Download LABORATORY #5: Genetics Lab

LABORATORY #5: Genetics Lab
Laboratory Exercise #5.3 Mendelian Genetics Problems
Complete all genetics problems. Do your work on a separate sheet of paper if more space is
needed.
1. A card is drawn from a deck of 52 cards. What is the probability the card:
a. will be the ace of hearts? 1/52
b. will be a king? 4/52 or 1/13
c. will be a king or a seven? 1/13 + 1/13 = 2/13
2. You roll a single dice (a die). What is the probability that you will roll:
a. a number divisible by 3? 1/3 (a number divisible by 3 = 3 and 6; each die is a 1/6
chance)
b. an even number? ½ (even numbers = 2, 4 or 6)
3. In a single throw of two dice, what is the probability that:
a. a seven will appear? 1/6
what are the ways to roll a seven?
= 6&1 OR 1&6 OR 2&5 OR 5&2 OR 3&4 OR 4&3
the chances of rolling a 1 AND a 6 = 1/6 x 1/6 = 1/36
the ways to roll a seven = 1/36 + 1/36+ 1/36 + 1/36 + 1/36 + 1/36 = 6/36 or 1/6
b. a doublet (two of the same number) will appear? 1/6
-say you roll a 1 and a 1, the chances of rolling a 1 AND another 1 = 1/6 X 1/6 = 1/36
-to roll a 2 and a 2 = 1/36
-to roll a 3 and a 3 = 1/36
-add them all up = 6/36 or 1/6
c. an even and an odd number will appear?
-what are the combinations of an odd and even number?
1 & 2 = 1/36
3 & 2 = 1/36 so that is 3/36 or 1/12
5 & 2 = 1/36
-the same is true for 1 & 4, 3 & 4, 5 & 4 AND 1 & 6, 3 & 6 , 5 & 6
-for each even/odd combination the chance is 1/12
-there are 3 sets = 3/12 or 1/4
d. two numbers over 4 will appear? (This means totaling over 4)
-two dice totaling over 4
- consider the first die is a one
1 & 4 = 1/36
1 & 5 = 1/36 that is 3/36 or 1/12 for that set
1 & 6 = 1/36
-consider the second die is a 2
2 & 3 = 1/36
2 & 4 = 1/36
2 & 5 = 1/36 that is 4/36 or 1/9 for that set
2 & 6 = 1/36
-for the first die being a 3 = total is 5/36
-for the first die being a 4, 5 or 6 – each will be 6/36
-so the probability is: 3/36 + 4/36 + 5/36 + 6/36 + 6/36 + 6/36 = 30/36 or 15/18
4. A bag contains 6 blue marbles, 4 red marbles and 2 green marbles (i.e. 12 marbles). If you select
a single marble at random from the bag, what is the probability that it will be:
a. red = 4/12 or 1/3
b. blue = 6/12 or 1/2
c. green = 2/12 or 1/6
d. white = 0
e. pink = 0
f. red or white = 1/3 + 0 = 1/3
g. green or white = 1/6 + 0 = 1/6
5. If you toss a coin six times, what is the probability you will:
a. get six heads = 1/64 = ½ x ½ x ½ x ½ x ½ x ½
b. get five heads and one tails – to do this problem, you need to determine all the way that
you can get five heads and one tail
e.g. HHHHHT or HHHHTH or HHHTHH or HHTHHH or HTHHHH ior THHHHH
each of these combinations is 1/64
-so you need to add these up to get 6/64 or 3/32
c. get three tails and three heads – same thing as above
d. get four heads and two tails - same thing as above
Monohybrid Problems
Do each of the following with a Punnett square or a Pedigree Chart. If needed, do your work on a
separate sheet of paper
1. In humans, six fingers (F) is the dominant trait and five fingers (f) is the recessive trait. Both
parents are heterozygous for six fingers.
What is the phenotype of the father and mother? Six fingered
What is the genotype for each parent? Ff and Ff
What is the probability of them having six-fingered children? 3/4
Five fingered children? ¼
2.
A mutant cucumber plant has flowers that fail to open when mature. Crosses can be done with the
plant by manually opening and pollinating the flowers with pollen from another plant. When closed X
open crosses were done, all F1 progeny were open. The F2 progeny after crossing F1 progeny were 145
open and 59 closed. A cross of closed X F1 plants gave 81 open and 77 closed. How is the closed trait
inherited?
