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Section
Studies in Genetics
Introduction
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which these plants are growing were produced
from a cross of hybrid parents (Gg x Gg).
Genetic studies show that characteristics are
passed on from parent to offspring in
reproduction.
Questions
Objectives
I. How many plants are growing?
Having completed the lab on genetics, the
student will be able to:
2. How many plants are green?
1. Define and apply terms such as genotype,
phenotype, homozygous, heterozygous,
dominant, recessive, codominant, and sex­
linkage.
_
3. How many plants are albino?
4. Using a Punnett Square, determine the
expected ratio in a cross (Gg x Gg) producing
green and albino plants:
G
2. Solve genetic problems involving
monohydrid cross with dominant and
recessive traits, codominant traits, and sex­
linked traits.
3. Solve dihybrid problems.
Monohybrid Cross in Corn
Introduction
When one pair of characteristics in an
individual is crossed, the individuals
possessing mixed genes are called
monohybrids.
The presence of chlorophyll in plants is due to
a dominant gene; the lack of chlorophyll
(albinism) is due to a recessive gene.
Materials
Tray of corn seedlings showing a hybrid cross
for green and albino plants.
Procedure
Examine a tray of corn seedlings showing
both green and albino plants. The seeds from
Green
Albino~
5. Was there a difference between the
expected ratio and the ratio of plants growing
in the tray?
Why? or why not?
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6. Why would albinism in plants be
considered a "lethal gene factor"?
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Monohybrid Genetic Problems
2. A man with normal pigmentation, whose
father was an albino, marries an albino. What
chance is there that their first child will be an
albino?
1. In humans, normal pigmentation (A) is
dominant to albinism (a). Set up a Punnett
Square to determine the phenotype and
genotype of the offspring for each of the
following:
a. AA x aa:
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Phenotype:,_If--:.....-_ _
Genotype:
3. A man with normal pigment marries a
woman with normal pigment. They have an
albino child. What are the genotypes of all
(three individuals?
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b. Aa x aa:
c. Aa x Aa:
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Woman
At{..
Child
aa
Cl~
-
11 : i
Phenotypic ratio:J
Genotypic ratio: 1.M:
11«..-.
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4. Some people are able to detect the bitter
taste of phenylthiocarbamide (P.T.c.) while
other people are unable to taste it. The
following diagram representing a family,
some of whom are "tasters" and some of
whom are "non-tasters". The squares
represent males; the circles represent females.
The shaded squares and circles represent
"tasters".
Phenotypic ratio.:_~1-f/ic...:......:.:~1~a..,.,-~_
Genotypic ratio: t Aa.,: L &<.-Q.....
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nty§
tt 11't
Man
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t It.t:t/
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d. aa x aa:
qq
Phenotype:,_t(,,
__
Genotype:
a. Under the symbols for each person write
their phenotype: "taster" or "non"
q q..,
T =T;;5 I EI(,
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c. Give the phenotype (Normal=N and
disorder=n) and genotype for each individual
in the following sets of parents:
b. Which is the recessive trait?
A/oN- rr17TI3[L
c. Above the symbol for each person, write
their genotype (IT, Tt, or tt)
A
Couple 1: Phenotype
Genotype
A
y
Aa
X
A C(.
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Couple 2: Phenotype CA:-- 4
Genotype aa. y4~
5. In chickens rose comb (R) is dominant over
single comb (r). A rose-combed male is
mated to a rose-combed female. Eighteen
chicks are produced, ten of which are rose­
combed and eight of which are single-combed.
What are the genotypes of the parents?
Couple 3: Phenotype ,4/".4
Genotype /lcrr X
dO\­
7. The diagram below represents a family
history for a particular genetic disorder. The
open squares and circles represent "nafflicted
individuals, the shaded squares and circles
represent afflicted individual .
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Male_R--'LL
_
Female
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6. Using the family history below, answer
these questions: (Shaded symbols represent a
genetic disorder)
generation?
