Question 1

In the Foofnagle family,the trait of being able to yodel affects members of both sexes in every generation.This trait could be

Accepted Answer

Since the trait affects both sexes and shows up in every generation, it is likely to be autosomal and dominant or recessive in nature. Option A) suggests that the trait could be either autosomal recessive or dominant.

Question 2

Destiny has distal symphalangism - her fingers and toes are stiff,with tiny nails.She looks it up and finds it is an autosomal dominant trait.Since her parents do not have it,she reasons that she is a new mutation.However,if she has children,each of them,assuming her partner does not have the condition,has a(n)_____ probability of inheriting the condition.

Accepted Answer

Since distal symphalangism is an autosomal dominant trait, each child of an affected individual has a 50% (1/2) chance of inheriting the condition, assuming the other parent does not have the condition.

Question 3

Ian and Bryony are both carriers for two autosomal recessive disorders,PKU (chromosome 12)and cystic fibrosis (chromosome 7).They are expecting a daughter.What is the probability that she will have PKU but not CF?

Accepted Answer

The probability of the daughter inheriting PKU is 1/4 (since both parents are carriers), and the probability of not inheriting CF is 3/4 (since both parents are carriers). The disorders are on different chromosomes, so the probabilities are independent. Multiply the probabilities: 1/4 * 3/4 = 3/16.

Question 4

Hillary is 8 years old and has neuronal ceroid lipofuscinosis,also known as Batten disease.It is autosomal recessive.She was fine until age five,when she began to lose her vision and stumble.She had learning difficulties in school and was diagnosed at age six.Hillary may only live into her twenties,experiencing dementia and seizures.Her younger brother Jaden,age six,is healthy,as are her parents.The probability that Jaden is a heterozygote is

Accepted Answer

Since Batten disease is autosomal recessive, both parents must be carriers (heterozygotes) for Hillary to be affected. Jaden, being healthy, has a 2/3 chance of being a carrier (heterozygote) because he cannot be affected (1/4 chance) and he cannot be homozygous normal (1/4 chance), leaving 2/3 as the probability of being a carrier.

Question 5

Which of the following genotypes is homozygous?

Accepted Answer

Homozygous refers to having two identical alleles for a particular gene. In this case, genotype RRYY has two dominant R alleles and two dominant Y alleles, making it homozygous dominant.

Question 6

Jerry is six years old.His sister Hanah has the autosomal recessive condition neuronal ceroid lipofuscinosis that affects learning and coordination.Their parents are healthy.The probability that Jerry is a homozygote for the wild type allele of the gene that causes the condition is
2-12-2013

Accepted Answer

Since the parents are healthy, we know that they are both carriers of the gene for neuronal ceroid lipofuscinosis. This means that each parent has one wild type allele and one mutant allele.

When the parents have children, there are four possible combinations of alleles that the child can inherit:

1. Wild type allele from the mother, wild type allele from the father 
2. Wild type allele from the mother, mutant allele from the father 
3. Mutant allele from the mother, wild type allele from the father 
4. Mutant allele from the mother, mutant allele from the father

Of these four possible combinations, we are interested in knowing the probability that Jerry is a homozygote for the wild type allele. This means that Jerry would have had to inherit a wild type allele from both his mother and his father.

To calculate this probability, we need to consider the frequency of the wild type allele in the population. Since the gene is autosomal and there are only two alleles (wild type and mutant), the frequency of the wild type allele is simply the complement of the frequency of the mutant allele.

Let's use q to represent the frequency of the mutant allele. Since the parents are carriers, we know that each of them has a 1/2 chance of passing on the mutant allele to their child. Therefore, the probability of a child inheriting the mutant allele from both parents is (1/2) * (1/2) = 1/4.

This means that the frequency of the genotype mutant/mutant is q^2 = 1/4. Since there are only two possible genotypes (wild type/wild type or wild type/mutant), the frequency of the wild type/wild type genotype must be 1 - 1/4 = 3/4.

Therefore, the probability that Jerry is a homozygote for the wild type allele is (3/4)^2 = 9/16, which simplifies to 1/4.

Question 7

A severe disease that is inherited as an autosomal recessive trait is more likely to skip generations in a pedigree than an autosomal recessive trait that causes a harmless phenotypic variant because

Accepted Answer

A person with an autosomal recessive disease is less likely to be well enough to reproduce than a person with a harmless trait variant. This means that individuals with a severe disease may not have as many opportunities to pass on the recessive allele as those with a harmless trait variant, reducing the frequency of the disease in some generations. However, the recessive allele is still present in the population and can be passed on to future generations.

Question 8

In a human pedigree that traces the inheritance of sickle cell disease,a half-filled circle represents a

Accepted Answer

In a human pedigree, a half-filled circle represents a female who is a carrier of the disease, meaning she is heterozygous for the sickle cell allele. This means she has one normal allele and one sickle cell allele.