P1 – closed cc X open CC
F1 all open (Cc)
Open Cc X Open Cc
F2 – Open C-- = 145
Closed cc = 59
3:1 ratio
Closed cc X Open Cc
Open Cc = 81
Closed cc = 77 1:1 ratio
Based on the 3:1 ratio when heterozygotes are crossed and the 1:1 ratio when a homozygous
recessive test cross is done – the Closed phenotype is inherited through a single gene with
standard complete dominance
3.
In humans, a dimple in the chin is a dominant characteristic.
a. A man who does not have a dimple has children with a dimpled woman whose mother lacked the
dimple. What proportion of their children would be expected to have a dimple?
Dad dd X Mother Dd
½ the children with dimples Dd
½ the children without dimples dd
b. A man with a dimple and a woman who lacks a dimple produce a child who lacks a dimple. What is the
man’s genotype?
Dd X dd  dd child
To make a dimple-less child the dimpled father must be a heterozygote
4.
You have just purchased a black stallion of unknown genotype. You mate him to a red mare and
she delivers twin foals, one red and one black. Can you tell from these results how color is inherited,
assuming that alternative alleles of a single gene are involved? What crosses would you perform to work
this out?
Let’s assume Black is dominant and red is recessive and see what happens
Black B-- X red bb  Bb and bb progeny
If you mate the F1 progeny – you should get a 1:1 ratio of red to black
You can also take the F1 progeny and mate them back to the parents
5.
In humans, true brown eyes, B, are dominant over true blue eyes, b. A brown-eyed man has 8
children with a blue eyed woman. They are all brown eyed. What are the possible genotypes of all
members of this family?
Brown B – X Blue bb  Bb brown-eyed children
Father can be BB or Bb
-if Father was BB – all kids are Bb
-if Father was Bb – kids are still Bb because they are brown-eyed
Mother can only be bb
6.
The parents of a blue-eyed man were brown eyed. He has children with a woman who had brown
eyes. Her father was brown eyed and her mother was blue eyed. Their one child has blue eyes. What are
the most probable genotypes of all the individuals mentioned?
Generation 1 – Bb and Bb/ B – and bb
Generation 2 – bb and Bb
Generation 3 - bb
7. George and his father have brown eyes (i.e. at least one B allele), while his mother and sister have
blue eyes (i.e. bb). George’s paternal grandparents both had brown eyes (i.e. at least one B allele),
while his maternal grandparents had blue eyes (i.e. bb). George has children with Helen. Helen has
brown eyes (i.e. at least one B allele). Her father had blue eyes (i.e. bb), while her mother had brown
eyes (i.e. at least one B allele). Helen’s paternal grandparents had brown eyes, while one of her
maternal grandparents had blue eyes. Draw a pedigree chart for this family. Determine all genotypes of
all individuals.
George = Bb
George’s sister = bb
George’s dad = Bb
George’s mom = bb
George’s dad’s dad = B-George’s dad’s mom = B-George’s mom’s dad = bb
George’s mom’s mom = bb
Helen = Bb
Helen’s dad = bb
Helen’s mom = Bb
Helen’s dad’s dad = Bb
Helen’s dad’s mom = Bb
Helen’s mom’s dad = Bb or BB
Helen’s mom’s mom = bb
8. Charlie decides to breed pigs. In pigs, hair color is either black, white, or tan. Tan hair coloring is a condition that
results from being heterozygous. HINT: assume black is the dominant color.
a. Isabel, Charlie’s black pig (i.e. BB) escapes from her pen one night. Sometime later, she gives
birth to a tan piglet (i.e. Bb) and two black piglets (i.e. BB). The color of what piglet tells the
father’s color? It is the Black piglets that are important. A tan piglet could have come from
Isabel (BB) mating with a white male pig (bb). But this mating could not produce black
piglets.
b. Charlies mates Matty, who is tan, with Butch, who is white? What color piglets can he expect?
He can expect tan and white piglets from the BbXbb cross.
c. Matty refuses to mate with Butch, so Charlies brings in Bas, who is tan. Matty ends up having
four piglets, two black and two white. What are the expected piglets in this cross? The
following piglets can be expected from this BBXBb cross: BB (1/4), Bb (1/2) and bb (1/4)
9. Piebald spotting is a condition found in humans in which there are patches of skin that lack
pigmentation. The condition results from the inability of pigment-producing cells to migrate properly
during development. Two adults with piebald spotting have one child who has this trait. Their second
child has normal pigmentation.
a. Is the piebald spotting trait dominant or recessive?
Dominant
If recessive – all kids would be piebald
b. What are the genotypes of the parents?