Couple -I
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Using the pedigree, answer the questions:
a. Is this disorder expressed in every
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b. Can two people phenotypically normal
produce a child that is afflicted with this
disorder?
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c. In the mating of two afflicted individuals
are there any children that are not afflicted
with this disorder? 11/.1 /I a..-- . /la."
a. Is this disorder carried as a dominant or
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recessi ve characteristic? ?
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d. Is this affliction a dominant or recessive
trait?
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b. What evidence supports your answer to
question a?
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8. Andalusian chickens may be either black,
white, or gray. The gene for black is not
dominant over the gene for white, nor is the
gene for white dominant over the gene for
black. When a black rooster is mated to a
white hen, all gray chicks are produced.
Materials
Metric ruler,
Procedure
Determine your phenotype and possible
genotype for the following characteristics by
reading the descriptions of the traits in the
following paragraphs:
a. What term is used to describe a condition
when neither gene shows dominance over its
homologous allele?
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Example:
Albinism (a) is caused by a recessive gene. If
you ARE albino your phenotype is .-1L and
your genotype is aa.
b. Using a Punnett Square, demonstrate the
inheritance of feather color in Andalusian
chickens by crossing a black rooster with a
white hen.
If you are NOT albino your phenotype is A
and your genotype could be AA or Aa. Since
you do not know use A-- for your genotype.
Trait
iv the col r(5) of the
ChiCkS:~)
c. Why would it be impossible to establish!
true breeding flock of gray Andalusian
chickens?
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BW
B
W
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8W
GbV
Phenotype
1. Attached
ear lobes
2. Widow's
Peak
3. Tongue
Roller
4. Hitchhiker's
Thumb
5. Bent little
finger
6. Long palmar
muscle
7. Pigmented
iris of eye
8. Mid-Digital
hair
9. Freckles
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Human Inheritance
Introduction
Many inherited human characteristics result
from the interaction of a number of genes, but
frequently the variation is due to a dominant
or recessive gene trait. Where possible,
determine your phenotype (expression) and
your genotype (genetic make up) for each trait
listed below. When you express a dominant
trait, it is not possible to determine whether
you carry two genes for this trait
(homozygous) or whether you carry one
dominant and one recessive (heterozygous).
10. Hair form
11. Long
eyelashes
4
Genotype
3. What is the ratio of appearance for these four
phenotypes derived above. (?PS:?Ps:?pS:?ps)
(Hint: To detennine a ratio divide all 4
numbers by the lowest number.)
Dihybrid Cross of Corn
Introduction
Two of the many genetic characteristics
observed in corn kernels are color (purple or
yellow) and texture (smooth or wrinkled).
Purple (P) is dominant to yellow (p) and
shows a 3: 1 ratio in a monohybrid cross.
Smooth (S) is dominant to wrinkled (s)
and shows a 3: 1 ratio in a monohybrid cross.
The quality of endosperm (food in the seed)
determines the texture of the seed. Wrinkled
kernels have a sweet endosperm; smooth
kernels have a starchy endosperm.
_ _PS:_ _PS:_--IpS:--p s
4. Using a Punnett Square, set up a dihybrid
problem to detennine the expected ratio from a
cross between ppSs x ppSs following these
steps:
a. For the F2 generation, parents with
the genotype ppSs are crossed. Each plant
would produce the following gametes:
Materials
One cob of com produced from parents with a
ppSs genotype.
Parent
Parent (PPSs):~'S)
Procedure
1. Before working with the cobs of com,
consider this problem:
A plant with genes homozygous for the purple
and smooth seeds (PPSS) is crossed with a
plant with yellow and wrinkled seeds (ppss).
Each plant would produce the following
gametes:
Parent
(frs,S):~,.;""r_'S
l"~
.. ~,
(~s):
" ?fs.j &~/ TQS'J>S
r ff-
1t)
fS)f.;w.S-­
b. The gametes from one parent are
placed on one side of the Purmett Square;
gametes from the other parent are placed on the
opposite side of the square. Complete the cross
to give the F2 generation.
_
..
