Genetics Problems

All problems are to be completed on a separate sheet of paper. Be sure to label each heading.

Monohybrid cross.

Use the information from the chart to solve the following problems

Trait / Dominant Allele / Recessive allele
Pod shape / Smooth (S) / Constricted (s)
Pod color / Green (G) / Yellow (g)
Flower position / Axial-on side (A) / Terminal- on ends (a)
Plant height / Tall (T) / Short (t)
Flower color / Purple (P) / White (p)

Complete a punnett square and determine the appropriate genotype and phenotype percent as well as the correct terms to describe the genotypes

  1. Example: Aa X aa

A / a / Genotype: 50% Aa; 50% aa
a / Aa / aa / Genotype terms: 50% Heterozygous; 50% homozygous terminal
a / Aa / aa / Phenotype: 50% Axial; 50% terminal
  1. Tt X Tt

T / t / Genotype: 25% TT; 50% Tt; 25% tt
T / TT / Tt / Genotype terms: 25% homozygous tall; 50% Heterozygous;
25% homozygous short
t / Tt / tt / Phenotype: 75% Tall; 25% short
  1. Two heterozygous are crossed for green pods.

G / g / Genotype: 25% GG; 50% Gg; 25% gg
G / GG / Gg / Genotype terms: 25% homozygous green; 50% Heterozygous;
25% homozygous yellow
g / Gg / gg / Phenotype: 75% Green; 25% yellow
  1. Hybrid for axial flowers is crossed with a plant that has pure-bred terminal flowering plant.

A / a / Genotype: 50% Aa; 50% aa
a / Aa / aa / Genotype terms: 50% Heterozygous; 50% homozygous terminal
a / Aa / aa / Phenotype: 50% Axial; 50% terminal
  1. Homozygous tall with a short plant.

T / T / Genotype: 100% Tt
t / Tt / Tt / Genotype terms: 100% Heterozygous
t / Tt / Tt / Phenotype: 100% tall
  1. Heterozygous smooth pods crossed with constricted pods.

S / s / Genotype: 50% Ss; 50% ss
s / Ss / ss / Genotype terms: 50% Heterozygous; 50% homozygous constricted
s / Ss / ss / Phenotype: 50% smooth; 50% constricted

Answer the following questions using a punnett square.

  1. When a tall plant is crossed with a short plant, some of the offspring are short. What are the genotypes of the parents and the offspring? What is the phenotypic ratio in the offspring?

T / t / Genotype of parents Tt & tt
t / Tt / tt / Genotype of offspring: Tt & tt
t / Tt / tt / Phenotype: 1:1
  1. 3 out of 4 plants produced by a cross between two unknown pea plants have axial flowers and 1 out of 4 have terminal flowers. What are the genotypes of the parent plants?

A / a / Genotype of parents are both Aa
A / AA / Aa
a / Aa / aa
  1. A student wants to figure out if the tall plant he bought was a pure bred. What plant would he cross with the tall plant to determine if he bought a pure bred? What is this called?

T / t / A homozygous short plant.
t / Tt / tt / Test cross.
t / Tt / tt
  1. What cross would result in ½ offspring having green pods and ½ offspring having yellow pods?

G / g / Genotype: Gg X gg
g / Gg / gg / Genotype terms: 50% Heterozygous; 50% homozygous yellow
g / Gg / gg / Phenotype: 50% Green ; 50% yellow

Calculations based on Generations

1.In tomatoes the texture of the skin may be smooth or peach (hairy). The Ponderosa variety has fruits with smooth texture. The red peach variety has fruits with peach texture. Crosses between the two varieties produce all smooth fruits. Crosses between these smooth fruited F1 plants produced 174 peach textured fruits and 520 smooth textured fruits. How are these skin textures inherited?

Parental (P) Cross: Smooth x Peach :SS x ss

F1 Generation: All (100%) Smooth: Ss (100% heterozygous)

F2 Generation: Punnett Square Illustrating F1 Generation Cross:

F2 Generation Genotype Ratio: SS : Ss : ss - 1 : 2 : 1

Note: 520:174 = 2.99:1

2.A brown mouse is mated is mated with two female black mice. When each female has produced several litters of young, the first female has had 48 black and the second female has had 14 black and 11 brown young. Deduce the pattern of inheritance of coat color and the genotypes of all of the parents.

First define alleles: B = black (the dominant trait) b = brown (the recessive trait) Brown male must by homozygous recessive (bb).