Father = Pp (piebald)
Mother = Pp (piebald)
Children = Pp (piebald) and pp (normal)
c. The first child with piebald spotting has a child with someone who has no history of the trait in their
family (i.e homozygous recessive). The second child has children with someone whose two parents
have the trait ((i.e heterozygous or homozygous dominant). What are the phenotypes and genotypes
of these “grandchildren”?
1st child: Pp X pp  pp and Pp children
2nd child: pp X PP or Pp  Pp and pp children
10. Albinism is a condition in which pigmentation is lacking. In humans, the result is white hair, nonpigmented skin and pink eyes. The trait in humans is caused by a recessive allele. Two normal parents
have an albino child. What are the parent’s genotypes? What is the probability that the next child will be:
a. an albino = ¼
parent’s are Aa and Aa
b. a normal girl = ½ (a girl) x ¾ (albino)= 3/8
¾ of kids are normal; ¼ are albino
c. an albino boy = ½ (a boy) x ¼ (albino) = 1/8
11. The common grandfather of two first cousins has hereditary hemochromatosis – a recessive condition
causing an abnormal buildup of iron in the body. Neither of the cousins has the disease, nor do any of
their relatives.
a. If the first cousins mate with each other and had a child, what is the chance that the child would have
the disease? Assume that the unrelated parents of these cousins are not carriers – 0%, 50% or 25%
b. How would your calculation change if you knew that 1 out of every 10 unaffected people in the
population (including the unrelated parents above) were carriers? – 25%
Generation #1: H – X hh
Generation #2: HH X Hh and Hh X HH
Generation #3: HH or Hh is first cousins with HH or Hh
If HH X HH cousins mate = 0% chance of disease
If Hh X HH cousins mate = 50% chance of disease
If Hh X Hh cousins mate = 25% chance of disease
12. Polycystic kidney disease is a dominant trait that causes the growth of numerous cysts on the kidneys.
The condition eventually leads to kidney failure. A child with the disease is born to a couple – neither of
which show the disease. What possibilities might explain this outcome?
The parents are heterozygous carriers
Dihybrid Problems:
Do each of the following with a Punnett square. If needed, do your work on a separate sheet of
paper
1. In horses, black color (B), dominates chestnut (b). The trotting gait (T) dominates the pacing gate (t). A
cross is made between a horse homozygous for black and pacing gait (i.e. BBtt) and a horse homozygous
for chestnut and trotting gait (i.e. bbTT).
a. List the probable genotypes and phenotypes from offspring resulting from this cross.
A BBttXbbTT cross will give you 100% BbTt, black and trotting offspring
b. List the probable phenotypes of offspring that would result if the offspring from the original cross were
crossed amongst themselves. A BbTtXBbTt dihybrid cross will give you a 9:3:3:1 ratio. 9 Black and
trotting: 3 Black and pacing: 3 chestnut and Trotting:1 chestnut and pacing
2. The following chart show the result of different matings between jimsonweed plants that either had
purple or white flowers and spiny or smooth pods. Determine dominant allele for these two traits and
indicate the genotypes of each of the parents in the crosses a through f
Offspring Phenotypes
Parents
purple
white
purple
white
spiny
spiny
smooth
smooth
a. purple spiny X
purple spiny
94
32
28
11
b. purple spiny X
purple smooth
40
0
38
0
c. purple spiny X
white spiny
34
30
0
0
d. purple spiny X
white spiny
89
92
31
27
e. purple smooth
X purple smooth
0
0
36
11
f. white spiny X
white spiny
0
45
0
16
a. is a 9:3:3:1 ratio which occurs when upon a dihybrid cross – PpSs X PpSs  P –S -, P – ss, pp S -- &
ppss
b. is a 1:1 ratio which occurs upon a test cross (homozygous recessive) of a heterozygote – the
purple trait is constant in this 1:1 ratio so it is the S allele that is heterozygous and test-crossed –
PPSs X PPss  PPSs and PPss
c. is a 1:1 ratio which occurs upon a test cross (homozygous recessive) of a heterozygote – the
spiny trait is constant in this 1:1 ratio so it is the P allele that is heterozygous and test-crossed –
PpSS & ppSS
d. is a 3: 1 ratio of spiny to smooth; but a 1:1 ratio of purple to purple and a 1:1 ratio of white to
white – PpSs X ppSs  PpS-, ppS-, Ppss & ppss
e. is a 3: 1 ratio purple to white (heterozygote cross); but all smooth (homozygous/true-breeding
for smooth) - Ppss X Ppss  Ppss and ppss
f. is a 3: 1 ratio spiny to smooth (heterozygote cross); but all white (homozygous/true-breeding
for white) ppSs X ppSs  ppSs & ppss
3. A third grader decides to breed guinea pigs for her school science project. She goes to a pet store and
buys a male with smooth black fur and a female with rough white fur. She wants to study the inheritance
of these traits and breeds them. She gets 8 pups that are all rough black animals. She breeds the parents
again and gets 7 rough black pups. The first F1 litter breeds among themselves and starts to produce
offspring. These F2 offspring have a variety of coat colors. Before long, the child has 125 F2 offspring.
