Prentice Hall Biology (California)
Prentice Hall Biology (California)
1st Edition
Kenneth R. Miller, Levine
ISBN: 9780132013529
Textbook solutions

All Solutions

Page 363: Chapter 14 Assessment

Exercise 1
Step 1
1 of 2
A human diploid zygote normally contains 46 chromosomes.
Result
2 of 2
B. 46 chromosomes
Exercise 2
Step 1
1 of 2
A pedigree chart shows how a trait is inherited by showing the occurrence of phenotypes of a gene in organisms all throughout the generations.
Result
2 of 2
A. pedigree
Exercise 3
Step 1
1 of 2
The term polygenic has the prefix poly- which means many. This is a type of inheritance that allows one organism to have a trait that is controlled by two or more genes.
Result
2 of 2
C. polygenic
Exercise 4
Step 1
1 of 2
When there are more than two possible alleles that can exist in a particular population, it is said to be that an individual gene has more than two variant forms of genes or alleles. This condition is known as multiple alleles. An example of this is the ABO blood type group. here are three alleles for blood type – ABO, which would result to four possible genotypes that include A, B, AB, and O.
Result
2 of 2
B. ABO blood groups
Exercise 5
Step 1
1 of 2
A sex-linked gene is usually located on the X chromosome and not the Y chromosome since the X chromosomes are larger and carries more genes.
Result
2 of 2
D. X chromosome
Exercise 6
Step 1
1 of 2
Hemophilia is a sex-linked genetic disorder in which a person suffers from a blood clotting problem.
Result
2 of 2
A. sex-linked
Exercise 7
Step 1
1 of 2
Normally, all X-linked alleles are only expressed in males. Since men pass their chromosomes to their daughters, for a woman to be colorblind, she must have gotten it from his father’s X-linked allele. In this case, both parents should be carriers of the disorder. A heterozygous normal-vision mother and a father with colorblindness will produce a female offspring with colorblindness.
Result
2 of 2
D. heterozygous normal vision, father with colorblindness
Exercise 8
Step 1
1 of 2
A sickle cell disease is characterized by a condition that affects the hemoglobin or the molecule found in the red blood cells. Those people who have this disorder have hemoglobin S, which is an abnormal hemoglobin molecule that can distort red blood cells into a sickle shape.
Result
2 of 2
C. sickle cell disease
Exercise 9
Step 1
1 of 2
In DNA fingerprinting, the sections of the DNA that have no function are the ones being studied to identify individuals.
Result
2 of 2
A. DNA fingerprinting
Exercise 10
Step 1
1 of 2
Gene therapy is the process of replacing an absent or a faulty gene with a normal functioning gene in order to get rid of the cause of some disorders.
Result
2 of 2
A. gene therapy
Exercise 11
Result
1 of 1
A karyotype is produced by halting or trapping a cell’s mitotic division during the metaphase stage. The cell is held and suspended by chemicals to show its chromosomes. Once the chromosomes are separated, they are stained and viewed under a microscope.
Exercise 12
Result
1 of 1
Autosomes are the chromosomes that do not carry the specific genes that determine the sex of an individual and they are homologous, whereas sex chromosomes carry the genes that determine an individual’s sex and they are not homologous because X chromosome is larger than Y chromosome.
Exercise 13
Result
1 of 1
A pedigree chart shows how a trait is inherited by showing the occurrence of phenotypes of a gene in organisms all throughout the generations. By using a pedigree, genetic disorders can be traced within the family and they can be helpful in knowing who will possibly inherit the disorder.
Exercise 14
Step 1
1 of 2
1 and 6 are definitely carriers because both females do not exhibit the phenotype, but pass the trait on to all of their sons.

The sons of person 3 did not inherit the trait because the mother must not be a carrier.

Hemophilia is an x-linked recessive trait, meaning that it is only exhibited in males where they have and X and a Y chromosome.

Females do not show the trait because they have two x chromosomes, where if they are a carrier, the regular one (the wild type allele) is dominant over the hemophilia allele.

