Mastering Genetics Unit 4

Which of the following statements describes metastasis?

The ability to form secondary tumors at another site. Cells from malignant tumors can break off from the main tumor, travel through the blood or lymph, and establish secondary tumors at another site.

Why is this discipline essential for studying genomes?

The efficiency of genome studying depends on the ability to store, efficiently share, and obtain the maximum amount of information from protein and DNA sequences.

What is the general term for a condition in which the chromosome number is not a multiple of a complete set?

Aneuploidy. The term aneuploid is derived from roots that mean "not true ploidy."

Why are most recombinant human proteins produced in animal or plant hosts instead of bacterial host cells?

Bacteria do not process eukaryotic proteins in the same manner as eukaryotes.

Which of the following events could result in a frameshift mutation?

Base deletion. A base deletion would shorten the DNA sequence and change the reading frame of the mRNA.

What is bioinformatics?

Bioinformatics is an interdisciplinary field that develops and improves upon methods for storing, retrieving, organizing and analyzing biological data.

Biotechnology has yet to produce a single useful product for a human health condition.

False.

Most mutations in a diploid organism are recessive. Why?

In most cases, the amount of product from one gene of each pair is sufficient for production of a normal phenotype.

Why are liver extracts used in the Ames test?

Liver enzymes may activate some innocuous compounds, making them mutagenic. Some compounds are innocuous until they are activated metabolically by liver enzymes.

Distinguish between oncogenes and proto-oncogenes.

Oncogenes are genes that induce or maintain uncontrolled cellular proliferation associated with cancer. They are mutant forms of proto-oncogenes, which normally function to regulate cell division.

For which codon(s) of isoleucine could a single base change account for an amino acid change to methionine? Select all that apply.

- AUU - AUA - AUC There are three codons for isoleucine (AUA, AUC, AUU) and one codon for methionine (AUG). Since positions #1 and #2 are identical in the two sets of codons, then a single base change in the #3 position of any of the isoleucine codons to a G would result in a methionine codon.

Skin cancer carries a lifetime risk nearly equal to that of all other cancers combined. Following is a graph [modified from K. H. Kraemer (1997). Proc. Natl. Acad. Sci. (USA) 94:11−−14] depicting the age of onset of skin cancers in patients with or without XP, where the cumulative percentage of skin cancer is plotted against age. The non-XP curve is based on 29,757 cancers surveyed by the National Cancer Institute, and the curve representing those with XP is based on 63 skin cancers from the Xeroderma Pigmentosum Registry. Provide an overview of the information contained in the graph. Select all that apply.

- By age 20, approximately 80% of the XP population has skin cancer. - For non-XP individuals, the chance of skin cancer increases as they age. - XP individuals almost always develop skin cancer by age 40. - Individuals with XP are more likely to develop skin cancer in their youth than non-XP individuals.

What are some of the reasons why GM crops are controversial? Check all that apply.

- Concerns exist regarding whether the design and patenting of GM crops might allow domination of the world food supply by a few companies. - Concerns exist regarding whether GM crops may reduce the world's supply of genetic variability. - Concerns exist regarding whether GM crops and foods have been properly tested for environmental impact. - Concerns exist regarding whether GM crops and foods have been properly tested for allergenicity.

The human insulin gene contains introns. Since bacterial cells will not excise introns from mRNA, can a gene like this be cloned into a bacterial cell that will produce insulin? If so, how? Select all that apply.

- The amino acid sequence of the protein can be used to produce the gene synthetically. - mRNA can be obtained and used to make cDNA, which can be used to make the desired product.

A couple with European ancestry seeks genetic counseling before having children because of a history of cystic fibrosis (CF) in the husband's family. ASO testing for CF reveals that the husband is heterozygous for the ∆508 mutation and that the wife is heterozygous for the R117 mutation. You are the couple's genetic counselor. When consulting with you, they express their conviction that they are not at risk for having an affected child because they each carry different mutations and cannot have a child who is homozygous for either mutation. What would you say to them? Select all that apply.

