Review Sheet answers

1. Garden peas are a great visible example of heredity. The plants have many characteristics (pea color, pea shape, pod color and flower color) which are all determined by a single gene and assort independently.
   
2. Only dominant phenotypes can be heterozygous. In order to show a recessive phenotype the genotype MUST be homozygous because with even one dominant allele, the phenotype will be dominant.
   
3. No. you must possess two recessive alleles to show the allele phenotype.
   
4. A test cross is a cross done in order to determine the genotype of a parent. It is used when the parent displays the dominant trait, but it is unknown whether or not the genotype is homozygous or heterozygous. If the unknown parent is crossed with a recessive phenotype parent and all progeny display the dominant genotype, the unknown parent is homozygous dominant. If the cross results with half of the offspring with the recessive phenotype and half with the dominant, then the unknown parent is heterozygous.
   
5. The gene for the flower color is showing incomplete dominance. In this situation the two flowers are both homozygous (one for blue phenotype and the other for yellow phenotype). When the grower crosses both flowers all of the offspring are heterozygous and are thus green.
   
6. Mendel's law of independent assortment
   
7. Each offspring will receive one allele, this follows mendel's law of segregation.
   
8. When you cut one plant and replant it, you are not changing any genetic material. A seed may yield a plant that varies from the parent because of crossing over. During meiosis homologous chromosomes align and genetic material exchanges between chromosomes. Ultimately at the end of meiosis, all sister chromatids separate and you end up with four daughter cells which genetically different material.
   
9. No, genes assort independently. Normally, the expression of one gene does not determine the expression of another.
   
10. The F1 generation would yield all heterozygous offspring, and so the F2 generation would be determined by crossing two heterozygous plants. When you cross two heterozygous parents, the offspring will have the phenotypic ratio of 3(dominant) to 1(recessive). Therefore, 25% of the offspring will show the recessive trait which is 8/32.
    
11. This varies, but typically one.
    
12. Seed color – Yy, Seed shape – Rr
    

    
    

    	YR	Yr	yR	yr
    YR	YYRR	YYRr	YyRR	YyRr
    Yr	YYRr	YYrr	YyRr	Yyrr
    yR	YyRR	YyRr	yyRR	yyRr
    yr	YyRr	Yyrr	yyRr	yyrr

    Yellow, round: 9, Green, round: 3, Yellow, wrinkled: 3, Green, wrinkled: 1
    

    
     

    Two heterozygous parents crossed in a dihybrid cross is always 9:3:3:1
    

     

13. Gametes produced are:
    
    1. E,F
       
    2. E,e,F
       
    3. e, f
       
    4. E,e,F,f
       
14. Grandparents = ¼, great grandparents 1/8
    
15. No. Hand cells are autosomal cells (non sex cells). Changes in cells in your hand will only affect future hand cells. Offspring are only affected by daughter cells produced in meiosis, which takes place in sex cells.
    
16. 50% chance. Each time you conceive you have a 50/50 chance of having a boy or a girl. The sex of previous children do not affect the sex of future children.
    
17. A roans genotype is heterozygous, Rr. Two roans cannot produce all roan colored offspring, the offspring have a 25% chance of being red, 50% chance of being roan and 25% chance. Roan cows are displaying codominance, when the cow has one allele for red, and another for white, both are expressed resulting in roan color. Since only 50% of the offspring will be heterozygous, only 50% will be roan.
    
18. For some traits, there are more than two possible alleles. An example of this is blood types and fruit fly eye color.
    
19. Don't answer this question.
    
20. Some blood types express co-dominance, and other complete dominance. There are multiple alleles, and blood type is made up of two genes (one for ABO, the other Rh- or +).
    
21. Both parents are heterozygous. Mary has AO, John has BO, Joan has OO, James has AO, Pete has BO.
    
22. 50% chance of B, 50% chance of O.
    
23. No there was not. It would be impossible for Mr Doe and Mrs Doe to have a child with O blood type because Mrs Doe can only produce gametes with A or B which will always be dominant over O.
    
