Mendelian (Dominant/ Recessive) Genetics Problems

  1. What ratio of phenotypes would you expect in each of the following pea crosses? Tall (T) dwarf (t)

a) heterozygous tall x homozygous tallb) heterozygous tall x dwarf

  1. In humans, brown eyes (B) are dominant to blue eyes (b). Indicate the phenotypes possible from each of the following crosses:

a)homozygous brown eyed father and blue eyed mother

b)heterozygous brown eyed mother and blue eyed father

c)two heterozygous parents

d)two blue eyed parents

e)What is the genotype of a brown eyed man whose father had brown eyes and mother had blue eyes?

f)What are the expected phenotypes of the children if the man in question e married a blue eyed woman?

  1. In squash, the allele for white colour (W) is dominant to the allele for yellow colour (w). A yellow squash was crossed with a white squash. The seeds were planted and produced 38 white squash and 42 yellow ones. State the genotypes of the parent plants.
  1. In cocker spaniels, black (B) is dominant to red (b). Two black cocker spaniels have 8 puppies - 5 black and 3 red. What is the genotype of each parent?
  1. In hogs, the allele that causes the black band (B) around the belly is dominant to the allele for a uniformly coloured hog (b). How many banded piglets should be expected in a litter of 12 if a uniformly coloured males is mated with a banded female whose mother was uniform?
  1. In some breeds of dog, the allele for erect ears (E) is dominant to the allele for drooping ears (e). State the genotypes of the parents and pups when an erect eared male mates with a droop-eared female and produces all erect eared pups.
  1. The black colour (B) is dominant in cats while the white coat is recessive (b). A certain black cat (Cat A) is mated to three other cats. Cat B, which is white, gives birth to a black kitten. Cat C, which is also white, gives birth to white kittens. Cat D, which is black, produces a white kitten. What are the genotypes of the four parents?

Answers

1a) 50% TT 50% Tt
1b) 50% Tt 50% tt
2a) 100% Bb
2b) 50% Bb 50% bb
2c) 25% BB 50% Bb 25% bb
2d) 100% bb
2e) Bb
2f) 50% Bb 50% bb / 3. Ww x ww
4. Bb x Bb
5. 6 banded and 6 uniform
6. P = EE x ee, F1 100% Ee
  1. Cat A - Bb Cat B - bb Cat C - bb Cat D - Bb

Dihybrid Problems

  1. In humans, brown (B) hair is dominant to blond (b) hair and curly (C) hair is dominant to straight (c) hair. A heterozygous brown, heterozygous curly haired man marries a blond straight haired woman. Show the possibilities for their offspring using a Punnet square.
  1. In chickens, red (R) coat is dominant to brown (r) and checkered (C) is dominant to plain (c). Conduct the following crosses and give the phenotypes of the F1 generation.

a)homozygous checkered red x plain brown

b)heterozygous checkered red x heterozygous checkered red

c)heterozygous checkered, brown x plain, heterozygous red

  1. In corn plants, tall (T) is dominant to short (t) and green (G) is dominant to yellow (g). A pure tall, heterozygous green corn plant is crossed with a short, yellow plant. Show the F1 cross then use those results for the F2 cross.
  1. In mice, the allele for coloured fur (F) is dominant to the allele for white fur (f). Also the allele for normal behaviour (N) is dominant to the allele (n) for waltzing. Diagram the results of the following crosses and list the genotypes and phenotypes for each.

a)FfNn x ffnn

b)ffnn x ffnn

c)FfNn x FfNn

  1. In some dogs the allele for erect ear (E) is dominant to the allele for droopy ear (e) and the allele for barking (B) is dominant to the allele for no barking (b). Diagram the following crosses and list the phenotypes and genotypes for each:

a)a pure bred erect eared silent dog is crossed with a droopy eared, pure bred barking dog

b)a heterozygous erect eared, pure bred barking dog is crossed with a droopy eared silent dog

c)two heterozygous, barking dogs are crossed

  1. A man who has blood type AB and is Rh negative married a woman who was type O, Rh positive. One of their children has Rh negative blood.

a)What are the parent’s genotypes? What are the possible blood types of any children they may have?

b)This couple is quite wealthy and when they die a man comes forward and claims to be their son. His blood type is O positive. Could he be correct?

c)What are the possible blood types of their children if a heterozygous Type A, negative man marries a heterozygous Type B, heterozygous positive woman?

Answers

  1. 25% brown curly, 25% blond curly, 25% brown straight, 25% blond straight
  1. a) CcRr
  1. 9 : 3 : 3 : 1 for checkered red, checkered brown, plain red, plain brown
  2. 25% checkered red, 25% checkered brown, 25% plain red, 25% plain brown
  1. F1 cross = 50% TtGg, 50% Ttgg

F2 cross = 3/8 tall green, 3/8 tall yellow, 1/8 short green, 1/8 short yellow

  1. a) 25% coloured, normal; 25% coloured, waltzing; 25% white, normal;

25% white, waltzing

b)100% white, waltzing

c)9 : 3 : 3 : 1 for coloured normal, coloured waltzing, white normal, white waltzing

  1. a) EEbb x eeBB F1 cross = 100% EeBb

b) EeBB x eebb F1 cross = 50% EeBb, 50% eeBb

c) EeBb x EeBb

F1 cross = 9 : 3 : 3 : 1 for erect barking, erect silent, droopy barking, droopy silent

  1. a) The parents are IAIBRh-Rh-and iiRh+Rh-

b)No he could not – there is no i allele for the father to pass on.

c) IAiRh-Rh- x IBiRh+Rh- possible outcomes – IAiRh+Rh-, IAiRh-Rh-, IAIBRh+Rh-,

IAIBRh-Rh-, IBiRh+Rh-, IBiRh-Rh-, iiRh+Rh-, iiRh-Rh-

Incomplete Dominance and Co-dominance Problems

  1. In cattle, there is an allele for red coat colour (CR) and an allele for white coat colour (CW). In a heterozygous individual the coat is mixture of red hair and white hair that results in “roan” hair/ colour.

a)Is this an example of incomplete dominance or co-dominance?

b)What phenotypes would be predicted if 2 roan cattle were mated?

c)Explain how the farmer could build a herd or red or white cattle from the cattle in b)?