First Cross (48 black offspring - Brown Male x Black Female (homozygous dominant) bb x BB All offspring will be heterozygous (Bb).

Second Cross (14 black, 11 brown offspring) - Brown Male x Black Female (heterozygous - bb x Bb 50% of the offspring will be brown (bb). 50% of the offspring will be black (Bb).

Dihybrid Cross

Solve the following problems. Use the FOIL technique to determine your gametes for the punnett square. You will need to use a 16 square punnett for each problem. See notes for the example of Dihybrid cross

  1. Two hybrid (heterozygous) pea plants for pod color and pod shape are crossed. What are the genotypic and phenotypic ratios in the offspring? What is the phenotype of the parents in this cross? Green, smooth (GgSs)

GS / Gs / gS / gs
GS / GGSS / GGSs / GgSS / GgSs
Gs / GGSs / GGss / GgSs / Ggss
gS / GgSS / GgSs / ggSS / ggSs
gs / GgSs / Ggss / ggSs / ggss
1 GGSS
2 GGSs
1 GGss / 2 GgSS
4 GgSs
2 Ggss / 1 ggSS
2 ggSs
1 ggss
9 Green, smooth
3 Green, constricted
3 Yellow, smooth
1 Yellow, constricted
  1. Mice running normally is dominant (R) while mice which run in circles (called waltzing) is recessive. Mice which have black hair are dominant (B) while having brown hair is recessive. If a homozygous running, black mouse is crossed with a heterozygous running, brown mouse – what are the phenotypic and genotypic ratios of the offspring? What are the genotypes of the parents? RRBB X Rrbb

RB / RB / RB / RB
Rb / RRBb / RRBb / RRBb / RRBb
Rb / RRBb / RRBb / RRBb / RRBb
rb / RrBb / RrBb / RrBb / RrBb
rb / RrBb / RrBb / RrBb / RrBb
8 RRBb
8 RrBb / or
1:1 RRBb:RrBb
16 Running, black mice
100% running, black

Incomplete Dominance

  1. In northeast Ohio there is a creature know as an oaks eagle. It comes in three colors, gold, green, and white. This trait is controlled by a single locus gene with incomplete dominance. A homozygous (GG) individual is gold, a homozygous (WW) individual is white, and a heterozygous (GW) individual is green. What would be the genotypes and phenotypes of the offspring if a gold eagle were crossed with a white one?

G / G
g / Gg / Gg / Genotype of offspring: Gg
g / Gg / Gg / Phenotype: All green
G / G
W / GW / GW / Genotype of offspring: GW
W / GW / GW / Phenotype: All green
  1. The lubber grasshopper is a very large grasshopper, and is black with red and yellow stripes. Assume that red stripes are expressed from the homozygous CRCRgenotype, yellow stripes from the homozygous CrCr genotype, and both from the heterozygous genotype. What will be the phenotypic ratio of the F1 generation resulting from a cross of two grasshoppers, both with red and yellow stripes?

R / r
R / RR / Rr / Genotype of offspring: RR, Rr, rr
r / Rr / rr / Phenotype: 1:2:1 Red to both to yellow
CR / Cr
CR / CR CR / CR Cr / Phenotype: 1:2:1 Red to both to yellow striped
Cr / CR Cr / Cr Cr
  1. Suppose you have two rose plants, both with pink flowers. You cross the two plants and are surprised to find that, while most of the offspring are pink, some are red and some are white. You decide that you like the red flowers and would like to make more. What cross would you perform to produce the most red flowered plants? Your mother decides she would like some of the pink flowered roses. Which cross would give you the most pink flowered plants?

R / W / RR x RR – All reds
R / RR / RW / RR X WW- All pinks
W / RW / WW

Multiple Alleles / Codominace

Human blood types genotypes can be expressed the following ways.

Phenotype (Blood type) / Genotype
O – universal donor / ii
A / IAIA or IAi
B / IBIB or IBi
AB + universal receiver / IAIB

1.Suppose a father of blood type A and a mother of blood type B have a child of type O. What blood types are possible in their subsequent children?

IA / i
IB / IAIB / IBi / Blood types: AB, A, B, or O
i / IAi / ii

2.Suppose a child is of blood type A and the mother is of type 0. What type or types may the father belong to?