Eight of them have smooth white coats, 25 have smooth black coats, 23 were rough and white and 69
were rough and black (i.e. a 9:3:3:1 ratio).
a. How are coat color and texture inherited? Dihybrid trait; complete dominance
Smooth black male rr B – X rough white female R – bb  rough black progeny RrBb
rough black RrBb X rough black RrBb  R – B – = 69 rough black
rr B – = 25 smooth black
R – bb = 23 rough white
rr bb = 8 smooth white
b. What phenotypes and proportions should the girl expect if she mates one of the smooth white F2
females to an F1 male?
rr bb X RrBb  1:1 ratio or RrBb rough black : rr bb smooth white
4. How many genetically different eggs could be formed by women with the following genotypes?
a. Aa bb CC DD = gametes are AbCD & abCB – 1 allele heterozygote = 21 or 2 gametes
b. AA Bb Cc dd - 2 allele heterozygote = 22 or 4 gametes
c. Aa Bb cc Dd = 3 allele heterozygote = 23 or 4 gametes
d. Aa Bb Cc Dd = 4 allele heterozygote = 24 or 4 gametes
5. A pea plant heterozygous for plant height, pod shape and flower color was “selfed”. The progeny
consisted of 272 tall, inflated pods with purple flowers; 92 tall, inflated pods with white flowers; 88, tall
flat pods with purple flowers; 93 dwarf, inflated, purple; 35, tall, flat, white; 31 dwarf, inflated white; 29
dwarf, flat, purple; 11 dwarf, flat, white. Which alleles are dominant?
Tall , Inflated and Purple are dominant
272 tall, inflated pods with purple flowers
92 tall, inflated pods with white flowers
88, tall flat pods with purple flowers
35, tall, flat, white
93 dwarf, inflated, purple
31 dwarf, inflated white
29 dwarf, flat, purple
11 dwarf, flat, white
Let’s write this another way to see the 9:3:3:1 ratios and their genotypes
272 tall, inflated pods with purple flowers = T – I – P –
88, tall flat pods with purple flowers = T – ii P-92 tall, inflated pods with white flowers = T – I -- pp
35, tall, flat, white = T – ii pp
93 dwarf, inflated, purple = tt I – P -29 dwarf, flat, purple = tt ii P -31 dwarf, inflated white = tt I—pp
11 dwarf, flat, white = tt ii pp
5. In dogs, black coloring is dominant and red color is recessive; solid coloring is dominant while white
spotting is recessive.
a. A homozygous black and white spotted male is crossed with a red and white spotted female. What is
the probability of them producing a solid black puppy?
There is no chance of producing a solid black puppy since the parents are BBss and bbss (i.e. both
are homozygous for the recessive spotted gene). The progeny will be Bbss which is black and
spotting
b. This black and white spotted male is mated with a solid black female. Can this mating produce solid
colored puppies? Why or why not?
Black and white spotted Bbss X solid black B –S –
Solid colored puppies are possible
Pedigree and X-linked Problems
For Questions 1-9, use the pedigree chart shown below. Some of the labels may be used more than
once.
C 1. A male
A 2. A female
B 3. A “marriage”
A 4. A person who expresses the trait
C 5. A person who does not express the trait
D 6. A connection between parents and offspring
3 7. How many generations are shown on this chart?
Assuming the chart above is tracing the autosomal dominant trait of "White Forelock (F)" through
the family. F is a tuft of white hair on the forehead.
Ff since some of the children do not have white forelocks 8. What is the most likely genotype of
individual “A”? (FF, Ff or ff?)
ff since they do not have the dominant white forelock 9. What is the most likely genotype of
individual “C”? (FF, Ff or ff?)
Complete the following problems. If necessary, draw a pedigree chart to help you work through
these problems
10. In humans, Hemophilia is a blood clotting disorder in which one of the proteins needed to form
blood clots is missing or reduced. Individuals have difficulty forming blood clots following injury
and may suffer significant blood loss from even minor cuts and bruises. The key clotting factor in
hemophilia is called Factor VIII. The gene for Factor VIII is found on the X chromosome. Hemophilia
is caused by a lack of Factor VIII and results from a recessive allele (h) found on the X chromosome.
