Download Part 1: Issues in Genetics and Cellular Biology

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BIOL 100 ALE #5
1.
What is a chromosome? A condensed molecule of DNA and its associated proteins. Carries genetic information
(genes)
2.
What are chromosomes made of? DNA and its associated proteins.
3.
What is a gene? A section of DNA (chromosome) that codes for a specific trait. A unit of inheritance
4.
Where are genes located? at specific locations (loci) on chromosomes)
5.
What is a genotype? particular alleles a person carries: AaBBccDd
6.
What is a phenotype? an individual’s observable traits (hair color, eye color, height,….
7.
What is an allele? alternative molecular forms of a gene
8.
What is the relationship between a gene and an allele? A gene has a specific location on a chromosome, and it has the
same location on its homologous pair. However, the homologous pair may have a different allele of that gene.
9. Define homozygous dominant, heterozygous, and homozygous recessive.
homozygous dominant: both homologous chromosomes carry the same dominant allele: AA
heterozygous: each homologous chromosome carries a different allele (typically one dominant, one recessive): Aa
homozygous recessive: both homologous chromosomes carry the same recessive allele: aa
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Monohybrid Crosses Clearly show/explain how you arrived at your answers to all of the questions below!
10. Jack and Jill are both heterozygous for cystic fibrosis, a recessive trait. What fraction of their offspring is predicted to
have cystic fibrosis? Answer: ¼. What fraction would be expected to be healthy? Answer: 3/4
Since both Jack and Jill are heterozygous, they would each have the genotype Aa. Punnett square – cross Aa with Aa,
yields ¼ offspring with CF, and ¾ offspring that are healthy, since the only way to have CG is to have the genotype aa.
A
a
A
AA
Aa
a
Aa
aa
11. Assume that a single gene determines hair color in humans as follows: BB individuals have black hair; Bb has brown
hair; and bb has blonde hair.
a. If two brown hair people have a child, what is the probability that their first child will have brown hair?
Answer: ½. This is a cross involving incomplete dominance. Punnett square – cross Bb with Bb, yields 2 out of 4
offspring with brown hair (Bb). Thus the probability is 1/2
b.
12. a.
B
b
B
BB
Bb
b
Bb
Bb
Assume that their first child did have black hair. What is the probability that the next child will also have black
hair? Answer: ¼. The probability of the first child’s hair color has nothing to do with the probability of the next
child’s hair color. Thus, use the same Punnett square as above, with shows that ¼ of the offspring can have black
hair (BB).
A red tomato plant is self-fertilized. 25% of the offspring are green. What is the genotype of the plant? (Indicate
the genotype and dominant allele.) Answer: Rr. In this problem we have to work backwards. If only 25% of the
offspring are green, then red must be dominant to green. Thus we can make R = Red and r = green. The only way
to have a green phenotype would be to have the genotype rr. The problem says that a red tomato plant is selffertilized. Most plants have egg and sperm on the same plant, so in this case we are crossing a plant with it self.
Our job is to determine whether the cross is RRxRR or Rr by Rr. Since 25% of the offspring are green, the
genotype of the red plant is Rr. To answer this question, you can draw a Punnett square and put “rr” in one of the
boxes. This shows you that each of the parents must at least have an r. Since they are both red, this means that
they must be Rr.
R
R
?
R
r
R
RR
Rr
r
Rr
rr
?
rr
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b.
If a plant heterozygous for the red/green alternative alleles is crossed to another tomato plant and all the progeny
are red, what is the genotype of the second plant? answer: RR. Again, create a Punnett square and work
backwards from what you know. The problem tells us that one parent’s genotype is Rr and we have no idea what
the other parent’s genotype is. Since all of the progeny are red, then we also know that all of the progeny have at
least one R.
?
?