- There is a 25% chance that their child will be a carrier of the R117 mutation. - There is a 50% chance that their child will be a carrier of a CF mutation. - There is a 25% chance that their child will develop CF. - There is a 25% chance that their child will be a carrier of the ∆508 mutation.

Now consider a single base mutation in a codon for leucine that creates a codon for phenylalanine. A true reversion changes the phenylalanine codon back to a codon for leucine.Which of the following leucine codon(s) could be mutated once to form a phenylalanine codon, and then mutated at a second site to restore a leucine codon? Select all that apply. (Note that two different positions in the codon must be mutated.)

- UUA - CUU - CUC - UUG There are six codons for leucine (UAA, UUG, CUA, CUC, CUG, CUU) and two codons for phenylalanine (UUC, UUU). Leucine codons (CUA, CUG) do not share two of the codon positions with phenylalanine. The remaining four leucine codons share two of the codon positions with phenylalanine, therefore they are capable of being mutated to a phenylalanine codon. A subsequent mutation at a second site restores a leucine codon in each case. For example, if the original leucine codon was UUA, then an A to C transversion mutation at position #3 would create a phenylalanine codon (UUC). A subsequent U to C transition mutation at position #1 would create a true reversion back to a leucine codon.

This woman is phenotypically normal. But some translocations result in an abnormal phenotype. Under what circumstances might you expect a phenotypic effect of such a rearrangement? Select the three correct answers.

- if translocation breakpoints occur within the genes - if functioning of the translocated genes depends on their neighboring genes - if translocation results in the loss of some genetic material

In what ways can proto-oncogenes be converted to oncogenes? Check all that apply.

- point mutations - translocations - repositioning of regulatory sequences - gene amplification

Suppose a diploid cell with three pairs of homologous chromosomes (2n = 6) enters meiosis. How many chromosomes will the resulting gametes have in each of the following cases? Drag one label into each space at the right of the table. Labels can be used once, more than once, or not at all.

1. 3 only 2. 2 or 4 3. 0 or 6 4. 2, 3, or 4 5. 0, 3, or 6 If one chromosome pair undergoes nondisjunction in meiosis I, half the gametes will have an extra chromosome (n +1), and half will be missing a chromosome (n - 1). If all chromosome pairs undergo nondisjunction in meiosis I, half the gametes will have twice the normal haploid number of chromosomes (2n), and half will have no chromosomes. If one chromosome undergoes nondisjunction in meiosis II, half the gametes will have the normal haploid number of chromosomes (n), one-quarter will have an extra chromosome (n +1), and one-quarter will be missing a chromosome (n - 1). If all chromosomes undergo nondisjunction in meiosis II, half the gametes will have the normal haploid number of chromosomes (n), one-quarter will have twice the haploid number (2n), and one-quarter will have no chromosomes.

Scientists carried out a microarray analysis to compare the gene expression of normal cells to that of cancer cells that arose from the same cell type. -The scientists labeled the cDNA from the normal cells with green fluorescent nucleotides. -They labeled the cDNA from the cancer cells with red fluorescent nucleotides. The two cDNAs were mixed and allowed to hybridize to a microarray, as shown in this diagram.When equal amounts of red and green fluorescence are measured, the spot on the microarray will appear yellow. When a gene is not expressed in either cell type, the spot on the microarray will appear black.In this example, assume that each spot on the microarray represents one gene. Predict what color each spot described below would be.

1. A spot containing a gene that is overexpressed in cancer cells: RED 2. A spot containing a housekeeping gene that is expressed equally in both cell types: YELLOW 3. A spot containing the p24 gene, which is present and expressed in normal cells, but absent in cancer cells: GREEN 4. A spot containing the p90 gene, which is normally repressed by the p24 gene: RED In this microarray experiment, cancer cell cDNAs were labeled with red fluorescence and normal cell cDNAs were labeled with green fluorescence. -A red spot represents a gene with increased expression in the cancer cell because more of the red-labeled cDNAs are present on the microarray. -A green spot represents a gene with decreased expression in the cancer cell because more of the green-labeled normal cell cDNAs are present on the microarray. -A yellow spot represents a gene that is expressed equally in both cell types. -A black spot represents a gene that is not expressed in either cell type.