24. Yes, both parents could be heterozygous (BO) and would then be able to have children that are OO (25% chance).
    
25. Because females have 2 X chromosomes. If one of the X chromosomes has a mutation they have another correct allele on their other X chromosomes. Males only have one shot with certain traits because they only have 1 X chromosome. If there is a problem with that allele they will express the recessive phenotype.
    
26. See 25
    
27. If parents know their genotype, which was probably determined using a pedigree, they can calculate the probability of their child inheriting a specific trait. For some parents this could deter them from having children.
    
28. In science a large sample size is always important. It is impossible to say that something is statistically correct without repeating it many times, you always have to account for randomness and error.
    
29. G
    
30. Farmers could selectively breed for sheep that can only produce white offspring, which would be sheep with the genotype WW.
    
31. The best option would be to cross the two together, if there is only one phenotype visible than this is the dominant trait, if both phenotypes are present, let's say two hairy and two naked, than create an F2 generation to see what the phenotypic ratio would be.
    
32. Inbreeding is discouraged because it limits genetic diversity and recessive traits (which many diseases are) tend to surface more often.
    
33. If the trait never appears in males than most likely having it is so lethal that the males don't even make it to birth.
    

Females can be carriers if one X chromosome carries the allele for the recessive trait and the other carries the allele for the dominant trait. Since females have 2 X chromosomes the dominant trait will be expressed and the recessive will do unnoticed, but the female still has a 50/50 chance of passing the recessive trait on to the offspring. If the offspring is a male, they will only receive 1 X chromosome in total (because their father gave them their Y chromosome), and the recessive trait will be expressed.

Sex Linked Problems

1. Mate a colorblind male and a carrier female.
   a) What are the chances of having a normal boy?
   b) What are the chances that their sons will be colorblind?
   c) What are the chances of having a normal girl?
   d) What are the chances that their daughters will be carriers?
   
2. Red-green colorblindness is caused by a sex-linked recessive allele. A color blind man marries a woman with normal vision whose father was color blind. What is the probability that they will have a color blind daughter? What is the probability that their first son will be color blind?
   
3. A man with hemophilia (a recessive, X-linked condition) has a daughter of normal phenotype. She marries a man who is normal for the trait.
   a) What is the probability that a daughter of this mating will be hemophiliac?
   b) What are the chances of having a son with the disorder?
   c) What are the chances of their daughters being a carrier?
   
4. A recessive sex-linked gene (h) located on the X chromosome increases blood-clotting time. This causes the genetic disease, hemophilia.
   

   a) Explain how a hemophilic offspring can be born to two normal parents.
   

    b) Can a female ever develop hemophilia?
   

   
    

Practice Problems (More) and Solutions

Learning Genetics Can Be Fun

1. A pea plant with round seeds is crossed with a pea plant that has wrinkled seeds. For the cross, indicate each of the following:

a) the genotype of each of the parents if the round seed plant is heterozygous.

b) the gametes produced by the parents

c) the genotypes and phenotypes of the F₁ generation

d) the F₂ generation if two round plants from the F₁ generation were crossed

2. A plant of genotype TTYy was crossed with one of genotype ttYy. If the cross yielded 800 seeds, what number of what phenotypes probably resulted? What were their genotypes?

3. In summer squash, white fruit color is dominant. Yellow is recessive. A squash plant that is homozygous for white is crossed with a homozygous yellow one. Predict the appearance of

a) the F₁ generation (F₁ = all white)

b) the F₂ (F₂ = 3/4 white and 1/4 yellow)

c) the offspring of a cross between an F₁ individual and a homozygous white individual. (A crossbetween an F₁ individual and a homozygous white individual wouldproduce only white offspring.)

4. For Dalmatian dogs, the spotted condition is dominant to non-spotted.

a) Using a Punnett square, show the cross between two heterozygous parents.

b) A spotted female Dalmatian dog is mated to an unknown male. From the appearance of the pups, the owner concludes that the unknown male was a Dalmatian. The owner notes that the female had six pups, three spotted and three non-spotted. What are the genotype and phenotype of the unknown male?