Result
2 of 2
Click to see the explanation
Exercise 15
Result
1 of 1
Alleles A and B are always dominant over allele O. As a result, a person who has an AO genotype will have an A blood type, whereas an individual who has a blood type O has OO genotypes. Therefore, if an allele A is inherited from one parent and an allele B from the other parent, the blood type of the offspring will be AB.
Exercise 16
Result
1 of 1
The RH blood group indicates that our blood can either be Rh-positive or Rh-negative. Rh-positive means that you have inherited a protein called an antigen that stays on the surface of your red blood cells, whereas Rh-negative means you didn’t inherit this kind of protein and you are lacking from it.
Exercise 17
Result
1 of 1
A mutation in the recessive allele causes the Tay-Sachs disease. This disease is characterized by the absence of a vital enzyme called hexosaminidase-A (Hex-A) which metabolizes lipid in the body. Due to this condition, a person who suffers from Tay-Sachs disease have lipid accumulation in the brain which results to mental deficiency and blindness.
Exercise 18
Result
1 of 1
Dominant alleles exhibit their effect in the form of a phenotype or a trait that is observed in an organism. On the other hand, recessive allele is masked by the dominant allele. It would only show if it is teamed up with another recessive allele. On the contrary, co-dominance happens when both alleles are expressed or manifested physically by the heterozygous organism.
Exercise 19
Solution 1
Solution 2
Step 1
1 of 2
$underline{nondisjunction}$: error in meiosis in which homologous chromosomes fail to separate
A chromosomal disorder is when there is an abnormal number of chromosomes in an individual. Down syndrome, Turner’s syndrome, and Klinefelter’s syndrome are chromosomal disorders that can result from nondisjunction.
Result
2 of 2
A chromosomal disorder is when there is an abnormal number of chromosomes in an individual. Down syndrome, Turner’s syndrome, and Klinefelter’s syndrome are chromosomal disorders that can result from nondisjunction.
Step 1
1 of 2
Chromosomal disorder is when there is an abnormal number of chromosomes in a gamete. Nondisjunction is when homologous chromosomes fail to separate in meiosis causing the abnormal number.
Result
2 of 2
see explanation for solution
Exercise 20
Result
1 of 1
In DNA fingerprinting, the sections of the DNA that have no function are the ones being studied to identify individuals.
Exercise 21
Step 1
1 of 4
Rapid sequencing generally involves identifying the sequence of DNA using labeled regions as markers. These markers are useful in locating and returning to certain regions in the genome.
Step 2
2 of 4
Rapid sequencing also involves a technique known as *shotgun sequencing*. This technique involves cutting a DNA sample into random fragments and then determining the sequence of bases in each fragment.
Step 3
3 of 4
In shotgun sequencing, a computer program would be used to locate overlapping regions in the DNA fragments. This would help in putting these fragments together by linking these overlapping sections.
Step 4
4 of 4
Aside from overlaps, the computer program may also use known markers in determining the sequence of the DNA sample. This results in the alignment of fragments, which would yield the actual sequence of the DNA sample.
Exercise 22
Step 1
1 of 3
Genes are generally described as DNA sequences that encode for a product such as RNA or polypeptide. Genes are typically arranged as a series of 3-letter sequences called *codons*.
Step 2
2 of 3
An open reading frame (ORF) consists of a sequence of codons that encode for a particular product. Therefore, the presence of an ORF in a DNA sequence would identify a particular gene.
Step 3
3 of 3
In summary, molecular biologists often look for ORFs since they will correspond to a particular gene. It is important to look at the correct “frame” or reference point when identifying the ORF.
Exercise 23
Step 1
1 of 2
A mutation in the recessive allele causes the Tay-Sachs disease. If a person carries one allele for this disease, it will not manifest in their phenotype since it is a recessive allele.

In this case, we can represent the recessive allele using letter a and the dominant allele, which is the normal allele, using capital letter A. If two parents carry one allele for the disease, their genotype is Aa. If you cross Aa with another Aa, the possible genotypes are AA, Aa, Aa, and aa.