Chromosomal mutations are changes in the normal structure or number of chromosomes. -Changes in chromosome structure can result from errors in meiosis or from exposure to radiation or other damaging agents. -Certain changes in chromosome number can result from nondisjunction during either meiosis or mitosis. Both structural mutations and nondisjunction can play a role in trisomy 21, commonly known as Down syndrome. The diagram below shows two normal chromosomes in a cell. Letters represent major segments of the chromosomes. The following table illustrates some structural mutations that involve one or both of these chromosomes. Identify the type of mutation that has led to each result shown. Drag one label into the space to the right of each chromosome or pair of chromosomes. You can use a label once, more than once, or not at all.

1. ABCDEFH = deletion 2. IJNKLMN = duplication 3. ABCLMN IJKDEFGH = translocation 4. AEDCBFGH = inversion 5. ABCDDCEFGH = duplication 6. ABCDEKLMNFGH IJ = translocation 7. IJMLKN = inversion A deletion is the loss of part of a chromosomal segment. A duplication is the repetition of a segment. The repeated segment may be located next to the original or at a different location, and its orientation may be the same as the original or the reverse. An inversion is the removal of a segment followed by its reinsertion into the same chromosome in the reverse orientation. A translocation is the transfer of a segment to a nonhomologous chromosome. Translocations may be reciprocal (two nonhomologous chromosomes exchange segments) or nonreciprocal (one chromosome transfers a segment without receiving one).

You would like to use a microarray to compare the gene expression of a normal cell to that of a cancer cell. Identify the steps involved in the microarray procedure. Drag the terms on the left to the blanks on the right to complete these sentences about the microarray procedure. Not all terms will be used.

1. Isolate MRNA from both the normal cells and the cancer cells. 2. Use the enzyme reverse transcriptase and fluorescently labeled nucleotides to make CDNA from the starting material isolated in Step 1. 3. Denature the fluorescently labeled molecule created in Step 2, and incubate it with the microarray. The fluorescently labeled molecules from the two different cell types will hybridize to SINGLE-STRANDED DNA on the microarray. 4. Wash and then scan the microarray to measure the FLUORESCENCE at each spot on the array. In a microarray experiment, mRNAs are isolated from different cell types. cDNAs are created from the mRNAs using fluorescently labeled nucleotides, and the cDNAs are incubated with the microarray. The amount of cDNAs present on the microarray can be determined by measuring the fluoresce at each spot.

Next, the scientists wanted to test the effect of a new cancer drug on gene expression in the cancer cells. Previous studies had shown that the drug has no effect on gene expression in normal cells.The scientists added the drug to both the cancer cells and normal cells, and collected cDNAs at specified time points. They performed a microarray experiment using samples from each time point. This microarray shows the results for those genes that showed a change after the addition of the drug.Analyze the results of this experiment by answering these questions.

1. Which group of genes showed increased expression in the cancer cells after the addition of the drug? C 2. Which group of genes showed repressed expression in the cancer cells after the addition of the drug? D 3. Which group of genes returned to baseline levels of expression in the cancer cells by the end of the experiment? A 4. Which group of genes remained highly expressed in the cancer cells throughout the experiment? C In this experiment, a drug was added to cancer cells and cDNAs were isolated at various time points. -The expression of genes in Group A was quickly repressed in the cancer cell (by the first time point after the drug was added), but returned to baseline levels of expression by the end of the experiment (indicated by the black color at the 11-hour time point). -The expression of genes in Group B was also repressed in the cancer cell, but remained repressed throughout the experiment (indicated by the green color at the 11-hour time point). -The genes in Group C were overexpressed in the cancer cell after the addition of the drug, and remained highly expressed throughout the experiment (indicated by the red color at the 11-hour time point).

A sequence of DNA was analyzed with a bioinformatics program that translated the sequence in all 6 possible reading frames. Below is the outcome of this analysis. If this sequence contains one gene, which frame is most likely the correct reading frame for this gene? (Remember that Met represents a potential start codon and Stop represents a stop codon.)

