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,; 1.
GENETIC PROBLEMS
't-
XI
which parental pair could produce a colorblind
\. .;
female?
\
A man heterorygous with blood type A marries
blood type of their offspring could not be:
1.A
3.
independent assortrnent
,t2)
do*io*""
25'/o
2.
50%
2.
4. AB
O
3. segregation
4. incomplete dominance
4.
S'; ts't"
incomPlete dominance
3.
100%
sex linkage
,:4. independent assorfinent
;
An organism possessing two identical genes for
l.
2.
7.
with blood type AB. The
A boy has brown hair and blue eyes, and his brother has brown hair and brown eyes. The
fact that they have different combinations of traits is best explained by the concept known
1. multiPle alleles
6.
(
ln pea plants, the trait for all tail stems is dominant over the trait for short stems. If nvo
be expected to
heterozygous tall plants are crossed, what percentage of the offspring would
-f ' t'
have the same phenotype as the tr arents?
l.
5.
'3.
(,,
\'
.1"*.'..'
white tulips.
Onty red tulips result from a cross between homozygous red and homozygous
This illustrates the PrinciPle of:
[.
4.
2.8
a woman
'i
)
Ho{nozygous normal-vision mother and colorblind father.
2. Colorblind mother and normal-vision father'
3. Heterozygous normal-vision mother and normal-vision father'
(+.; Heterorygot t norrnal-vision mother and colorblind father-
l.
Z.
l-*\-
heterozYgous for the trait
hybrid for the trait
a
trait is said to be:
i3.: homozygous for the trait
4 incompletely dominant for the trait
When two four-o'clock plants are crossed a8 pink four-o'clocks and 52 white fouro'clocks are produced. The phenotypes of the parents are:
f
.
2.
pink and white
pink and red
3. pink and pink
4. red and white
as:
GENETIC
g.
PROBLEMS
polydactyly is a human characteristic in which a person has six fingers per hand. The trait
for polydactyly is dominant over the trait for five fingers. If a man who is hetero4rgous
for this trait manies a woman with the normal number of fingers, what are the chances that
their child would be PolYdactYl?
'2.'soyo
I.0Y"
g.
75Yo
4.
t000A
TI
O
Oo, is homozygous dominant and the other heterorygous.
goth are hetero4/gous for the dominant trait.
3.
4.
One is homozygous dominant and the other homozygous recessive
Both are homozygous for the dominant tait.
T-r
tlrye O has a mother with blood type A and a father with blood brye B
The parental genotypes for blood types must be:
A child with blood
l.
2.
11.
3.
\.,
Which statement describes how two organisms may show the same traig yet have different
genotyPes for that PhenotYPe?
Z
10.
2
3. IAIB and IAIB
?-)loi and IBi
IAIA and IBIB
fi and IBIB
',
n,
agt#^'
'i?'-;,
Two pea plants, hybrid for a single trait, produce 60 pea plants. Approximately how many
'1-tt
of the pea plants are expected to exhibit the recessive tait?
L
tl
);., i.
,
2. 4s
lY
12.
1
-.', .,'
(2,.-'zs"l"
'v-
A colorb-lind woman marries a man who
of having a colorblind daughter
floz
t4.
3.
30
ll-','--:
4.
60 7,-i:a
( +:._::
_l -_:_
Three brothers have blood types A" B, and O. What are the chances that the parents
these three will produce a fourth child whose blood type is AB?
l. 0%
13.
,'
2.
250h
_r-{l_
3.50oA
has
4. to}%
'
T"
: in,: i-
--**,
li:.,.11*.
vvv
X
hormal color vision. What are their chances
3.750A
4.
1000A
A student crossed wrinkled-seeded (rr) pea plants with round-seeded (RR) pea plants.
Only round seeds were produced in the resulting plants. This is caused by:
l.
2.
independent assortrnent
segregation
of
3.'dominance
-4. incomplete dominance
GENETIC
3
PROBLEMS
end of
ln peas, flowers located along the stem,(*ig) are dominant to flowers located at the
th; stem (terminal.) Let A represent the allele for axial flowers and a represent the allele
for terminal flowers. When plants with axial flowers are crossed with plants having
terminal flowers, all of the offspring have axial flowers. In the cross the genotypes of the
parent Plants are most likelY:
15.
t.
16.
aa
x aa
2- Aa
i'i.-)'a * *
x Aa
4' AA x AA
ln cabbage butterflies, white color flMJ is dominant and yellow color (w) is recessive. If a
pure white cabbage butterlly mates with a yellow cabbage butterfly, all the resulting (F1)
tutterllies are heterozygous white. Which cross represents the genotype of the parent
generation?
2. WW x
l.Wwxww
Ww
tl.)WW x
ww
4. Ww x Ww
17.
When a mouse with black fur is crossed with a mouse with white fir, all F1 generation
results can be expected in the F2 generation?
offspring
- trave gray fiu. Which phenotype
',,
/'Jt
'.1,
-.-t=l
t. t00o/o gray
_3. s}yoblac( 50% white
';
Z. 2so/oUlack Tl%ow\ute .'",'-.
,'-4.-125Y"blac( 50%o gray,25%ow\ttte
18.
There are multiple alleles for the ABO blood goup. Why are there only trvo of these
alleles normally present in any one individual?
l.
