chromosomes. A DdRRbb individual is crossed with a DdrrBb
individual. Give the expected proportions of phenotypes among the progeny of this cross. (show your work
either by Punnett square or branch diagram)
Punnett square
DrB
Drb
drB
drb
DRb
dRb
DDRrBb
Dull,
orange,
bitter
DDRrbb
Dull,
orange,
non-bitter
DdRrBb
Dull,
orange,
bitter
DdRrbb
Dull,
orange,
non-bitter
DdRrBb
Dull,
orange,
bitter
DdRrbb
Dull,
orange,
non-bitter
ddRrBb
glossy,
orange,
bitter
ddRrbb
glossy,
orange,
non-bitter
Branch
¾ D_
½ B_ =
3/8 D_R_B_ dull, orange, bitter
½ bb
3/8 D_R_bb dull, orange, non-bitter
1 R_
¼ dd
=
½ B_ =
1/8 ddR_B_ glossy, orange, bitter
½ bb
1/8 ddR_bb glossy, orange, non-bitter
1 R_
=
41. Some flies have white eyes and others red eyes. The difference in eye color is due to inheritance of one
gene. Two true-breeding flies are mated, with the female white-eyed and the male red-eyed. The F1 are ½
red eyed and female, and ½ white-eyed and male. (There are no white-eyed females, or red-eyed males.)
The F2 includes:
Phenotypes
53 red-eye females
45 white-eyed females
42 red-eyed males
60 white-eyed males
Observed
Number (o)
Expected
Number (e)
d
(o – e)
d2
d2 / e
53
50
3
9
0.18
45
50
-5
25
0.5
42
50
-8
64
1.28
60
200
50
10
100
2.0
3.96
Red-eye
female
White-eye
female
Red-eye male
White-eye
male
Total
2 = _3.96_______
Degrees of freedom (df) = __3____
_0.5___ > P > _0.1___
a. Complete the 2 table.
(For the exam, you would be provided with a copy of the 2 probability table.)
Based on the information – there are no white eyed females and no red eyed males in F1. Any
differences seen between the sexes such as these indicate that this is an X-linked trait. Let XR = red and
XW = white. The F1 females are XRXW, and the F1 males are XWY. Such a cross will yield the following
expected ratio for the F2 generation: 1 red eyed female:1 white eyed female: 1 red eyed male: 1 white
eyed male.
XW
Y
XR
XRXW
Red female
XRY
Red male
XW
XWXW
Whitefemale
XWY
White male
b. Write the null hypothesis (H0).
The is no difference between the observed ratio and the expected ratio of 1 red eyed female: 1 white eyed
female: 1 red eyed male; 1 white eyed male.
NOTE: The null hypothesis is NOT that observed is due to random chance (this is a genetics class,
afterall) – it is that the DIFFERENCE between observed and expected values is due to chance.
c. Briefly interpret of the results of the 2 test. In your answer, be certain that you identify the mode of
inheritance for the white eye color of flies.
The null hypothesis can be accepted since the P value falls between 0.1 and 0.5. This means that
there is a 10%-50% probability differences observed is just due to random chance. White eyes is an X
-linked recessive trait, with red eyes being the dominant form.
42. Pink toe pads (T) is an autosomal trait that is DOMINANT to black toe pads (t). Round ears (XE) is an Xlinked trait that is dominant to pointy ears (Xe). You cross a female rat with pink toe pads and pointy ears to
a male rat with black toe pads and round ears. The F1 progeny all have pink toe pads. What is the
genotype of parental generation? What is the genotype of the F1 progeny? If the F1 are crossed to produce
F2 progeny, what proportions of phenotypes is seen? Show your work.
