Question 1(Multiple Choice Worth 2 points)

Question

(03.07 MC)

This pedigree chart tracks the inheritance of a recessive trait that is not sex-linked. Based on the information in the chart, which of the following statements is true?

This pedigree chart shows three generations. The individuals are numbered from left to right, starting at the top left of the chart. Squares represent males and circles represent females. If the symbol is gray, that individual exhibits the recessive trait. If the symbol is white, the individual exhibits the dominant trait. On the first row there are two couples mated together. The first couple is made up of individuals 1 (recessive) and 2 (dominant). They have four offspring on the second row that all exhibit dominant traits: 5 (dominant), 6 (dominant), 7 (dominant), and 8 (dominant). The second couple in generation 1 is made up of individual 3 (dominant) and individual 4 (dominant). They have three offspring in generation 2: 9 (dominant), 10 (recessive), and 11 (dominant). Individual 8 from the first family and individual 9 from the second family mated together and produced three offspring in generation 3. These offspring are 12 (recessive), 13 (dominant), and 14 (recessive).

Individual 8 must be heterozygous for the trait.
Individual 10 must be heterozygous for the trait.
Individual 13 must be homozygous dominant.
Individual 3 must be homozygous dominant.
Question 2(Multiple Choice Worth 2 points)
(03.07 MC)

A cat is blinded in one eye in a fight. When this cat later sexually reproduces with another cat, how will the blindness affect the offspring?

All of the offspring will be blind because blindness is a dominant trait.
Some of the offspring will be blind if the blind cat passes on the allele for blindness.
Most of the offspring will be blind if the other parent cat carries the same allele for blindness.
None of the offspring will inherit blindness from this cat, because it is not a genetic trait.
Question 3(Multiple Choice Worth 2 points)
(03.07 MC)

Hemophilia is a rare bleeding disease that is carried on the X-chromosome. More than 400,000 people worldwide live with this disease, 20,000 of those within the United States. It is commonly called the "royal disease" because Queen Victoria was known to be a carrier of the disease and passed the gene to other members of the royal family.

If a healthy individual who does not have hemophilia gets a small cut on their arm, it takes two to eight minutes on average for clotting to occur and for the bleeding to stop. People suffering from hemophilia lack an essential blood-clotting protein in their blood plasma. If they get cut, it can take 30 minutes to 22 hours for the bleeding to stop. Excessive bleeding is a constant danger to individuals with hemophilia.

Figure 1 shows the pedigree of hemophilia in four generations of Queen Victoria's family.

Pedigree showing four generations of Queen Victoria's family. One of Queen Victoria's sons, Leopold, has hemophilia. One of Queen Victoria's daughters, Alice, is a carrier of the gene for hemophilia. Alice has children with a man that does not have hemophilia. Two of Alice's daughters, Alexandra and Irene, carry the gene for hemophilia, and her son has hemophilia. Alexandra has a son, Alexis, with a man that does not have hemophilia. Alexis has hemophilia. Irene has two sons with a man that does not have hemophilia. Both sons have hemophilia.

Based on the information provided, which of the following statements is true about the 400,000 people living with hemophilia?

They are all males because females can only be carriers of the trait.
The males with the disorder can't pass the trait to their daughters.
Due to the method of inheritance, the majority of those with the disorder are males.
The females affected with the disorder inherited the trait from their father.
Question 4(Multiple Choice Worth 2 points)
(03.07 MC)

Two traits are being analyzed in a dihybrid cross. If an organism has the genotype EEGg for these traits, which of the following best demonstrates the possible genetic combinations that could be present in a gamete cell of the organism?

EE, Gg
EG, Eg
EE, GG, gg
EE, EG, Eg, Gg
Question 5(Multiple Choice Worth 2 points)
(03.07 LC)

Which of the following genotypes is homozygous dominant?

tt
Ff
XY
SS
Question 6(Multiple Choice Worth 2 points)
(03.07 MC)

A man and woman have a son who is color blind, a recessive sex-linked trait carried on the X chromosome, but neither parent is color blind. The woman is now pregnant with their second child. Which of the following statements is true about the second child?

A daughter will be color blind because the father is a carrier of the recessive allele.
A son will be color blind because the mother is a carrier of the recessive allele.
A daughter will not be color blind because the father is not a carrier of the recessive allele.
A son will be color blind because the father's mother must have been a carrier of the recessive allele.
Question 7(Multiple Choice Worth 2 points)
(03.07 MC)

In one type of dog, black spots (B) are dominant over brown spots (b), and long tails (L) are dominant over short tails (l). Use the dihybrid cross for parents with the genotypes BBLl and BbLl and determine the likelihood of offspring with the genotype BBLl.

BL Bl BL Bl
BL BBLL BBLl BBLL BBLl
Bl BBLl BBll BBLl BBll
bL BbLL BbLl BbLL BbLl
bl BbLl Bbll BbLl Bbll
1/16
2/16
3/16
4/16
Question 8 (Essay Worth 4 points)
(03.07 HC)

In Persian cats, there is a hereditary condition called polycystic kidney disease (PKD), which can lead to kidney failure and death. This is a dominant disorder.