There are not enough nucleotides in a red blood cell to produce a third allele.
,./f)tEu"h parent contributes only one allele for the ABO blood goup to the offspring.
'-3. Each allele in the ABO goup must be either dominant or recessive.
4. Blood goup alleles are not segregated during meiosis.
19.
Which will most probably
XY
be true of the children of a iolorblind man and a woman with
normal color vision?
\,
l.
Atl males will carry the gene for color blindness-
2. Nl males will be colorblind.
3. All females will be colorblind.
7|'Ntfemales will carry the gene for color
\-/
20.
A father has type A blood and his child has rype O blood. Which blood goup genotype
does the child's mother probably have?
l.
2t.
blindness.
I"I"
2.
The offspring of a cross between
l.
2.
dominance
segregation
a
r3'
J. I I
I"lb
Bb
x
4.
Iblb
Bb illustrates the pattern of heredity known
3.
4.
independent assorftnent
sex linkage
as:
lr
4
GENETIC PROBLENIS
On your
al^wer sheet, write the best atawer to each question.
for questions
Use the
following information
22 - 28'
b, i ,- -
,
humans, the gene for brown eyes, B, is dominant to the gene for blue eyes,
l"
brownwas
father
whose
and
A bro*o-eyed man, whose mother *ui blu.-eyed
eye{ and a blue-eyed womalL both of whose parents were brown-eye4 marry'
I'
22.
What is the woman's
23.
What is tle,man's genotYPe?
genotYPe?
't"
;'i.
":
24.
What is the genot)?e of the man's mother?
25.
What is the genotype of the woman's parents?
.a
l,
.,.
',
7,6. If the man and woman have four children, how many would be expected to be
heterozYgous for eYe color?
L
27.
How many of the children would be expected to have blue eyes?
28.
How many of the-phildren would be expected to have brown eyes?
t!
t-
For questions 29 & 30 refer
to
the
follov'ing tnformation:
ln pigeons, checkered pattern is dependent on a dominant gene C, and plain
putt.rn is dependent on recessive allele c. Red color is controlled by a dominant
gene
B and brown
color is controlled by recessive gene b.
'', -
29.
What is the result of a cross between a homozygoul checkered, redpigeon and
homozygous Plain' brown Pigeon?
'
30.
a
t'
A male pigeon who's genotype
is CcBb
is mated with the following female cCbB
How many pigeons will be plain and brown?
31.
In rabbits, black frr is dependent on dominant gene B and brown fi.r on the recessive
allele b. Normal length fiu is determined b1' a dominant gene R and short fi,u on recessive
gene
r'
:,
Show the results of a cross befween rwo heterozygous black ,,orrnil'l.ngth fur rabbits.
How many offspring will probably be homozygous for both traits? What is the ratio of
phenotypes? , \ I
rl
,
!t', 1,., i/'''t
it'
i'
tt,r..
"fi-.r,ftf--ir
:t,i
i-.
,Lr-,
i
)
5
GEI{ETIC PROBLEMS
l1
32.
Rolling the tongue is dominant over non-rolling and rurattached ears are dominant over
attached ears.
'' ''
'.)
f.-
If a tongue rollin! woman with attaihed ears has a child who is a non-ioller with
,"r.. unattached ears, what are the probably genotypes of the woman, her husban4 and
'
the child?
i'.. I i ,) ,;
f
l"\'
- ".
33.
In sheep, white coud ii dominant to black. Occasionally, a black sheep appears in the
flock. The farmer cannot sell black wool. How can the gene for black coat be eliminated
from the flock?
34.
In guinea pigs with colored hair, the hair may be black (B) or brown (b). Black is
dominant to brown hair. Also, the hair may be short (S) or long (s). Short hair is dominant
to long hair. ln each cross described so far, we have begun with homozygous individuals.
This time we will cross animals of heterozygous genotypes: BbSs x BbSs.
What is their phenotype? What will be the genotypes and phenotypes of the F1 offspring?
35.
So far, we have spoken of traits as dominant or recessive. But there are other possibilities.
In shorthorn cattle, for instance, when a red bull is crossed with a white cow, the
heterozygous offspring are neither red nor white. lnstead, they are roan, which means they
have mixed red and white hairs. In a cross between a roan bull and a roan cow, what ate
are the chances it will be ioan or red?
the chances that the calf will be white?
*ul
36.
37
.
Work out a cross of roan
x
ri,:i
red.
If a four-o'clock plant with red flowers is crossed with a white-flowered four-o'clocl<,
F1 plants
all have pink flowers.
the
From this evidence alone, what ratios would you expect to
frnd in these crosses:
1. red
38.
white?
Suppose two newborn babies were mixed up in the hospital. From the following blood
t)?es, determine which baby belongs to which parents:
Baby l:
Baby 2
39.
2. Piok x
x pink?
'
.
Type
Type
O
A
Mrs. Brown
Mr. Brown
Type
Type
B
AB
Mrs-
Smith
Mr. Smith
Type B
Type B
If a woman who is not a carrier of genes for hemophilia is married to a man who is a
hemophiliac, what percentage of their male offspring could be expected to be
hemophiliacs? WhY?
40.
a woman whose father was colorblind, what is the probability
child being a colorblind boy?
If a colorblind man marries
of their
frst
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