5. In horses, the trotter characteristic is dominant to the pacer characteristic. A male trotter mates with three different females, and each female produces a foal. The first female, a pacer, gives birth to a foal that is a pacer. The second female, also a pacer, gives birth to a foal that is a trotter. The third female, a trotter, gives birth to a foal that is a pacer. Determine the genotypes of the male, all three females, and the three foals sired.

6. Imagine for hair color that B gives brown hair and b gives blonde hair. Use a Punnett square to determine the following in a cross of two heterozygous parents.

a) What are the chances of the offspring being homozygous brown haired?

b) What are the chances of the offspring having blonde hair?

c) What are the chances of the offspring being heterozygous brown haired?

d) What is the genotypic ratio?

e) What is the phenotypic ratio?

f) Is there a heterozygous blonde haired offspring? Why?

g) If curly hair is dominant to straight hair, what letters will we use to show these genes?

h) A heterozygous curly haired male marries a straight haired female. What are their genotypes?

i) What would be the gametes for the male parent?

j) What would be the gametes for the female parent?

k) What are the chances of the offspring being homozygous curly haired?

l) What are the chances of the offspring having straight hair?

m) What are the chances of the offspring being heterozygous curly haired?

n) What is the genotypic ratio?

o) What is the phenotypic ratio?

p) Is there a heterozygous straight haired offspring? Why?

7. Blonde hair is due to a recessive gene, while brown hair is dominant. Agnes has brown hair, but

her mother was a blonde. Ralph, her husband, has brown hair also, and his father was blonde. Is it

possible for Ralph and Agnes to have blonde children? If so, what percentage?

8. Thalassemia is a serious human genetic disorder that causes severe anemia. The homozygous condition (T^mT^m) leads to severe anemia. People with thalassemia die before sexual maturity. The heterozygous condition (T^mTⁿ) causes a less serious form of anemia. The genotype TⁿTⁿ causes no symptoms of the disease. Indicate the possible genotypes and phenotypes of the offspring if a male with the genotype T^mTⁿ marries a female of the same genotype.

9. For guinea pigs, black fur is dominant to white fur color. Short hair is dominant to long hair. A guinea pig that is homozygous for white and homozygous for short hair is mated with a guinea pig that is homozygous for black and homozygous for long hair. Indicate the phenotype(s) of the F₁ generation. If two hybrids from the F₁, generation are mated, determine the phenotypic ratio of the F₂ generation.

10 Black coat color (B) in cocker spaniels is dominant to white coat color (b). Solid coat pattern (S) is dominant to spotted pattern (s). The pattern arrangement is located on a different chromosome than the one for color, and its gene segregates independently of the color gene. A male that is black with a solid pattern mates with three females. The mating with female A, which is white, solid, produces four pups: two black, solid, and two white, solid. The mating with female B, which is black, solid, produces a single pup, which is white, spotted. The mating with female C, which is white, spotted, produces four pups: one white, solid; one white spotted; one black, solid; one black, spotted. Indicate the genotypes of the parents.

11. For human blood type, the alleles for types A and B are codominant, but both are dominant over the type O allele. The Rh factor is separate from the ABO blood group and is located on a separate chromosome. The Rh+ allele is dominant to Rh-. Indicate the possible phenotypes from the mating of a woman, type,O, Rh-, with a man, type A, Rh+.

12. In a disputed paternity case, a woman with blood type B has a child with type O, and she claimed that it had been fathered by a man with type A. What can be proved from these facts? (man could be father but this is not proof)

13. Multiple alleles control the intensity of pigment in mice. The gene D¹ designates full color, D² designates dilute color and D³ is deadly when homozygous. The order of dominance is D¹ > D² > D³. When a full color male is mated to a dilute color female, the offspring are produced in the following ratio: two full color to one dilute to one dead. Indicate the genotypes of the parents.