At one point, as many as 50% of Persian cats had PKD. Recently, however, the prevalence of PKD has dropped dramatically as DNA testing that can detect affected cats has been developed. As a result, breeders have become aware of this genetic disorder and have been careful not to breed affected cats.

tan colored, fluffy Persian cat
© Takashi(aes256) 2012
Creative Commons, CC BY-SA 2.0

Part 1: A cat that is heterozygous for PKD is bred with a cat that does not have PKD. Use a Punnett square to help determine the probability of the possible genotypes and phenotypes in the offspring. Explain your results in paragraph form.
Part 2: If a cat does not have PKD, what can you conclude about the genotypes and phenotypes of its parents? Explain your answer in paragraph form.

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Question 9 (Essay Worth 4 points)
(03.07 HC)

All modern domesticated pigeons have distinct traits that have been chosen by breeders, including the trait for feathered feet. Although most birds have scaly feet (without feathers), some species, including domesticated pigeons, possess a genetic mutation that allows them to have feathered feet, which is a non-Mendelian trait.

A breeder crosses a homozygous bird with long feathers on the feet (FLFL) with a homozygous bird with scaly feet and no feathers (FNFN), and all of the offspring have feet with unusually short feathers, unlike either parent.

Parent pigeon (F N F N) shows a foot with scaly feet (nonfeathered). Parent pigeon (F L F L) shows a foot fully covered with long feathers. Offspring pigeon (unknown) shows short feathers covering the foot.
Public Domain

Part 1: Based on the information provided, what can you conclude about the pattern of inheritance for the feathered foot gene?
Part 2: Explain to the breeder what happened in this cross. Be sure to include both phenotypes and genotypes in your explanation.

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Question 10 (Essay Worth 3 points)
(03.07 HC)

A couple has two children, one with type B blood and the other with type O blood. The father is type O, but the mother doesn't know her blood type. What can you infer about the genotypes for both parents? Explain your answer.

Answer 1

Here are the answers to your questions one by one:

Question 1: The correct statement is: Individual 8 must be heterozygous for the trait.

• Individual 8 must be heterozygous because they exhibit the dominant trait but have a recessive child (12).

Question 2: The correct answer is: None of the offspring will inherit blindness from this cat, because it is not a genetic trait.

• The cat's blindness was due to an injury and not a genetic factor.

Question 3: The correct statement is: Due to the method of inheritance, the majority of those with the disorder are males.

• Hemophilia is X-linked recessive, and males are more likely to express the trait since they have only one X chromosome.

Question 4: The best demonstration of possible genetic combinations in a gamete cell for the genotype EEGg is: EG, Eg.

• Gametes can only carry one allele for each gene.

Question 5: The genotype that is homozygous dominant is: SS.

• Homozygous dominant means both alleles are the same and dominant.

Question 6: The correct statement is: A daughter will not be color blind because the father is not a carrier of the recessive allele.

• The father can only pass on a normal X to daughters.

Question 7: The likelihood of offspring with the genotype BBLl is: 4/16.

• There are 4 unique combinations that yield BBLl in the dihybrid cross.

Question 8: Part 1: To understand the potential genotypes and phenotypes from a cat that is heterozygous for PKD (Pp) and a cat that does not have PKD (pp), we can set up a Punnett square:

       P  |  p
    ---------------
 p |   Pp  | pp
    ---------------
 p |   Pp  | pp

From the Punnett square, the probabilities of the offspring genotypes are:

• 50% will be Pp (carriers who exhibit the phenotype of no PKD).
• 50% will be pp (normal cats, without PKD).

This means half of the offspring are carriers, and the other half do not carry the gene.

Part 2: If a cat does not have PKD, it must have the genotype pp (homozygous recessive). For a cat to not have the disorder, it can only receive the recessive allele from each parent. This concludes that at least one parent must carry the dominant gene for PKD (Pp or PP), but if they do not have the disorder, the parent with PKD could not be homozygous dominant (PP). Therefore, it is conclusive that if at least one cat does not show PKD, then they are homozygous recessive (pp).

Question 9: Part 1: The inheritance pattern for the feathered foot gene appears to be an example of incomplete dominance. Neither parent phenotype is fully expressed in the offspring, suggesting that the short feather trait is a blend or intermediate phenotype between the feathered and non-feathered feet.

Part 2: In this cross between the homozygous long-feathered (F_L_F_L) and homozygous scaly-footed (F_N_F_N) pigeons, the resulting offspring had shorter feathers than either parent. This suggests that both alleles (feathered feet and no feathers) are for some dominant interaction that leads to a new phenotype of short feathers. This indicates that the trait doesn't follow Mendelian dominance but perhaps an incomplete dominance or co-dominance pattern where the resulting phenotype is a blend or a third phenotype.

Question 10: The father being type O means his genotype must be ii. The child with type B blood has a genotype that can be either IBi (type B) or IBIB (type B). This means the mother must have at least one IB allele, making her likely type B (genotype can be either IBi or IBIB). However, since they have a child with type O blood, it indicates that the mother cannot have the IBIB genotype, so she must be IBi. Therefore, the mother's genotype is IBi because they have one type O child (ii from the father).

Let me know if you need further explanations or have any other questions!