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Anonymous Carolina Castillo2
wrote...
2 months ago
1) If two offspring result from the cross between a heterozygous hen and a heterozygous rooster, what is the probability both chicks will have the same homozygous dominant genotype (BB)?

If a heterozygous black hen mated with a white rooster, what is the probability an offspring would be a black rooster?

a. 0.50

b. 0.25

c. 0.75

d. 0.13




2) Sickle cell anemia is a disease that is caused by a mutation in the gene that produces hemoglobin.  Hemoglobin carries oxygen in red blood cells.  The HbA allele produces normal hemoglobin and the HbS allele produces hemoglobin that sticks together and causes red blood cells to sickle.  Heterozygous indivuals (HbAHbS) produce both normal and "sickle" hemoglobin so the HbA and HbS alleles are codominant.  Heterozygotes do not develop sickle cell anemia and are described as having the sickle cell trait.  Individuals that are homozygous for the sickle allele (HbSHbS) only produce "sickle" hemoglobin and develop sickle cell disease.

Note:  Individuals that have sickle cell trait and sickle cell disease have protection against malaria so the HbS allele is higher in areas where malaria is common. 

A man who is homozygous for the normal allele married a woman with the sickle cell trait. What is the expected probability of this couple having a boy with the sickle cell trait?

Record your answer as a value between 0 and 1 rounded to two decimal places

ANSWER





3) A man with the sickle cell trait married a woman with the sickle cell trait. Determine the probability that they will have children with the sickle cell trait or sickle cell disease.

Record your answer as a value between 0 and 1 rounded to two decimal places.

ANSWER




4) In horses, tobiano is a white spotting pattern. The tobiano allele (T) is dominant over the non-tobiano (t) allele.

What is the probability, expressed as a percentage, of two heterozygous horses producing two non-tobiano foals in succession?

Express your answer as a percentage rounded to two decimal places.

ANSWER %



5) What is the probability of the third foal being tobiano?

Record your answer as a value between 0 and 1 rounded to two decimal places.

ANSWER




6) Three different genes determine the coat colour of cats. One of the genes controls the black-based coat colour. There are three alleles for black-based coat colour genes with the following order of dominance: black (B) > chocolate (bch) > cinnamon (bc).

Two black cats were crossed. The offspring produced were three black kittens and one chocolate kitten.

Which of the following rows identifies the possible genotypes of the parents?

a. Parent 1          Parent 2
       BB                    Bb^ch

b. Parent 1          Parent 2
       Bb^c                   Bb^c

c. Parent 1         Parent 2
       Bb^c                 BB

d. Parent 1          Parent 2
      Bb^ch                 Bbc






7) What is the probability of having a male chocolate kitten?

a. 0.25

b. 0.06

c. 0.13

d. 0.50




8) In rabbits, white fur colour (W) is dominant over black fur colour (w), and upright ears (U) are dominant over floppy ears (u).

A white rabbit with upright ears, heterozygous for both traits, mated with a homozygous white rabbit with floppy ears.

What are the possible phenotypes of the offspring?

a. White floppy and black upright

b. Black floppy and black upright

c. Black floppy and white upright

d. White floppy and white upright





9) A heterozygous white rabbit with floppy ears mated with a black rabbit with heterozygous upright ears.

What is the probability of having a black kit with upright or floppy ears?

a. 0.50

b. 0.25

c. 0.06

d. 0.13





10) Two white rabbits with upright ears, heterozygous for both traits, mated and produced offspring.

What is the probability the first offspring has black fur with upright ears and the second offspring has white fur with floppy ears?

a.  9 / 256

b.  3 / 8

c.  3 / 16

d.  3 / 128

e.  1 / 16



11) In Labrador retrievers, two genes work together to determine coat colour.The first gene affects the colour of the dark pigment, eumelanin. A genotype containing at least one dominant allele of this pigment gene results in black pigmentation (B_). A homozygous recessive genotype of this gene results in brown pigmentation.

The second gene affects whether the pigment eumelanin is present in the fur. If a dog has one dominant allele for this expression gene (E_), fur colouration is determined by the pigment gene. If a dog is homozygous recessive for the extension trait, the dog will have a yellow coat.

 Description                                         Genotype(s)

Black coat                                    BBEE / BbEE / BBEe / BbEe

Chocolate coat                                  bbEE / bbEe

Yellow coat                                   BBee / Bbee / bbee


What percentage of chocolate labs would be produced from the mating of a yellow lab who is homozygous recessive for both genes to a black lab who is heterozygous for both coat colour and expression?

a. 25%

b. 50%

c. 33%

d. 75%






12) A male and female have three children, all girls. The couple is expecting their fourth child. The probability their fourth child will be female is __i__, and the probability the couple will have four female children is __ii_.