Results:
AA = 25%
Aa = 50%
aa = 25%

The results indicate that there is a 25% chance that the parents will have a healthy offspring. There is also a 50% chance that the recessive allele will be carried by the offspring and there is 25% chance that the offspring will have the Tay-Sachs disease.Exercise scan

Result
2 of 2
Results:
AA = 25% (healthy)
Aa = 50% (carries the allele for the disease)
aa = 25% (with Tay-Sachs disease)

The results indicate that there is a 25% chance that the parents will have a healthy offspring. There is also a 50% chance that the recessive allele will be carried by the offspring and there is 25% chance that the offspring will have the Tay-Sachs disease.

Exercise 24
Step 1
1 of 2
Red blood cells that contain the sickle cell allele would usually have a weaker or more brittle cell membrane. When the protist infects this cell, the cell would burst. This prevents the protist from living in the cell and reproducing.
Step 2
2 of 2
On the other hand, normal red blood cells would usually have an intact and stronger cell membrane. In this case, the protist would be able to infect the cell without causing it to burst. Hence, the protist can live in the cell and reproduce.
Exercise 25
Step 1
1 of 2
Color blindness is considered an X-linked recessive disorder. If a man is a carrier of the defective X-chromosome, he will be colorblind. However, if he marries a woman who is a carrier of the disorder, the husband will not pass the colorblindness into his sons. Instead, sons will inherit the disorder from their mother, which is the carrier of the disorder.

aa = affected father
Aa = carrier mother

aa x Aa = Aa, Aa, aa, aa

Aa = 50%
aa = 50%

The results show that there is a 50% chance that the sons will be colorblind
and 50% chance that the girls will carry the defective gene.Exercise scan

Result
2 of 2
The results show that there is a 50% chance that the sons will be colorblind
and 50% chance that the girls will carry the defective gene.
Exercise 26
Result
1 of 1
Based on the data shown on the graph, 3% out of the 100 children, whose mother was under 30, was born with down syndrome. To the women of age 35, there was 4% of the 100 children was born with down syndrome, while there was 10% when the woman was aged 40 and 30% at the age of 50. This indicates that the as the age of woman increases, they have a greater risk of birthing to babies with down syndrome.

Normally, a woman has eggs between 500,000 to 2 million eggs in her entire lifetime but as she ages, this number declines. By the time that a woman reaches the age of 40, her eggs might not work the same way it was when she was younger and the rate of protein production in her egg cells might have decreased. As a result, abnormalities in the chromosome may occur.

Exercise 27
Result
1 of 1
A karyotype can only detect the chromosome number and its abnormalities. Since cystic fibrosis occurs when there is a mutation in the gene, it will not show in the karyotype. Therefore, the answer is no.
Exercise 28
Step 1
1 of 4
Since the child is male, he should have both X and Y chromosomes. The X chromosome should be inherited from the mother, while the Y chromosome should be inherited from the father.
Step 2
2 of 4
Since colorblindness is linked to the X chromosome, the occurrence of colorblindness in the child should be dependent on the genotype of the mother since she provides the chromosome.
Step 3
3 of 4
Since the father will not influence the occurrence of colorblindness, he can have all possible genotypes. He will either have a genotype of $X^NY$ (normal vision) or $X^CY$ (colorblind); where $X^N$ is the normal allele, while $X^C$ is the colorblind allele.
Step 4
4 of 4
On the other hand, the mother should have at least one allele in her X chromosomes with the colorblind trait in order for colorblindness to be inherited by the male child. Therefore, she should have a genotype of either $X^CX^N$ (carrier) or $X^CX^C$ (colorblind).
Exercise 29
Step 1
1 of 4
According to the given karyotype, the individual has 22 normal pairs of chromosomes. However, it appears that there is a missing sex chromosome since there is only one X chromosome.
Step 2
2 of 4
Therefore, the chromosomal disorder that is associated with the given karyotype is **Turner’s syndrome**. This disorder occurs only in females, and it is characterized by a missing or partially missing X chromosome.
Step 3
3 of 4
Turner’s syndrome would therefore give a genotype of $XO$ with regards to the sex chromosomes. This disorder can cause a variety of developmental problems such as short height, defective ovaries, and heart abnormalities.
Result
4 of 4
Turner’s syndrome
unlock
Get an explanation on any task
Get unstuck with the help of our AI assistant in seconds
New