5'- 3' direction, Frame 3A E Met R D G G A F S G P L L L E T P G P E P G Q Q E P L V F G S G D T V E L T C P S P I D S P T G P S I W V K D G V G L V P S D R I L V G P R R L Q V L N A S H E D A G T Y S C R Q R L T Q R I L C H F S V Stop In a computer translation of a gene sequence, the frame most likely to be the correct one is the frame that contains a start codon (Met) with the longest sequence of uninterrupted amino acids before a stop codon. In this example, 5'-3' Frame 3 has a start codon near the beginning of the sequence and a stop codon near the end of the sequence with no other stop codons in between.

The first step in annotating a gene is identifying the open reading frame. In bacteria, all genes that code for a protein have a start and stop codon with an open reading frame between them. A bioinformatics program can translate all the possible reading frames (series of 3 codons) in an attempt to find the longest open reading frame. How many possible open reading frames can a bioinformatics program translate from one DNA sequence? Enter your answer numerically (for example: 2).

6 Each strand of DNA has six possible reading frames. A reading frame is a series of 3 bases (codons) that can be used to predict an open reading frame. There are three reading frames in the 5'- 3' direction and three reading frames in the 3'- 5' direction. You can also think of it as three possible frames in one direction on each strand of DNA.

Which of the following statements is TRUE regarding heritability of cancer?

At least two mutational events are required for a cell to become cancerous. One copy of a mutated gene is not enough to trigger development of cancer. Mutation in the other copy of the gene is also required. In addition, mutations in other genes are often required to fully express the cancer.

A diploid cell with 4 chromosomes (one pair metacentric, the other telocentric) begins meiosis, and nondisjunction occurs during the first meiotic division. Drag each cell to the correct location in the chart. Make sure the chromosomes correspond to the ploidy indicated for the four daughter cells. The terminology "n - 1" and "n + 1" describes the variations in numbers from the haploid set of chromosomes.

F1 generation: - double blue, double pink, double dark green - double light green F2 generation: n+1= single blue, single pink, single dark green n-1= single light green If nondisjunction occurs during meiosis I, homologous chromosomes fail to separate. This produces abnormal gametes that contain two members of the affected chromosome or none at all.

A diploid cell with 4 chromosomes (one pair metacentric, the other telocentric) begins meiosis, and nondisjunction occurs during the second meiotic division. Drag each cell to the correct location in the chart. Make sure the chromosomes correspond to the ploidy indicated for the four daughter cells. The terminology "n - 1" and "n + 1" describes the variations in numbers from the haploid set of chromosomes.

F1 generation: - double blue, double dark green - double pink, double light green F2 generation: n+1= double blue, single dark green n-1= single dark green n= single pink, single light green

The Human Genome Project seeks to rid the human population of genetic disease.

False.

An individual possessing a risk allele for a genetic disease will develop the disease.

False. A risk allele may increase or decrease an individual's risk for a given disease. However, many other factors, such as other genes and environmental influences, also affects whether or not an individual actually develops the disease.

In order to create the possibility of generating a trisomy, nondisjunction must occur during meiosis II.

False. Correct. Nondisjunction during either meiosis I or meiosis II creates gametes that will generate trisomies if fertilized.

Females with only one X chromosome do not develop; this condition is lethal.

False. Females with only one X chromosome are viable but have Turner syndrome, which is characterized by underdeveloped ovaries. Males that lack an X chromosome do not develop; this condition is lethal.

All compounds that have been found to be mutagenic in the Ames test are also carcinogenic.

False. The Ames test is used as a preliminary screening tool. Not all compounds that give a positive Ames test are carcinogenic.

How do translocations such as the Philadelphia chromosome contribute to cancer?

Genetic mapping established that certain genes were combined to form a hybrid oncogene ( BCR/ABL ) that encodes a 200kDa protein that has been implicated in the formation of chronic myelogenous leukemia.

Which nucleotide will base‑pair with the enol form of 5‑bromouracil?

Guanine. The enol form of 5‑bromouracil forms a base pair with guanine.

Which of the following accurately describes a possible meiotic nondisjunction event?