The row that indicates the correct probabilities is

a.            i                           ii
           1 /4                      1 / 8

b.           i                           ii
          1 / 16                    1 /4

c.           i                          ii
          3 / 4                    1 /4

d.          i                           ii
         1 /2                       1 / 16




13) For each of the following traits, select whether it is a continuous trait or a discrete trait in humans.

         Trait                     Discrete trait / Continuous trait

       Height                                      Answer
 

    Albinism                                     Answer
 

  Learning ability                          Answer
 

   Body weight                              Answer
 

  Number of fingers                     Answer





14) Which of the following scenarios best describes an epistatic gene?

a. Fur pattern is controlled by two genes such that one gene inhibits the expression of the other gene.

b. Mouse fur colours are expressed as a blend of the dominant black colour and the recessive white colour, resulting in a heterozygous brown colour.

c. Individuals with sickle cell anemia and carriers of the sickle cell allele have some resistance to malaria.

d. Chicken combs follow a dominance hierarchy whereby a walnut comb is the most dominant and a single comb is the least dominant.
 

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wrote...
Staff Member
2 months ago
Quote
1) If two offspring result from the cross between a heterozygous hen and a heterozygous rooster, what is the probability both chicks will have the same homozygous dominant genotype (BB)?

If a heterozygous black hen mated with a white rooster, what is the probability an offspring would be a black rooster?

a. 0.50

Source: https://biology-forums.com/index.php?topic=1815094.0

Quote
2) Sickle cell anemia is a disease that is caused by a mutation in the gene that produces hemoglobin.  Hemoglobin carries oxygen in red blood cells.  The HbA allele produces normal hemoglobin and the HbS allele produces hemoglobin that sticks together and causes red blood cells to sickle.  Heterozygous indivuals (HbAHbS) produce both normal and "sickle" hemoglobin so the HbA and HbS alleles are codominant.  Heterozygotes do not develop sickle cell anemia and are described as having the sickle cell trait.  Individuals that are homozygous for the sickle allele (HbSHbS) only produce "sickle" hemoglobin and develop sickle cell disease.

Note:  Individuals that have sickle cell trait and sickle cell disease have protection against malaria so the HbS allele is higher in areas where malaria is common.

A man who is homozygous for the normal allele married a woman with the sickle cell trait. What is the expected probability of this couple having a boy with the sickle cell trait?

Record your answer as a value between 0 and 1 rounded to two decimal places

ANSWER

To summarize, HbA:Normal

HbS: Sickle cell

Thus, homozygous HbA/HbA will be normal; HbA/HbS (heterozygous) will have sickle cell trait and HbS/HbS (hmomzygous) will have sickle cell disease (note the difference between sickle cell trait and sickle cell disease.

Man is homozygous (HbA/HbA) and woman has sickle cell trait which means she is heterozygous, HbA/HbS. Following will be the cross

HbA   HbA
HbA   HbA/HbA (Normal)   HbA/HbA (Normal)
HbS   HbA/HbS (Sickle Cell trait)   HbA/HbS (Sickle Cell trait)
Thus 50% of the offspring will suffer from sickle cell trait.

3)

If both parents have sickle cell trait, there is a 25% (1 in 4) chance with each pregnancy that the baby will have sickle cell anemia. A child with sickle cell anemia appears normal at birth.

 • 25% chance of two normal hemoglobin genes (normal hemoglobin- AA), OR
 • 50% chance of one normal hemoglobin gene and one sickle cell gene (sickle cell trait- AS), OR
 • 25% chance of two sickle cell genes (sickle cell anemia- SS)

Quote
4) In horses, tobiano is a white spotting pattern. The tobiano allele (T) is dominant over the non-tobiano (t) allele.

What is the probability, expressed as a percentage, of two heterozygous horses producing two non-tobiano foals in succession?

Express your answer as a percentage rounded to two decimal places.

ANSWER %

Answer: https://biology-forums.com/index.php?topic=1815096.0

PS: I hate having to go up-and-down each time to answer countless questions per thread. The posting guidelines clearly state you are allowed a maximum of one question per thread.

https://biology-forums.com/index.php?action=help;page=guidelines
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Anonymous Author
wrote...
2 months ago
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