Parent (ppss):,_~(f-5+-=_ _~_ _
The Fl generation of seeds produced from this
cross would have thl~ following:
genotype:
"GPp S$
T_S
phenotype:_.....
_
2. For the F2 generation, parents with the
genotype ppSs are crossed, producing four
different phenotypes. Using the traits listed in
the table below, classify and tally the kinds of
kernels in fOUT rows of the com cob.
Phenotypes
Purple-smooth
CPS)
Purple-wrinkled
CPs)
Yellow-smooth
(pS)
Yellow-wrinkled
5. Give the expected phenotypic ratio
from the Punnett Square:
Number Tallied
6. Compare the expected phenotypic ratio
with the actual ratio produced in answer #3.
(ps)
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d. BBRr x BbRr: (;,ffl''1';
Genetic Problems: Dihybrid and
Sex-linked
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br
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e. BbRr x BbRr:
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Phenotype
Genotype·
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b. BBRR x bbRR:
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Bit. Bil RPt
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r;, /ltt!! 'I s is,\()
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Phenotype
Genotype
f3,-y---;;;~=-=-
Phenotypic Ratio
~'IJR. 3: 13r 3: b R. ;. !b r
Genotypic Ratio
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c. ~r x ~: r;,/Ifl~i£' J}.r ~'r ~} ~
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2. A black, rough male guinea pig is mated to
a white, rough female. Out of several litters
totaling 27 individuals, 9 are black, rough; 7
are black, smooth; 8 are white, rough; 3 are
white, smooth. What are the genotypes of the
parents? (Hint: List what is given for the
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parents and offspring first.)
br
Phenotypic Ratio
t BJr: 1. Pr : t b~;
Genotypic Ratio_----.--
br
--­
-
a. BBRR x bbrr:
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1. In guinea pigs, black coat color (B) is
dominant to white (b), and a rough coat (R) is
dominant to smooth (r). Using a Punnett
Square, determine the expected genotypic and
phenotypic results of the following crosses.
.~
~k
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~ 86Kr ~:J. brr: i b'bR~ ~ t 6~,..r
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8
g
~ ~b
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Male: ab RrFemale: hl;.R. Y'
/ J3- R_
"1­ 75 _ r-r
e b6R-­
3 hb
rr
5. A roan bull which is heterozygous for the
polled condition is mated to several cows of
identical genotype to his. Using a Punnett
Square, determine how many roan, polled
animals should be produced out of 16.
3. In rabbits, black fur (B) is dominant to
brown (b), and long hair (L) is dominant to
short hair (I). A male is mated to several
brown, short-haired females. These matings
result in the following offspring: 11 brown,
long-haired; 16 black, long-haired; 12 brown,
short-haired; 15 black, short-haired rabbits.
Express the genotype and phenotype of the
male. (Hint: List what is given about the
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female and the offspring first.)
g :: ;SLf.JCk
1- -; Lv N6
____
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bb II
I') B- "
/1,
Male:
b I
Phenotype:_O_h
I ':. S
Iff)
66 "
73ffC/C ~l'l t,
H~
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11 HfllAj
IfhX'f­
Hhll~J
HII'i-W HltlJ~
11 h Jl. 'tI
IIh w'W
HhK/L (111 ,-11/
f//l1-/t
hAR.W
flhRVJ /fhw,J tAil "tJ
Ah wi\!
HI(ilfL
b = g~uJfV
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Genotype:&
LI
b_
Number of Roan, polled calves_ _
4. In cattle, the polled condition (H) is
dominant to the horned condition (h). A cross
between an individual with red coat (R) and
white coat (W) results in roan (RW). A
polled, red bull is mated to three cows. With
cow A, which is horned and white, a polled
roan calf is produced. With cow B, which is
horned and roan, a horned, red calf is
produced. With cow C, which is polled and
red, a horned calf is produced. Give the
genotypes of all individuals.