  1. Two short tailed (Manx) cats are bred together. They produce three kittens with long tails, five with short tails, and two without any tails. From these results, how would you think that tail length in cats is inherited? Show genotypes to support your explanation.
  1. Pooh bear has a colony of Tiggers whose stripes go across the body (SA). His American cousin Yogi Bear sent him a Tigger whose stripes run lengthwise (SL). When Pooh crossed this Tigger with one of his own he obtained plaid Tiggers. Interbreeding among the plaid Tiggers produced litters of a majority of plaid but some crosswise and lengthwise stripes as well.

a)What type of inheritance pattern is involved?

b)Use a Punnet square to support these results.

  1. A pure bred red snapdragon (CRCR) is cross-pollinated with a pure bred white snapdragon (CWCW). The F1 generation all have pink phenotypes. What is the F1 genotype? Outline the results you would expect if the F1 snapdragons were allowed to self-pollinate. What is the phenotype ratio? The genotype?
  1. When two Mexican Hairless dogs are mated, about 1/3 of the pups have hair, 2/3 of the pups have no hair and some deformed pups are stillborn. Explain this situation.
  1. In 4 o’clock flowers, there is a red allele (CR) and a white allele (CW). Individuals that are homozygous for red allele are red; those homozygous for the white allele are white. Heterozygotes appear pink. State the phenotypes expected for each of the following crosses:

a)red x white

b)pink x white

c)pink x pink

Answers

1a) codominance

b) 1 red 2 roan 1 white

c) breed together only the red to red for a red herd or white to white for a white herd

2. The short tailed cats are the heterozygotes so this would be incomplete dominance because the short tail is a blend of long tail and no tail. Let TL be long tail and TN be no tail, then the parents (short tails) are TLTN

3a) co-dominance because both traits are expressed

b) P = SASA x SLSL

F1 = 100% SASL

F2 cross: genotypesphenotypes

SA / SL
SA / SASA / SA SL / ¼ SASA / 25% Stripes across
SL / SA SL / SLSL / ½ SA SL / 50% Plaid
¼ SLSL / 25% Stripes lengthwise
  1. F1 genotype = CRCW

F1 cross = CRCW x CRCW

F2 results = 25% CRCR, 50% CRCW, 25% CWCW

  1. This is an example of a lethal allele combination. The actual cross results are 1:2:1 but since ¼ die that makes it seem as if the ratio is 1/3 and 2/3.
  1. a) 100% pink
  2. 50% red, 50% white
  3. 25% red, 50% pink, 25% white

Multiple Allele Problems

  1. The human blood groups A, B, AB, and O are determined by interaction of three possible alleles for blood characteristics: A, B, or O. What blood types could be expected from the following crosses?

a) IAIB x ii b) IAIB x IAIAc) IAIB x IBid) IAi x IBi

  1. Two mothers delivered babies in a small British hospital during an air raid in World War II. In the confusion, no one remembered which baby belonged with which mother. Therefore all the individuals were asked to submit to a blood test. Baby 1 was found to be Type O and baby 2 was Type A blood. One married couple (the mother and her husband) was Type A. The other as type AB. Because her husband was a prisoner of war, he could not be tested at that time. With this information, is it possible to correctly identify the real mother for each baby? Explain your answer.
  1. A man has Type A blood and his wife has Type B blood. A physician types the blood of their four children and finds one of each of the four blood types among them. Show how this is possible.
  1. A wealthy elderly couple dies together in a car accident. Soon a man shows up claiming to be their only son who ran away from home when a boy. Other relatives dispute his claim. Hospital records show that the deceased couple was Types AB and O respectively. The claimant to their fortune was Type O. Do you think the claimant was an imposter? Will he inherit? Explain.

Answers

  1. a) 50% IAi, 50% IBi

b) 50% IAIB, 50% IAIA

c) 25% IAIB, 50% IBIB, 25% IBi

  1. 25% IAIB, 25% IAi, 25% IBi, 25% ii
  1. Baby 1 must belong to the first couple (type A) because if they are both heterozygotes for Type A then they can have a type O baby. The other couple cannot because the mother will contribute either an A or B allele so the child can’t be type O.
  1. They were both heterozygotes.
  1. No, because any child they had would have to be either type A or B.

Sex linked trait Problems

Remember that these types of problems often require a pedigree chart to help demonstrate your answer. Also, results must be gathered by gender.

  1. What are the chances that a colour blind man and a carrier woman could have a colour blind son? Is there a chance of having a colour blind daughter?
  1. What are the chances that a normal woman who had a hemophiliac father could have a hemophiliac son if her husband is normal?
  1. A hemophiliac male marries a normal woman whose father was a hemophiliac. What are the chances of having a hemophiliac daughter?
  1. A normal woman had a sister who had a hemophiliac son. If the woman marries a normal male, what would be the chances of having a hemophiliac son as well?
  1. A colour blind woman marries a colour blind man. Is there any chance they could have a normal child?
  1. Given the genotypes for each person of the following pedigree for colour blindness. Mark in the carriers.

Answers

  1. P = XbY x XbX 50% chance of a colour blind son and 50% chance of a carrier daughter
  1. 50% chance
  1. 50% chance
  1. 50% chance
  1. no
  1. I2, I3 and II6 are carriers