Homologs fail to separate during meiosis I. Correct. Ordinarily, homologs separate during meiosis I. Failure of this separation is one of the ways in which nondisjunction can occur.

What distinguishes paracentric inversions from pericentric inversions?

Inclusion of the centromere in the inversion. Paracentric inversions do not include the centromere; pericentric inversions do include the centromere.

Annotation of the human genome sequence reveals a discrepancy between the number of protein-coding genes and the number of predicted proteins actually expressed by the genome. Proteomic analysis indicates that human cells are capable of synthesizing more than 100,000 different proteins and perhaps three times this number. What is the discrepancy, and how can it be reconciled?

Increased protein production from approximately 20,000 genes is probably related to alternative splicing and various posttranslational processing schemes. In addition, a particular DNA segment may be read in a variety of ways and in two directions.

What was the first human protein to be produced by recombinant DNA technology?

Insulin. Humulin (recombinant human insulin) was first produced by cloning the human gene for insulin and expressing it in bacterial cells. Humulin was licensed for human use by the FDA in 1982.

What is the role of the p53 protein in the cell cycle in normal cells?

It temporarily arrests the cell cycle in G1 before entering S. The p53 protein apparently arrests the cell cycle until any DNA damage has been repaired.

Choose the correct image of how these chromosomes would pair during meiosis.

Looks like a + (plus symbol) where every letter is across from itself on each arm

How can mutations in noncoding segments of DNADNA contribute to the development of cancers?

Mutations in noncoding regulatory sequences (such as promoters, enhancers, and transcription binding sites) of genes such as proto-oncogenes, tumor-suppressor genes, or cell cycle genes could alter the timing or level of their expression leading to cancer.

What is a spontaneous mutation?

Mutations that occur as a result of natural biological and/or chemical processes are considered spontaneous.

Genetic tests that detect mutations in the BRCA1 and BRCA2 oncogenes are widely available. These tests reveal a number of mutations in these genes - mutations that have been linked to familial breast cancer. Assume that a young woman in a suspected breast cancer family takes the BRCA1 and BRCA2 genetic tests and receives negative results. That is, she does not test positive for the mutant alleles of BRCA1 or BRCA2. Can she consider herself free of risk for breast cancer?

No, she will still have the general population risk of about 10 percent.

The incidence of Down syndrome, also known as trisomy 21, increases with increasing maternal age. Which of the following errors most likely produces this condition?

Nondisjunction during either meiosis I or II in the female gamete. Nondisjunction during either meiosis I or II can produce a gamete that will result in a trisomic zygote. Since the incidence of Down syndrome increases with maternal age, it is likely that this error occurs in the female gamete.

A true reversion occurs when the wild-type DNA sequence is restored to encode its original message by a second mutation at the same site or within the same codon.Which of the following isoleucine codon(s) could be mutated once to form a methionine codon, and then mutated at a second site to restore an isoleucine codon? Select all that apply. (Note that two different positions in the codon must be mutated.)

None of these codons. In order to generate a methionine codon from an isoleucine codon, the #3 position must be changed to a G, giving AUG. A subsequent mutation would then be required at a second site in the AUG codon, but since no isoleucine codons (AUA, AUC, AUU) contain a G in the #3 position, a true reversion can't occur without mutating the #3 position again.

How do normal cells protect themselves from accumulating mutations in genes that could lead to cancer? How do cancer cells differ from normal cells in these processes? Drag the terms on the left to the appropriate blanks on the right to complete the sentences.

Normal cells are often capable of withstanding mutational assault because they have checkpoints and DNA REPAIR mechanisms in place. When such mechanisms fail, CANCER may be a result. Through MUTATION, such protective mechanisms are compromised in cancer cells, and as a result, they show HIGHER than normal rates of mutation, chromosomal abnormalities, and GENOMIC INSTABILITY.

Researchers have identified some tumors that have no recurrent mutations or deletions in known oncogenes or tumor-suppressor genes and no detectable epigenetic alterations. However, these tumors often have large chromosomal deletions. What is the possible explanation that could account for the genetic causes behind these tumors?

The deleted chromosomal regions detected in these tumors could have contained genes whose products regulate the expression of tumor-suppressor genes and/or oncogenes. Such a loss of regulation would lead to tumor formation.