The condition for normal blood clotting dominates over non-clotting. A hemophiliac male has a child
with a woman who is a carrier for this condition. Using the letter H for normal Factor VIII and h for
abnormal Factor VIII, what are the chances that they have a male child who clots normally?
From a Punnett square, there is a ¼ chance for this.
Alternatively, there is a ½ chance the child will be a boy AND a ½ chance the child will clot
normally = ½ X ½ = 1/4
11. In humans, the condition for normal vision dominates color blindness. Both are linked to the X
chromosome. Using C for normal vision and c for the color-blind disorder, what are the chances that a
color-blind female and color-blind male will have a daughter with normal vision?
There is no chance of this since the color blind female carries two recessive X chromosomes and
the father carries one recessive X chromosome.
XcXc x XcY
12. Remember that because hemophilia is an X-linked disorder, when you identify genotypes in this
pedigree, you must use the XX/XY notation and use superscripts with each X chromosome to indicate
whether the “H” (normal) or “h” (hemophilia) allele is present. (Ex. XHY = normal male).
Hemophilia became known as the “Royal disease” after it suddenly cropped up in some of the descendants
of Great Britain’s Queen Victoria and spread through the royal families of Europe. Queen Victoria and her
husband Prince Albert had 9 children – 5 girls (Beatrice, Victoria, Alice, Helena, and Louise – none of whom
were hemophiliacs) and 4 boys (Edward, Alfred and Arthur had normal blood clotting; their son Leopold,
however was a hemophiliac). Beatrice married a man named Henry and they had four children (sons
Leopold and Maurice who were hemophiliacs, daughter Eugenie who was not a hemophiliac, and another
son who was also not a hemophiliac). Eugenie married Alfonso XIII of Spain (non-hemophiliac) and they
had 6 children (2 normal sons, 2 normal daughters and 2 hemophiliac sons). One of those normal sons
married a non-hemophiliac woman and gave birth to one son – a non-hemophiliac they named Juan Carlos
(the reigning King of Spain).
Using this information, draw the pedigree chart for this Royal family
Draw your own pedigree for the Royal family in your lab notebook
1. Draw the marriages and children – make the men boxes and the women circles
2. Color in all affected individuals – i.e. the ones with hemophilia
3. Fill in the X and Y chromosomes
4. Add the H and h alleles to the appropriate chromosome
HINT: Do the males first since you know if they have hemophilia they are XhY and if they don’t they
are XHY
Victoria = XHXh
Henry = XHY
Albert = XHY
Sons with hemophilia = XhY
Sons with hemophilia = XhY
Eugenie = XHXh
Beatrice = XHXh
Alfonso XIII = XHY
Sons with hemophilia = XhY
A little bit more info – for fun
Another of Victoria and Albert’s daughters was Alice (XHXh ) who married Louis (XHY). Their daughter
Alexandria (XHXh ) married Csar Nicholas of Russia (XHY). They had one hemophiliac son (Alexander=
XhY) and what we think were normal daughters. We don’t know since they along with their entire
families were murdered during the Bolshevik revolution.
One of Victoria and Albert’s sons, Edward Louis (XHY) married and had a son names George V (XHY).
George V married Mary and had a son, George VI. George the VI married Elizabeth and had the current
Queen Elizabeth II. Because George V was normal and married a non-hemophiliac women who wasn’t a
carrier, the current Royal family of Britain is now hemophilia free
Another child of Alice and Louis was the grandmother of Prince Philip, the current consort of Queen
Elizabeth II
Dominance Problems
1. In four o-clocks, the allele for red flowers is incompletely dominant over the allele for white flowers, so
that heterozygotes have pink flowers. What ratios of flower colors would you expect among the offspring
of the following crosses?
a. pink x pink = Rr X Rr = 1 red; 2 pink, 1 white
b. white x pink= rr X Rr = 1 pink, 1 white
c. red x red= RR X RR = all red
d. red x pink= RR X Rr = 1 red; 1 pink
e. white x white= rr X rr = all white
f. red x white= RR X rr = all pink
g. If you wanted to produce pink flowers, which of these crosses would be most efficient?
RR X rr
2. A cross between two plants that both have yellow flowers produces 80 offspring plants of which 38
have yellow flowers, 22 have red flowers and 20 have white flowers. If one assumes that this variation in
color is due to a single gene, what is the genotype associated with each flower color? How is this flower
color inherited?