Question 9

A very narrow pedigree,such as that for Egypt's Ptolemy dynasty,reflects

Accepted Answer

A very narrow pedigree, such as that of Egypt's Ptolemy dynasty, indicates a high rate of consanguinity, i.e. marriage between blood relatives. This is because the narrow pedigree means that there are limited individuals in the family tree, indicating that many relatives have married within the family. This has been documented in the Ptolemy dynasty, where intermarriage between siblings and cousins was common in order to maintain the dynasty's power and wealth.

Question 10

Jermaine and Shikia are healthy parents of three young children.The oldest two,Bethany and Amber,are healthy,and they are six and four,respectively.Their youngest child,Roscoe,was born with Peeling Skin Syndrome two years ago.Roscoe's skin constantly peels,and is red and itchy.If Jermaine and Shikia have another child,the probability that he or she will inherit the skin condition is

Accepted Answer

Peeling Skin Syndrome is a rare genetic disorder that follows a autosomal recessive pattern of inheritance. This means that for a child to inherit the disorder, both parents must be carriers of the gene mutation. 
Given that neither Jermaine nor Shikia have Peeling Skin Syndrome, but their child Roscoe does, it is likely that they are both carriers of the mutated gene. 
Therefore, if they were to have another child, there is a 25% chance that the child will inherit the mutated gene from each parent (as there is a 50% chance of inheriting one mutated gene from a carrier parent, and a 50% chance of inheriting a normal gene from the other parent). 
Multiplying these probabilities together gives us a 1/4 or 25% chance that the child will inherit Peeling Skin Syndrome.

Question 11

One reason that diseases caused by recessive alleles remain in populations is that

Accepted Answer

Diseases caused by recessive alleles often do not produce symptoms until a person has already had children, so the individual may unknowingly pass on the allele to their offspring. This allows the allele to remain in the population.

Question 12

The phenotypic ratio expected from a dihybrid cross involving complete dominance is

Accepted Answer

In a dihybrid cross involving complete dominance, the phenotypic ratio expected is 9:3:3:1, representing the combinations of dominant and recessive traits for two different genes.

Question 13

The wild type eye color for humans is

Accepted Answer

Brown is the most common eye color in humans and is considered the wild type. Green and hazel are variations that result from genetic mutations, while pink is not a natural eye color in humans.

Question 14

The original pedigrees,used in France in the fifteenth century,resembled

Accepted Answer

The bird's foot was the original design for pedigrees used in France in the fifteenth century.

Question 15

Sickle cell disease is inherited as an autosomal recessive trait.The genotype of an individual with sickle cell disease is

Accepted Answer

Sickle cell disease is an autosomal recessive trait, which means an individual must inherit two copies of the abnormal hemoglobin gene (HbS) to have the disease. Therefore, the only genotype that can result in sickle cell disease is homozygous recessive (ss).

Question 16

Using the product rule,the probability that parents heterozygous for two traits (AaBb)produce an offspring homozygous recessive for the same two traits (aabb)is

Accepted Answer

The probability of producing an offspring homozygous recessive for one trait (aa) from heterozygous parents (Aa) is 1/4. Similarly, the probability for the second trait (bb) is also 1/4. Using the product rule, multiply these probabilities: 1/4 * 1/4 = 1/16.

Question 17

In a human pedigree that traces the inheritance of albinism,a filled square represents a

Accepted Answer

Albinism is a recessive genetic disorder, which means that an individual must inherit two copies of the mutated gene (one from each parent) to show the trait. Since squares represent males in a pedigree, a filled square indicates an individual who expresses the trait of albinism, which can only happen if he inherits two copies of the mutated gene for albinism. Therefore, the correct answer is A - male with albinism.

Question 18

Eric and Janelle are carriers of the recessive allele causing Tay-Sachs disease,which is lethal in early childhood.If they have a healthy child,what is the probability that this child would NOT be a carrier of Tay-Sachs disease?

Accepted Answer

If Eric and Janelle are both carriers, their genotype is Tt. A healthy child could either be TT (not a carrier) or Tt (a carrier). The possible genotypes for their children are TT, Tt, Tt, and tt. Since tt results in Tay-Sachs, it is not considered healthy. Among the healthy genotypes (TT, Tt, Tt), only TT is not a carrier. Therefore, the probability that a healthy child is not a carrier is 1 out of 3, or 1/3.

Question 19

Mode of inheritance reflects

Accepted Answer

Mode of inheritance is determined by whether a gene is located on an autosome or sex chromosome, and whether the allele is dominant or recessive.

Question 20

For two traits A and B with two alleles each,the genotypes of the gametes that a heterozygote would produce are

Accepted Answer

For two traits with two alleles each, a heterozygote would produce gametes with all possible combinations of the two alleles. In this case, the possible Alleles for Trait A are A and a, and for Trait B are B and b. Therefore, a heterozygote would produce gametes with the A and a alleles as well as with the B and b alleles. Thus, the possible genotypes of the gametes are AB, Ab, aB, and ab. These are all included in option A.

Quiz 4: Single-Gene Inheritance