Which of the following statements about genomes is true?

The number of genes in eukaryotic genomes is not necessarily correlated with the amount of DNA in the genome. Eukaryotic genomes contain both coding and noncoding sequences. Consequently, most eukaryotic genomes contain more DNA than would be predicted by the number of genes.

Why are spontaneous mutations rare?

They are relatively rare in comparison to induced mutations that are more directed to the physical or chemical properties of DNA.

Which of the following statements does not accurately describe gene microarrays?

They contain thousands of different spots of protein.

Below is a partial DNA sequence (the original sequence with no mutations); only the coding strand is shown.Assume the sequence is transcribed and translated from left to right with the reading frame as indicated. This sequence represents a single mutation.Mutation #1: 5' - | GGC | GCG | GTA | TTA | GCG | - 3' Complete these sentences about this mutated DNA sequence.

This mutant DNA sequence is the result of a TRANSITION mutation. The effect of this base substitution on the amino acid sequence results in a MISSENSE mutation. The second codon changed from GTG to GCG as the result of a transition mutation (pyrimidine to pyrimidine). This point mutation results in a single amino acid change from valine to alanine, causing a missense mutation.

This sequence represents a single mutation.Mutation #2: 5' - | GGC | GTG | GTA | TAA | GCG | - 3' Complete these sentences about this mutated DNA sequence.

This mutant DNA sequence is the result of a TRANSVERSION mutation. The effect of this base substitution on the amino acid sequence results in a NONSENSE mutation. The fourth codon changed from TTA to TAA as the result of a transversion mutation (pyrimidine to purine). This point mutation results in a change from a codon for leucine to a stop codon, causing a nonsense mutation.

This sequence represents a single mutation.Mutation #3: 5' - | GGC | GTG | GTC | TTA | GCG | - 3' Complete these sentences about this mutated DNA sequence.

This mutant DNA sequence is the result of a TRANSVERSION mutation. The effect of this base substitution on the amino acid sequence results in a SILENT mutation. The third codon changed from GTA to GTC as the result of a transversion mutation (purine to pyrimidine). Although this point mutation results in a codon change, both codons still code for valine, causing a silent mutation.

Which term describes plants or animals that carry a foreign gene?

Transgenic. An organism whose genome has been modified by the introduction of foreign DNA into the germ line from another organism is termed a transgenic organism.

Which of the following terms can be used to describe Down syndrome?

Trisomic and aneuploid. Down syndrome individuals have 3 copies of chromosome 21, and therefore an abnormal number of individual chromosomes making both terms correct.

Glyphosate (a herbicide) inhibits EPSP, a chloroplast enzyme involved in the synthesis of several amino acids.

True.

The study of orthologs would be useful to determine the function of a specific gene in a species.

True. Because orthologs are descended from a common ancestral gene, they tend to have the same function in different species. If the function of a gene in one species were known, then its ortholog in a different species would most likely have that same function.

Many chemicals are more mutagenic after being processed in the liver.

True. Correct. Many potential mutagens are poorly mutagenic until passing through the liver.

Tautomers of nucleotide bases are isomers that differ from each other in the location of one hydrogen atom in the molecule.

True. Nucleotide tautomers differ only in the bonding location of one hydrogen atom.

Which of the following statements about the products produced when nondisjunction occurs during meiosis I is true?

Two products of the second meiotic division have both the maternal and paternal chromosomes of a set and the other two products have none for that set. Nondisjunction during meiosis I yields two gametes that are missing a particular chromosome after meiosis II.

How can duplications arise?

Uneven crossing over during meiotic prophase. Synapsis of homologs and unequal crossing over of chromatids during meiosis I can produce one chromatid with a duplication and one with a deletion.

Why are X rays more potent mutagens than UV radiation?

X rays are of higher energy and shorter wavelength than UV light. They have greater penetrating ability and can create more disruption of DNA.

Explain why individuals with XP show such an early age of onset.

XP individuals lack one or more genes involved in DNA repair.