IA / i
i / IAi / ii / Genotypes: IAIA or IAi or AB Phenotype: A blood type
i / IAi / ii

3.Suppose a father and mother claim they have been given the wrong baby at the hospital. Both parents are blood type A. The baby they have been given is blood type O. What evidence bearing on this case does this fact have?

IA / i
IA / IA IA / IA i / The parents would be IAi. It may not be theirs. They could have gotten the child
from both IBi parents as well and had an O type child.
i / IAi / ii

Polygenic Inheritance

1. Melanin is coded for when the gene L, M, or N is present. The more of these genes are present, then the darker the skin the individual. The absence of melanin is coded by X, Y, or Z alleles. The more of these genes are present, then the lighter the skin the individual. List the following individuals in order of the darkest to lightest skin color phenotypes.

LLMMNN (George) / LXMYYZ (Betty)
LLMXNN (Jane) / XXMMZZ (Barney)
LXMMYY (Fred) / LXMMNZ (Wilma)

Who would probably have darker skinned children- Fred and Betty or Fred and Wilma? Why?

Darkest to lightest: George, Jane, Wilma, Fred, Betty = Barney.

Fred and Wilma – More allele possibilities for darker skin.

Sex-Linked

In humans, the genes for colorblindness (Xb) and hemophilia (Xh) are both located on the X chromosome with no corresponding gene on the Y. These are both recessive alleles.

1.A human female "carrier" who is heterozygous for the recessive, sex-linked trait causing hemophilia, marries a normal male. What proportion of their male progeny will hemophiliac?

XH / Y
XH / XH XH / XH Y / 50% males will be hemophiliacs
Xh / XH Xh / Xh Y

2.If a man and a woman, both with normal vision, marry and have a colorblind son, draw the Punnett square that illustrates this. If the man dies and the woman remarries to a colorblind man, draw a Punnett square showing the type(s) of children could be expected from her second marriage. How many/what percentage of each could be expected?

XB / Y / Xb / Y / 25% Color blind female
25% Carrier female
25% Normal male
25% Color blind male
XB / XB XB / XB Y / XB / XB Xb / XB Y
Xb / XB Xb / Xb Y / Xb / Xb Xb / Xb Y

3.A boy, whose parents and grandparents had normal vision, is color-blind. What are the genotypes for his mother and his maternal grandparents. Use XB for the dominant normal condition and Xb for the recessive, color-blind phenotype.

XB / Y / Boy: Xb Y Mother: XB Xb
XB / XB XB / XB Y / Grandfather: XB Y Grandmother: XB Xb
Xb / XB Xb / Xb Y

4.Clouded leopards are a medium sized, endangered species of cat, living in the very wet cloud forests of Central America. Assume that the normal spots (XN) are a dominant, sex-linked trait and that dark spots are the recessive counterpart. Suppose as a Conservation Biologist, you are involved in a clouded leopard breeding program. One year you cross a male with dark spots and a female with normal spots. She has four cubs and, conveniently, two are male and two female. One each of the male and female cubs have normal spots and one each have dark spots. What is the genotype of the mother?

Xn / Y
XN / XN Xn / XN Y / Mother is XN Xn
Xn / Xn Xn / Xn Y

8. Examine the original chromosomes order. Compare the corresponding damaged one to the original and then correctly identify what abnormality is present.

ABCDEFGHIJK
PQRSTUVWXYZ / 1. ABCDEHIJK- Deletion FG / 4. PQSTUVWXZ- Deletion of R and Y
2. PQRSTUVHIJK- Translocation / 5. ABCGFEDHIJK- Inversion DEFG
3. ABCDECDEFGHIJK- Duplication of CDE / 6. PQTSRUVWXWXYZ-
Inversion RST and duplication WX
/ Pedigree #1
Blonde is an autosomal recessive trait, while black hair is a dominant trait (B).
  1. What are the possible genotypes of each individual?
A: BB B:bb C-F: Bb
  1. If person F marries someone who is blonde, what percent chance will their children have of being blonde? Of being black haired?
b / b / 50% blonde (bb)
B / Bb / Bb / 50% black (Bb)
b / bb / bb / ratio 1:1
  1. What family relationship is C to F? Brother and sister