38 yellow
Written another way:
22 red
22 red = RR (1)
20 white
38 yellow = Rr (2)
20 white = rr (1)
Rr X Rr = monohybrid cross with incomplete dominance
Multi-Allelic and Gene Interaction Problems
Complete the following. Use a Punnett square or a Pedigree chart to help you determine your
answers.
1.
In humans, the trait for type A blood and type B blood show incomplete dominance, so that a
person with both alleles as blood type AB. Both A and B are dominant over the O allele. Therefore type A
blood can have two genotypes: AA and AO. A person with type A blood has a child with a person with
type O blood. List the types of offspring they could have and the probability for each blood type.
AA or AO X OO  AO and OO children –each with a ½ chance
2.
There are several genes in humans in addition to the ABO gene that give rise to antigens on the
surface of an RBC. The MN and Rh genes are two such examples. The Rh locus can either contain a
positive or a negative allele, with positive being completely dominant over the negative allele. M and N
are actually co-dominant alleles for the MN gene. The following chart shows several mothers and their
children. For each mother-child pair, choose the father of the child from among those listed in the right
column. Assume one child per father.
Mother
Child
Father
a. O M Rh+
B MN Rh-
O M Rh-
b. B MN Rh-
O N Rh-
A M Rh+
c. O M Rh+
A M Rh-
O MN Rh+
d. AB N Rh-
B MN Rh -
B MN Rh +
Let’s write out everyone’s genotype
a. OO MM Rh+ Rh+ X B --MN Rh+ Rh-  BO MN Rh- B MN Rh + father
b. BO MN Rh- Rh- X OO MN Rh+ Rh-  OO NN Rh- Rhc. OO MM Rh+ Rh- X A – MM Rh+ Rh-  AO MM Rh- Rhd. AB NN Rh- Rh- X OO MM Rh- Rh-  BO MN Rh- Rh-
O MN Rh+ father
A M Rh+ father
O M Rh- father
3.
In radishes, the shape may be long, round or oval. Crosses between long and oval gives 159 oval
and 156 long. Crosses between oval and round produce 203 oval and 199 round. Crosses between long
and round give 576 oval. Crosses between oval and oval give 121 long, 243 oval and 119 round. Which of
the shapes are defined by a homozygous dominant genotype? Homozygous recessive? Heterozygous?
For this problem, use the letters L and l
Long is LL
Round is ll
Oval is Ll – If a cross between long and round give 100% oval offspring, you know that long and
round MUST be homozygous and the offspring must be heterozygous because only a cross
between two homozygous genotypes will give offspring all with the same phenotype.
To check: the LL and Ll cross will give LL and Ll in a 1:1 ratio; the Ll and ll cross will give Ll and ll
in a 1:1 ratio; crosses between Ll and Ll give a 1:2:1 ratio of long:oval:round
Of course there is nothing saying that Long can’t be ll and Round be LL!
4. Alleles of the gene that determines seed coat patterns in lentils can be organized in a dominance series:
marbled > spotted = dotted (co-dominant) > clear. A lentil plant homozygous for the marbled seed
pattern allele was crossed to one homozygous for the spotted pattern allele. In another cross, a
homozygous dotted lentil plant was crossed to one homozygous for clear. An F1 plant from the first cross
was then mated to an F1 plant from the second.
a. What phenotypes and in what proportions are expected?
b. What are the expected phenotypes of the F1 plants from the two original parental crosses?
Let’s use m and marbled, s for spotted, d for dotted and c for clear
Marbled mm X spotted ss  ms
Dotted dd X clear cc  dc
since m>s the lentil is marbled
since d > c the lentil is dotted
Marbled ms X dotted dc  md = marbled (1/4 of progeny)
mc = marbled (1/4 of progeny)
sd = spotted/dotted (co-dominant) (1/4 of progeny)
sc = spotted (1/4 of progeny)
5. In some types of wheat, color is caused by two sets of genes. The dominant genes, R and B, are needed
for red wheat. Red wheat has the genotypes RRBB, RrBB, RRBb or RrBb. White results when both genes
are homozygous and recessive – i.e. rrbb. A strain with the genotype Rrbb is crossed with a strain of
wheat with the genotype rrBb. What is the color of each of the parent strains? Using a Punnett square,
what are the possible genotypes and colors that can result from this cross?
The important thing to realize is that if you have a dominant R and a dominant B allele in the
genotype– you get red wheat, but if you have only ONE of these alleles in dominant form (e.g. R-bb
or rrB-) you get brown wheat. If you have both in recessive forms – rrbb – you get white wheat.