R
R
R
r
R
R
R
R
R
R
R
r
Rr
Rr
c. In a cross between a red tomato plant and a green tomato plant, if 50% of the progeny
(offspring) are red and 50% are green, what are the genotypes of the two parents? answer : rr
and Rr. The first parent is red, so its genotype can be RR or Rr. The second parent is green, so
its genotype is rr. Of the progeny, 50% are red, so they at least each have one R. 50% are
green, so they have rr. Fill in what you know:
R
?
r
R
rr
r
R
rr
R
r
r
Rr
rr
r
Rr
rr
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13. In cattle the polled (hornless) trait is dominant and is determined by the dominant allele D. The horned trait is
recessive and is determined by the allele d. A certain polled bull is mated to three cows. Cow A, which is horned,
gives birth to a polled calf. Cow B, also horned, produces a horned calf. Cow C, which is polled, produces a horned
calf—this data is summarized in the table below. Determine the genotypes of the four parents. Clearly show/explain
how you arrived at your answers!
Critter
Phenotype
Genotype
Phenotype of Calf produced with Bull
Cow A
Horned
dd
Polled
Cow B
Horned
dd
Horned
Cow C
Polled
Dd
Horned
Bull
Polled
Dd
First, we already know the genotypes of Cow A and Cow B: dd because the horned trait is recessive. Cows C & the Bull are
polled, so they can be either Dd or DD. To determine the exact genotypes, start by creating a Punnett square for each cross
and fill in what you know:
Cow C x Bull
Cow A x Bull
D
?
d
Dd
?d
d
Dd
?d
Cow B x Bull
D
d
D
d
D
DD
Dd
d
Dd
dd
d
Dd
dd
d
Dd
dd
The cross of Cow A by the bull doesn’t yield any new information, but the cross of Cow B by the bull tells us that the bull is
Dd. The cross of Cow C by the bull tells us that Cow C must also be heterozygous, to produce a horned calf
14. In summer squash, white fruit color is dominant; yellow is recessive.
a. If a squash plant that is homozygous for white is crossed with a homozygous yellow, what will be the phenotype of
the F1 generation? Answer: all of the F1 will be White
The corss here is: WW x ww This yields all Ww offspring, thus all of the F1 will be White
b.
W
W
w
Ww
Ww
w
Ww
Ww
What phenotypes and their fractional amounts would be expected in the F2 generation?
Answer: 3/4 whites and 1/4 yellow. To make the F2 generation, we cross an F1 offspring with another F1 offspring:
Ww x Ww. This yields the F2 generation yields 3/4 whites and 1/4 yellow (or 3:1 ratio of white to yellow)
W
w
W
WW
Ww
w
Ww
ww
c.
What would be the appearance and fractional amount of the offspring of a cross between an F2 individual and a
homozygous yellow individual? It depends on which F2 individual is use in the cross. There are 3 possible
crosses:
ww x ww yields all yellow offspring
ww x Ww yields ½ white and ½ yellow
ww x WW yields all white
d.
What is the name of the type of cross that was performed in c?Test cross
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Dihybid and Sex-Linked Crosses Clearly show/explain how you arrived at your answers to all of the questions below!
15. In the land of Magumba there grows the fabled Bungula. In Bungulas, Red Fur color, A, is incompletely dominant over
Purple Fur, a, the heterozygous condition being Green Fur. Long Wings, B, are dominant over short wings, b. This
question involves 2 different genes, one for fur color and one for wing length. Notice also that the alleles for fur color
involve incomplete dominance
a. If a pure breeding Red, short-winged Bungula (AAbb) is mated with a pure breeding Purple, long-winged one
(aaBB), what will be the Phenotypes and their expected occurrences in the F1 generation?
Answer: all will have green fur and long wings.
The cross is AAbb x aaBB. Only one type of gamete is possible for the first parent: Ab, and only one type is possible
for the second parent: aB. Thus all offspring in the F1 generation will be AaBb, and this gives a phenotype of
green fur and long wings.
Ab
aB
b.
AaBb
What will be the phenotypes and their expected occurrences in the F2? This cross will be: AaBb x AaBb. The four
possible gametes for each parent are: AB, Ab, aB, ab. Construct a Punnett square:
AB
Ab
aB
ab
AB
AABB
AABb
AaBB
AaBb
Ab
AABb
AAbb
AaBb
Aabb
aB
AaBB
AaBb
aaBB
aaBb
ab
AaBb
Aabb
aaBb
aabb
I have color-coded the genotypes for their fur color phenotype.