14. Multiple alleles control the coat color of rabbits. A gray color is produced by the dominant allele C. The C^ch allele produces a silver-gray color when present in the homozygous condition, C^chC^ch, called chinchilla. When C^ch is present with a recessive gene, a light silver-gray color is produced. The allele C^h is recessive to both the full color allele and the chinchilla allele. The C^h allele produces a white color with black extremities. This coloration pattern is called Himalayan. An allele C^a is recessive to all the other alleles. The C^a allele results in a lack of pigment, called albino. The dominance hierarchy is C > C^ch > C^h > C^a. The table below provides the possible genotypes and phenotypes for coat color in rabbits. Notice that four genotypes are possible for full color but only one for albino.

Phenotypes Genotypes

Full color CC, CC^ch, CC^h, CC^a

Chinchilla C^chC^ch

Light gray C^chC^h, C^chC^a

Himalaya C^hC^h, C^hC^a

Albino C^aC^a

a) Indicate the genotypes and phenotypes of the F₁ generation from the mating of a heterozygous Himalayan coat rabbit with an albino coat rabbit.

b) The mating of a full color rabbit with a light gray rabbit produces two full color offspring, one light gray offspring, and one albino offspring. Indicate the genotypes of the parents.

c) A chinchilla color rabbit is mated with a light gray rabbit. The breeder knows that the light grey rabbit had an albino mother. Indicate the genotypes and phenotypes of the F₁ generation from this mating.

d) A test cross is performed with light gray rabbit, and the following offspring are noted: five Himalayan color rabbits and five light gray rabbits. Indicate the genotype of the light-gray rabbit.

15. A geneticist notes that crossing a round shaped radish with a long shaped radish produces oval shape radishes. If oval radishes are crossed with oval radishes, the following phenotypes are noted: 100 long, 200 oval, and 100 round radishes. For the cross between a pure-breeding round and a pure-breeding long radish, use symbols to show the F₁ and F₂ generations.

16. Palomino horses are known to be caused by the interaction of two different genes. The allele C^r in the homozygous condition produces a chestnut, or reddish color, horse. The allele C^m produces a very pale cream color, called cremello, in the homozygous condition. The palomino color is caused by the interaction of both the chestnut and cremello alleles. Indicate the expected ratios in the F₁ generation from mating a palomino with a cremello.

17 In mice, the gene C causes pigment to be produced, while the recessive gene c makes it impossible to produce pigment. Individuals without pigment are albino. Another gene, B, located on a different chromosome, causes a chemical reaction with the pigment and produces a black coat color. The recessive gene, b, causes an incomplete breakdown of the pigment, and a tan, or light-brown, color is produced. The genes that produce black or tan coat color rely on the gene C, which produces pigment, but are independent of it. Indicate the phenotypes of the parents and provide the genotypic and phenotypic ratios of the F₁ generation from the following crosses:

a) CCBB x Ccbb (b) ccBB x CcBb (c) CcBb x ccbb (d) CcBb x CcBb

18. For ABO blood groups, the A and B genes are codominant but both A and B are dominant over type O. Indicate the blood types possible from the mating of a male who is blood type O with a female of blood type AB. Could a female with blood type AB ever produce a child with blood type AB? Could she ever have a child with blood type O?

19. Baldness (H^B) is dominant in males but recessive in females. The normal gene (Hⁿ) is dominant in females, but recessive in males. Explain how a bald offspring can he produced from the mating of a normal female with a normal male. Could these parents ever produce a bald girl? Explain your answer.

20. Use the phenotype chart (pedigree) at the right to answer the following questions.

a) How many children do the parents A and B have?

b) Indicate the genotypes of the parents.

c) Give the genotypes of M and N.

21. For the family outlined, draw a pedigree

a) Generation I : A male with blood type O marries a female with blood type A.

b) Generation II : They have 4 children. The first born is a male with A type blood, the second born is a girl with O type blood, and the last two children were identical twin boys.

c) Generation III : The last boy married a woman with type B blood. They had two girls, one with AB blood type and one with B blood type.

SOLUTIONS:

http://kvhs.nbed.nb.ca/gallant/biology/genetics_learning_problems_121_solutions.html