Bull
JI.h- "RJZ
COW
Allh 11/
Calf A
JIb ".j
Cow B
AARWI
Calf B
h" RB
Cow C
HA "jl J!-
Calf C
6. In human, the determination of sex depends
upon whether the male sperm carries an "X"
chromosome (resulting in a daughter) or a "y"
chromosome (resulting in a son). In other
words, body cells of females carry two X
chromosomes, and those of the males carry
one X and one Y. During meiotic division,
the egg of the female must of necessity carry
one X, whereas segregation of the X and Y in
spermatogenesis results in half of the sperm
that are Y bearing, and half that are X bearing.
One human abnormality, called red-green
color blindness, is the result of a recessive
gene carried on the X chromosome. It has no
allele on the Y. Consequently, the genotypes
CC, Cc and cc are possible in females, but a
male must be either CY or cY.
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Answer the following questions:
What proportion of their daughters may be
a. Is it possible for a female to be color-blind?
expected to
y.l.5
XeX C
If one of their sons whose vision is normal
marries a woman of the genotype CC, can
they have any color-blind children?
Dem~trate with a ~ett Square.
b. Can two persons with normal vision
produce a color-blind daughter? a color-blind
son? Demonstrate using a Punnett Square.
xC--;. tTIovY't1 {,y
be;:'~dIlG4'M;
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X
X
L
c
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DaUghter:----'&b~~_ _
Son:
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7. In Drosophila frui flies, sex determination
is the same as in the human. Normal flies
have bright red eyes, a certain recessive sex­
linked gene is responsible for white eyes.
From a pure-breeding strain of red-eyed flies,
a female is selected and bred to a male from a
pure-breeding strain of white-eyed flies. Set
up a Punnett Square to answer the following
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questions.
'fR.4-­
c. Can two color-blind parents produce a child
with normal vision? Demonstrate using a
Punnett Square.
X X -:.
X Y -;.
F£/lIIU
/hllile
X\
'ACJ) -=-
R
Xr-=-
/,JlfiTt.--...
r
r
d. Suppose that a woman with normal vision,
whose father was color-blind, marries a color­
blind man. Using a Punnett Square, determine
the fOll0c..ng:
f(
~r
/ It 111 1
y
R Rr
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a. Will all their offspring have eyes of the
same color?
c
y Xy
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b. What will be the color of the male flies?
What proportion of their sons may be
expected to be color-blind?
hoj{5 Jv5
c. What will be the color of the female flies?
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Gametes produced by man:
8. The males and females produced in the
above problem mate at random to produce the
F2 generation. Exactly 100 of their offspring
are collected (50 males and 50 females). Set
up a Punnett Square for this problem and
answer the following questions:
~y
Rn
'Rv-­
rYJ
II/I)
Gametes produced by woman:
a
HI htA./
Punnett Square:
KY
Y
RyJ
R
,
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.
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a. How many of the 50 females should have
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white eyes?
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b. How many of the "50 males should have
white eyes?
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c. How many of the 50 females should have
red eyes?
d. How many of the 50 males should have red
eyes?
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a. Assuming that four sons are born to this
union, what different combinations of traits
might they exhibit in regard to aniridia and
hemophilia?
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9. In humans, there is a type of blindness
known as aniridia, which is due to a dominant
gene. Another gene, hemophilia is a disease
in which the blood does not clot properly, and
results from a sex-linked recessive gene. A
non-hemophilic man who is blind from
aniridia, but whose mother was not blind,
marries a non-hemophilic woman who is not
blind, but whose father was afflicted with
hemophilia. Set up a Punnett Square
(dihybrid) after determining these answers:
Genotype of man:__
Genotype of woman:_
(A)
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w,Tt-I lIU11d9~/~/'tt0j
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is ~CU1t?L
b. Assuming that four daughters are born to
this union, what different combinations of
traits might they exhibit in regard to aniridia
and hemophilia?
c. What percentages or ratios of the children
would exhibit the following combinations of
traits:
Aniridia
Only_---,~,L~_yLL-----
Hemophilia only_
I
Both aniridia and hemophilia,_YfL-Neither aniridia or hemophilia
_
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