Here are the phenotypes and their expected
ratios: red fur, long wings : 3/16. Red fur short wings: 1/16. Green fur long wings: 6/16, green fur short wings: 2/16.
purple fur Long wings: 3/16, purple fur short wings: 1/16.
16. In Drosophila (fruit fly) yellow body color is sex linked—yellow is recessive to normal body color. If a yellow bodied
female is crossed with a normal male and (a) an F1 female from this cross is mated with her father and (b) an F1 male is
mated with his mother, what will be the phenotypes (as to body color) and their expected occurrences in the offspring
of cross (a) and cross (b)? Don’t Panic! State results for the two sexes separately
The original cross will be: XaXa (yellow female) x XAY (normal male). I am using the superscript “a” here to denote the
recessive yellow allele. The offspring of this cross are either carrier females (X AXa) or affected males (XaY)
Xa
Xa
XA
XAXa
XAXa
Y
X aY
X aY
Cross “a” : XAXa x XAY
XA
Xa
XA
XAXa
XAXa
Y
XAY
X aY
Cross “b” : XcY x XcXc
phenotypes:
females all
normal
body color,
half the males
yellow and
half normal
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Xa
Xa
Xa
X aX a
X aX a
Y
X aY
X aY
phenotypes:
all men & all
women are
yellow
Multiple Alleles and Blood Groups Clearly show/explain how you arrived at your answers to all of the questions below!
17. What are the possible blood types of the children in the following families?
a. Mother: Type A blood and Father: Type A. Answer: A or o. You need to do Punnett squares for all of the
following crosses to get the answer: AA x AA, AA x Ao, and Ao x Ao. The possible blood types of the children are A
or o. The third cross, Ao x Ao, yields a child with blood type O:
b.
A
o
A
AA
Ao
o
Ao
oo
Mother: Type B blood and Father: Type AB. Answer: A, B, AB . Since the mother is blood type B, her genotype
could be BB or Bo. Since the Father is blood type AB, his genotype is AB. Make Punnett squares for the
following crosses: BB x AB, Bo x AB. These two Punnett squares yield the blood types (phenotypes) A, B, and
AB.
B
B
A
AB
AB
B
BB
BB
B
o
A
AB
Ao
B
BB
Bo
Mother: Type A blood and Father: Type O. Answer: A,o The mother’s genotype is either AA or Ao. The
father’s genotype has to be oo. Thus we will have two possible crosses: AA x oo and Ao x oo.
c.
A
A
o
Ao
Ao
o
Ao
Ao
A
o
o
Ao
oo
o
Ao
oo
18. A man is suing his wife for divorce on the grounds of infidelity. Their first child and second child, whom they both
claim to be their children, are blood groups O and AB respectively. The third child, whom the man disclaims, is blood
type B. Can this information be used to support the man’s case? Why? Answer: No. It is possible for this couple too
have a child that is blood type B. Make a Punnett Square and fill in what you know.
A
o
B
AB
Bo
o
Ao
oo
Since they have children with types O and AB, then the blood types of the parents must be A
and B. Filling in the rest of the square, we see that the couple can indeed have a child with
blood type B, so the fact that the third child has blood type B cannot be used to support the
man’s case.
19. A mother has Type A, Rh- blood and the father has A, Rh+ blood.
a. What are all the possible genotypes of the offspring these two could produce? Answer: AARr, AoRr, AArr, Aorr,
ooRr, oorr,
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First complete the crosses for the blood types, then complete the crosses for the Rhesus factor. Combine the results.
The mother and Father are both blood type A, so their genotypes can be AA or Ao. This means we have to do
three crosses: AA x Ao, AA x AA, and Ao x Ao.
A
A
A
AA
AA
A
AA
AA
A
o
A
AA
Ao
A
AA
Ao
A
o
A
AA
Ao
o
Ao
oo
From this we see that the three possible genotypes for the children, in terms of blood groups, are AA, Ao and oo.