P generation ♀ TTXeXe (must be homozygous TT since all offspring have pink toes) x ♂ ttXEY
F1 generation ♀ TtXEXe (pink toes round ears) x ♂ TtXeY (pink toes pointy ears)
F2 generation
Punnett square
TXe
tXe
TY
tY
For daughters:
3/8 pink round
3/8 pink pointy
1/8 black round
1/8 black pointy
TXE
TXe
tXE
tXe
TTXEXe
Pink
round
female
TtXEXe
Pink
round
female
TTXEY
Pink
round
male
TtXEY
Pink
round
male
TTXeXe
Pink
pointy
female
TtXeXe
Pink
pointy
female
TTXeY
Pink
pointy
male
TtXeY
Pink
pointy
male
TtXEXe
Pink
round
female
ttXEXe
black
round
female
TtXEY
Pink
round
male
ttXEY
black
round
male
TtXeXe
Pink
pointy
female
ttXeXe
back
pointy
female
TtXeY
Pink
pointy
male
ttXeY
black
pointy
male
For sons:
3/8 pink round
3/8 pink pointy
1/8 black round
1/8 black pointy
Branch diagram ♀ TtXEXe (pink toes round ears) x ♂ TtXeY (pink toes pointy ears)
¼ XEXe
3/16 pink round female
¼ Xe Xe
3/16 pink pointy female
¼ XEY
3/16 pink round male
¼ XeY
3/16 pink pointy male
¼ XEXe
1/16 black round female
¼ Xe Xe
1/16 black pointy female
¼ XEY
1/16 black round male
¼ XeY
1/16 black pointy male
¾ T_
¼ tt
43. In Drosophila, a mutant strain has plum-colored eyes. A cross between a plum-eyed male and a plum-eyed
female gives 2/3 plum-eyed and 1/3 red-eyed (wild type) progeny flies. A second mutant strain of
Drosophila called stubble, has short bristles instead of the normal long bristles. A cross between a stubble
female and a stubble male gives 2/3 stubble and 1/3 normal bristle flies in the offspring. Assuming that the
plum gene assorts independently from the stubble gene, what will be the phenotypes and their relative
proportions in the progeny of a cross between two plum-eyed, stubble-bristled flies? Describe the
inheritance pattern of plum eyes and stubble bristles. (Both genes are autosomal). Show your work.
With the information given, both plum-colored eyes and stubble bristles are lethal alleles in the
homozygous condition(which is why the offspring are in ratios of thirds instead of quarters). Therefore,
you would not see homozygotes for either plum eyes or stubble bristles in surviving offspring.
Let P = plum eyes, p = red/wild type eyes
S = stubble bristles, s = normal (long bristles)
P generation PpSs x PpSs
Either a Punnett square or branch diagram could be used (personally, I would have gone the branch
diagram route – a lot faster to do, and since the ratios of offspring are given, you could have done the
calculations even if you didn’t realize the lethal pattern!) The problem states that both traits are
AUTOSOMAL, so NO sex chromosomes should be used.
Branch Diagram
2/3 Ss
= 4/9 PpSs plum, stubble
1/3 ss
= 2/9 Ppss plum, normal
2/3 Ss
= 2/9 ppSs red, stubble
1/3 ss
= 1/9 ppss red, normal
2/3 Pp
1/3 pp
Punnett Square
PS
Ps
pS
ps
PS
PPSS
Lethal
PPSs
Lethal
PpSS
Lethal
Ps
PPSs
Lethal
PPss
Lethal
pS
PpSS
Lethal
Ps
PpSs
Plum
stubble
PpSs
Plum
stubble
Ppss
Plum
normal
PpSs
Plum
stubble
ppSS
Lethal
PpSs
Plum
stubble
Ppss
Plum
normal
ppSs
red
stubble
ppss
red
normal
ppSs
red
stubble
44. You are studying body color in an African spider and have found that it is controlled by a single gene with
four alleles: B (brown), br (red), bg (green), and by (yellow). B is dominant to all the other alleles, and by is
recessive to all the other alleles. The bg allele is dominant to by but recessive to br.
a. You cross a pure-breeding brown spider with a pure breeding red spider. Diagram the cross and
predict the genotype and phenotype of the progeny.
BB x brbr
B
B
br
br
Bbr
Bbr
Bbr
Bbr
100% brown offspring
b. You cross a pure-breeding brown spider with a pure-breeding green spider. Diagram the cross and
predict the genotype and phenotype of the progeny.
BB x bgbg
B
B
bg
bg
Bbg
Bbg
Bbg
Bbg
100% brown offspring
c. You cross the progeny from parts (a) and (b), Diagram the cross and predict the genotype and
phenotype of the progeny.
Bbr x Bbg
B
br
B
bg
BB
Bbr
Bbg
brbg
75% brown offspring:25% red
d. You cross the non-brown progeny from part (c) to a pure-breeding yellow spider. Diagram the cross
and predict the genotype and phenotype of the progeny.
brbg x byby
br
bg
50% red offspring:50% green
by
brby
bgby
45. In some goats, the presence of horns is produced by an autosomal gene that is dominant in males and
recessive in females. A horned female is crossed with a hornless male. The F1 offspring are then
intercrossed to produce the F2 generation. What are the phenotypic proportions of the F2 generation? Show
your work.
Since horns is dominant in males, a male needs only one copy of the allele to have horns; in females,
horns is recessive, so a female must be homozygous for the horn allele to exhibit horns. (The problem
states that the trait is AUTOSOMAL, so NO sex chromosomes should be used)
“Horn” = horns
“Less” = no horns
P generation ♀horns/horns x ♂ less/less
F1 generation 100% horns/less (100% of sons will have horns; 100% of daughters will be hornless)
F2 generation genotypes:
25% horns/horns
50% horns/less
25% less/less
Phenotypes ♀ 75% hornless:25% horns
♂ 75% horns:25% hornless