Rrbb crossed with rrBb will give the following genotypes in equal numbers: RrBb, rrBb, Rrbb,
rrbb
The rrbb offspring will be white, the genotype RrBb will be red. All other genotypes will give
brown wheat.
6. In a population of rabbits, you find three different coat color phenotypes: chinchilla (C), Himalaya (H)
and albino (A). To understand the inheritance, you perform the following crosses and note the results in
the table below.
a. What can you conclude about the inheritance of coat color?
Chinchilla > Himalaya > albino
b. Ascribe genotypes to the parents in each of these 10 crosses
see below
c. What kinds of progeny would you expect, and in what proportions, if you crossed the chinchilla parents
in crosses 9 and 10? ¾ chinchilla (CC, CH, CA) and ¼ himalaya (HA)
CA X CH  CC, CH and CA (all Himalaya) & HA himalaya
Cross #
Parental Phenotypes
And Genotypes
Progeny phenotypes
Progeny Genotypes
1.
H X H = HA X HA
¾H:¼A
Himalaya > albino  HH, HA and AA
2.
H X A = HA X AA
½H:½A
HA and AA
3.
C X C = CH X CH
¾C:¼H
Chinchilla > Himalaya  CC, CH and HH
4.
C X H = CC X HH
All C
CH
5.
C X C = CA X CA
¾C:¼A
Chinchilla > albina  CC, CA and AA
6.
H X A = HH X AA
All H
HA
7.
C X A = CA X AA
½C:½A
CA and AA
8.
A X A = AA X AA
All A
AA
9.
C X H = CH X HH
½ C: ½ H
CH and HH
10.
C X H = CA X HA
½C:¼H:¼A
CA, CH (chinchilla), HA (Himalaya) and AA
(Albino)
7. In humans, the allele for brown eyes (B) is dominant over the allele for blue eyes (b). The allele for
normal melanin production and deposition in the iris (M) is dominant over the allele for albinism (m). If
the gene for melanin production and deposition is homozygous recessive (i.e. mm) it results in the
appearance of a red iris, as it “vetoes” the phenotype that would result from the B and b alleles.
a. Give the expected genotypes and their ratios for the children of two individuals who are heterozygous
for brown eyes and melanin production.
So B> b
And M > m
B—mm = red iris
B – mm and bb mm progeny in equal ratios
b. How many of these children will be albinos and what are their genotypes?
BbMm X Bb Mm  B – M-bb M -B-- mm
bb mm
9 brown
3 blue
3 albino
1 albino
9:3:4 ratio which is characteristic of recessive epistasis
c. Upon studying the phenotypes of numerous families, you discover that a co-dominance relationship
exists between the B and M loci. You specifically discover that if the offspring has both dominant alleles,
their eyes will be half-brown and half-blue. Being homozygous or heterozyogous for these dominant
alleles does not matter. How many different phenotypes should you see in the children after the
heterozygous cross? What are they?
B – M-- 9 half-brown; half-blue irises
bb M -- 3 blue
B-- mm 3 albino
bb mm
1 albino
8. “Secretors” (genotypes SS and Ss) secrete their A and B antigens into their saliva and other body fluids,
while “non-secretors” do not. What would be the apparent phenotypic blood group proportions among
the offspring of a type AB/heterozygous secretor woman and a homozygous type A/ heterozygous
secretor man?
IAIB Ss X IAIA Ss
Gives the following children:
1 – AASS
2 – AA Ss
1 – Aass
1 ABSS
2 ABSs
1 ABss
9. In zebra finches, two pairs of non-allelic genes control body color. The dominant alleles of each gene
(Q and Z) produce normal body color and the recessive alleles of each pair (q and z) produce albinos. If
either body color gene is homozygous for albinism, the finch is an albino, no matter what the genotype for
the other gene is. Do the following crosses and determine the expected genotypic and phenotypic ratios
of the possible offspring
a. Cross between two normal colored finches who are heterozygous for both genes
This is a standard dihybrid cross between QqZz X QqZz finches
But we are looking for either qq or zz genotypes. This means that in a qqZZ finch, you
might expect a normal color because of the ZZ part of the genotype. But the qq “vetoes”
this genotype and the bird ends up an albino.