Pedigree #2

High Cholesterol (H) is a dominant trait which results in high levels of LDL cholesterol. High LDL levels put an individual at risk for heart attacks at a young age.
  1. What relation is Generation III – individual 1 to Generation I Individual 1? What relation is Generation I – individual 2 to Generation III – individual 2?
Grand-daughter; Grandfather
  1. What happened to the Generation II, Individual #10? He died.
Create a punnett square from the parents of Generation I to answer the following questions.
h / h
H / Hh / Hh
h / hh / hh
  1. What is the genotype of the couple in Generation I? Hh X hh
  2. What are the possible genotypes of Generation II offspring? What percent chance will the offspring have of having high cholesterol?
50% Hh & 50% hh; 50% high cholesterol
  1. What percent chance does Generation II – couple 3 & 4 have of passing on the high cholesterol trait?
How do you know? 0% both are homozygous recessive.
h / h
h / hh / hh
h / hh / hh
  1. What are the possible genotypes of the offspring of Generation II – couple 7 & 8?
h / h
H / Hh / Hh / 50% Hh & 50% hh
h / hh / hh
  1. Generation III – Individual 1 marries a homozygous male for the high cholesterol trait. Using a punnett square, What percent chance will the offspring be normal (no high cholesterol)?
H / H
H / HH / HH / 0 % All will have high cholesterol
h / Hh / Hh

Pedigree # 3

Case: Dennis 25 and Lucy 22 are married and going through pre-pregnancy genetic counseling. Dennis is the youngest child. He has two normal brothers, Jack 30 and Frank 28 and a normal sister, Cindy 32. He also had a brother, Ted the oldest and a sister, Julie, the second oldest who died at 22 and 18 years, respectively, of cystic fibrosis (CF). It is a recessive trait causing poor absorption do to pancreatic insufficiency and recurrent pulmonary infection with progressive respiratory insufficiency, usually leading to death in early adulthood. Dennis is in good health (normal for both alleles CC). Lucy is in good health. Dennis' parents are alive and well in their late 50's and none of their other relatives have died young or had similar problems.

  1. Draw and label the pedigree with names of those involved. Be sure to differentiate carriers, those who had CF, and those who are deceased.
  2. Why did some of Dennis’s siblings have CF and other did not? Parents both had recessive trait.
  3. What terms could you use to describe Dennis’s parents? Carriers, Heterozygous, hybrids.
  4. If Lucy is a normal, what will be the percent of genotypes and phenotypes of the children?

C / c / If Dennis a carrier / C / C / Dennis normal
C / CC / Cc / 50% Carrier / C / CC / CC / 100% normal
C / CC / Cc / 50% normal / C / CC / CC
  1. If Lucy is a carrier of CF, what will be the percent of genotypes and phenotypes of the children?

C / c / If Dennis a carrier / C / C / Dennis normal
C / CC / Cc / 25% normal / C / CC / CC / 50% normal
c / Cc / cc / 50% carrier / c / Cc / Cc / 50% carrier
25% cystic fibrosis
  1. Based on #4 and #5 would you recommend them having more children of their own? Why or why not?

Yes but their offspring will possibly be carriers.

Pedigree #4

Red-green color blindness is an X-linked, recessive trait (b). In this problem set we will establish the pedigree of Audrei's family and see how the color perception defect is passed on from one generation to the next, but first let's look at a brief introduction to sex-linked inheritance.

Construct a pedigree based on the following information.

Audrei's family

There are 7 children in Audrei's family, three girls and four boys. Two of the girls, Audrei and Liz, are red-green color blind. Caroline has normal color vision. Only two of the boys have been tested. Paul is color blind and David has normal color perception. Andrew and Jason, who have not been tested, may or may not have normal color perception.
Barbara, the mother of the seven children, has normal color vision, but Sidney, the father, has the red-green color perception defect. Audrei also has a half brother Stephan. Audrei and Stephen have the same mother, but a different father. Stephan is also red green color blind.
  1. Audrei is the family member who contacted us. She and her father Sydney are color blind, but her mother, Barbara, has normal vision. What is Audrei's genotype?

Xb / Y
XB / XB Xb / XB Y
Xb / Xb Xb / Xb Y
  1. Now that we determined that Audrei is homozygous recessive for the red-green color blind allele. What is her father's genotype? Xb Y
  1. We have determined that Audrei is homozygous recessive for the red-green color blind allele and that her father is hemizygous with a recessive allele. What is the mother's genotype? Remember that Audrei's mother does not have a color perception defect. XB Xb
  1. What is the genotype of Audrei's sister Caroline, who has normal vision? XB Xb
  1. In her message, Audrei also tells us that her sister Liz and brother Paul are also color blind, but that her brother David has normal vision. How will Liz be represented on the family pedigree? Solid circle.