The mother is Rh-, which means her genotype is rr. The father is Rh+, which means his genotype is RR or Rr. Thus we
have two possible crosses: rr x RR or rr x Rr.
r
r
R
Rr
Rr
R
Rr
Rr
r
r
R
Rr
Rr
r
rr
rr
From this we see that the two possible genotypes for the offspring are Rr and rr. Now if we combine the genotypes
from the blood groups with the genotypes from the Rhesus factor, we get the following possible genotypes for the
offspring: AARr, AoRr, AArr, Aorr, ooRr, oorr,
b. What are all the possible phenotypes of their potential offspring? Answer: A+, A-, o+, o-. We get this info from
our genotypes that we listed above. AARr is A+, AoRr is also A+, AArr is A-, Aorr is also A-, ooRr is O+ and , oorr is O-.
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20. A mix-up happened in the maternity ward of a hospital. Baby X (Blood type A) and Baby Y (Blood type O) lost their
ID tags!! If the suspected parents have the following blood types, match the babies with the correct parents.
Of course you can construct Punnett squares to answer this question, but since you are getting so good at those, you could
also figure this out logically. For Baby Y to have blood type O its genotype must be oo. Thus it needs to get an O allele
from each parent. Since one of the parents in Couple 2 has type AB blood, we know that parent has no O allele. Therefore
couple 2 could never produce a baby with blood type O.
Couple 1: Type B and Type A – Baby Y
Couple 2: Type A and AB Baby X
Probability Method Clearly show/explain how you arrived at your answers to all of the questions below!
21. How many different types of gametes could be generated from individuals with the following genotypes?
a. AaBb - AB, Ab, aB, ab 4 different types.
b.
AaBbCc – ABC, AbC, ABc, Abc, aBC, abC, aBc, abc. 8 different types. This is equal to 2 x 2 x 2. We can have
two different gametes from Aa: A and a, and two from B: B and b, and two from C: C and c. Thus to figure out the
# of different gamete types, you can use the formula: 2n where n = the # of different genes. However, you can only
use this when the gene is heterozygous.
c.
AABBCC – one type: ABC
AABBCc –two types: ABC and ABc
d.
22. Given AaBbCC x AabbCc what are the chances of producing the following genotypes? First, you should construct
Punnett squares for each individual cross and then use the multiplication rule to get the probabilities
A
a
A
AA
Aa
a
Aa
aa
a.
b.
c.
d.
b
b
B
b
Bb
bb
Bb
C
C
C
CC
CC
c
Cc
Cc
bb
AabbCC = ½ x ½ x ½ = 1/8
aaBBCc = no chance – no BB combination
aabbCc = ¼ x ½ x ½ = 1/16
AaBbCc = ½ x ½ x ½ = 1/8
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23. Suppose A = Red and a = White. B = Tall and b = Short. Given the following cross AaBb x aaBb what are the
chances of producing the following phenotypes? First determine the possible gametes for each parent, and place them
into a Punnett square. Remember, you can also do two separate 4-square crosses and multiply the probabilities!!!
AB
Ab
aB
ab
aB
AaBB
AaBb
aaBB
aaBb
ab
AaBb
Aabb
aaBb
aabb
aB
AaBB
AaBb
aaBB
aaBb
ab
AaBb
Aabb
aaBb
aabb
Or, you can do two separate 4-squares and multiple to get the answer:
A
a
a
Aa
aa
a
Aa
aa
B
b
B
BB
bb
b
Bb
bb
a.
Red Tall – In other words, how many out of the 16 genotypes above have a big “A” and a big “B”? 6/16 or
b.
Red Short – need a big “A” and “bb” 2/16 or 1/8
c.
White Short – need “aabb” – 2/16 or 1/8
24. Suppose A = Red and a = White; B = Tall and b = Short: C = Round seed and c = wrinkle seed.
Given the cross of AaBbcc x AabbCc what are the chances of producing the following phenotypes? Since this
one involves 3 genes, lets do three separate 4-squares:
A
a
A
AA
Aa
a
Aa
aa
B
b
b
Bb
bb
b
Bb
bb
a.
Red, Tall, and Round Seed : ¾ x ½ x ½ = 3/16
b.
White, Short, and Wrinkled Seed : ¼ x ½ x ½ = 1/16
c.
Red, Short and Wrinkled Seed : ¾ x ½ x ½ = 3/16
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c
c
C
Cc
Cc
c
cc
cc