The following genotypes fulfill the condition of either homologous chromosome being
homozygous for albinism (i.e. qq or zz): QQzz, qqZZ, qqzz will be albinos and you will get
1 of each (i.e. 1/16 chance); but 2 finches will be Qqzz and 2 finches will be qqZz. These
will also be albinos and each of their probabilities will be 1/8.
b. A QQzz x qqZZ cross
A QQzz x qqZZ cross between two albino finches produces QqZz offspring – all of which will be
normal colored
10. You pick up two mice (one male, one female) that have escaped from a lab. One mouse is yellow is
yellow in color the other is brown agouti. You know that the mouse colony has animals with different
alleles at only three coat color genes. The A gene has three alleles – yellow, agouti and non-agouti. The B
gene has two alleles: black or brown. The C gene has two alleles – albino or non-albino. However, you
don’t know which alleles of these genes are actually present in the mice you have captured. So you breed
them together. The first litter has three pups – one is albino, one is brown (non-agouti) and the third is
black agouti.
a. What alleles of the A, B and C genes are present in the two mice you caught?
b. After raising several litters from these parents, you have many offspring. How many different coat
colors do you expect to see expressed in these offspring? What are these phenotypes?
This is a tough one and we will go over this in lecture
11. Complementary gene action produces a ______9:7__________ ratio in the F2
Dominant Epistasis produces a _____12:3:1 OR 13: 3______ ratio in the F2
Recessive Epistasis produces a ______9:3:4________ ratio in the F2
12. The garden flower Salpiglossis sinuate comes in many different colors. The following table shows the
crosses and the resulting phenotypes.
a. State a hypothesis explaining the inheritance of flower color in this plant – two genes, gene Y is
required in dominant form for the first gene to be visible; if the second gene is recessive, the color
is yellow
b. Assign genotypes to the parents, F1 and F2 progeny
Parents
F1 Phenotypes
F2 Phenotypes
red x blue
All red (P1 are
probably true
breeding)
102 red, 33 blue
Red > blue; 3: 1
ratio
lavender x
blue
All lavender
149 lavender, 51
blue
Lavender > blue; 3:
1 ratio
lavender x
red
All bronze
84 bronze, 43 red, 41
lavender
Lavendar = red 
bronze (incomplete
dominance)
red x yellow
All red
133 red, 58 yellow,
43 blue
1:2: 1 ratio
YELLOW x
blue
All lavender
183 lavender, 81
yellow, 59 blue
1:2: 1 ratio
Let’s see if this is due to inheritance of One gene
Red X blue = RR X rr  Rr X Rr  RR, Rr and rr
How do you get lavender?? And yellow? And Bronze?
Probably two genes being inherited
Let’s make the 1st gene for the color
CR = red
CL = lavender
CB = Blue
Let’s make the 2nd gene required for color deposition
YY or Yy = the color is visible
yy = the color is yellow no matter what the 1st gene is
a. red X blue
CRCR YY (red) X CBCB YY (blue) CRCB YY (red) X CRCB YY (red)  CRCR YY red
CRCB YY red (because red > blue)
CBCB YY blue
b. lavender x blue
CLCLYY (lavendar) X CBCBYY (blue)  CLCB YY (lavender) X CLCB YY (lavender)  CLCL YY
lavender
CLCB YY lavender (because lavender > blue)
CBCB YY blue
c. lavender x red
CLCL YY (lavendar) X CRCR YY (red) CRCL YY (bronze) X CRCL YY (bronze) CLCL YY lavender
CLCR YY bronze (incomplete
dominance)
CRCR YY red
1 lavendar: 2 bronze: 1 lavendar
d. red x yellow
CRCR YY (red) X CBCB yy (yellow)  CRCB Yy (red) X CRCB Yy (red)  CRCB Yy red
CRCR yy yellow
CBCB YY blue
2 red: 1 yellow: 1 blue
e. yellow X blue
CLCLyy (yellow) X CBCB yy (blue)  CLCB Yy (lavender) X CLCB Yy (lavender)  CLCL yy yellow
CLCB Yy lavender
1 yellow: 2 lavendar: 1 blue
CBCB YY blue
c. In a cross between true-breeding yellow and true-breeding lavender plants, all F1 are bronze. If you
cross two bronze plants, what phenotypes and what ratio would you expect?
CRCR yy X CLCL YY  CRCL Yy X CrCL Yy
Gametes will be:
CRY
CRy
CLY
CLy
CRY
CRy
CLY
CLy
4X4 cross = 16 progeny
1 of these
CRCR YY
[CRCL YY
CLCL YY
2 of these
CRCR Yy
CRCL Yy
CLCL Yy
3 red: 3 lavender: 6 bronze: 4 yellow
1 of these
CRCR yy] X 1 set
CRCL yy] X 2 sets
CLCL yy] X 1 set