Bioinformatics programs can search DNA for specific sequences to help identify potential protein-coding regions. Label the sequences in the eukaryotic gene below that could be used by a computer program to identify a gene.

a. transcription regulatory sequences b. promoter c. start codon d. splice sites e. stop codon f. transcription termination sequences Bioinformatics programs can use several consensus sequences found in most eukaryotic genes to identify the location of genes in a DNA sequence. Some sequences are critical in transcription; others are critical in translation. Transcription regulatory sequences, such as enhancers or silencers, are found upstream of the promoter. The promoter sequence is found upstream of the ATG start codon and is important for the initiation of transcription. Start and stop codons signal the initiation and termination of translation. Splice site sequences are found in introns and are important for splicing, an important step in mRNA processing that occurs before the mRNA is translated. Transcription termination sequences are found downstream of the stop codon.

Why are frameshift mutations likely to be more detrimental than point mutations, in which a single pyrimidine or purine has been substituted?

because they are likely to change more than one amino acid.

Down syndrome is caused by trisomy 21, the presence of three copies of chromosome 21. The extra copy usually results from nondisjunction during meiosis. In some cases, however, the extra copy results from a translocation of most of chromosome 21 onto chromosome 14. A person who has had such a translocation in his or her gamete-producing cells is a carrier of familial Down syndrome. The carrier is normal because he or she still has two copies of all the essential genes on chromosome 21, despite the translocation. However, the same may not be true for the carrier's offspring.The diagram shows the six possible gametes that a carrier of familial Down syndrome could produce. Suppose that a carrier of familial Down syndrome mated with a person with a normal karyotype. Which gamete from the carrier parent could fuse with a gamete from the normal parent to produce a trisomy-21 zygote? Drag the appropriate labels to their respective targets. Drag one of the gametes to the target of Group 1 in the diagram. Drag one of the zygotes to the target of Group 2.

gamete from carrier of familial Down Syndrome: 14-21 21 trisomy 21 zygote: 14 14-21 21 21 A carrier of familial Down syndrome has two copies of chromosome 21 and a normal phenotype. However, one of those copies has been translocated to another chromosome, often chromosome 14. Some of the carrier's gametes will contain both the normal and the translocated chromosome 21. If one of those gametes fuses with a gamete from a person with a normal karyotype, a zygote with trisomy 21 will result.

Cancer is best described as a ________.

genetic disorder at the cellular level. The location of genetic mutations and how they occur vary greatly. However, all cancers result from a genetic disorder at the cellular level.

Which bacteria grow on the agar plate if the Ames test is positive?

his + prototrophs The bacteria used in the Ames test to evaluate mutagenicity are his− auxotrophs. If the Ames test is positive, these bacteria have reverted back to wild type and are his + prototrophs

A 3-year-old child exhibited some early indication of Turner syndrome, which results from a 45,X chromosome composition. Karyotypic analysis demonstrated two cell types: 46,XX (normal) and 45,X. Propose a mechanism for the origin of this mosaicism.

mitotic nondisjunction of X chromosomes.

A woman who sought genetic counseling is found to be heterozygous for a chromosomal rearrangement between the second and third chromosomes. Her chromosomes, compared to those in a normal karyotype, are diagrammed here: What kind of chromosomal aberration is shown?

reciprocal translocation.

A DNA microarray (also called a DNA chip) can be used to ________.

scan a population of nucleic acids for abundance and mutations.

In the Ames Test, the appearance of his+ revertants in the presence of a non-mutagenic control compound indicates that _______.

some of the reversion mutations are not caused by the mutagen being tested. Correct. His+ revertants on the control plate are the result of spontaneous mutation.

The purpose of the Ames Test is to _______.

test the mutagenic effects of chemicals. Correct. The Ames test detects whether a given chemical can cause a reversion mutation in his- bacteria.

The Human Genome Project began in 1990 with fifteen year plan and a proposed budget of three billion dollars. Which one of the following statements was not a major goal of the Human Genome Project?

to map and sequence the Neanderthal (Homo neanderthalensis) genome.

All of the following events occur during normal meiosis except _______.

two haploid gametes fuse to form a diploid cell. Correct. Fusion of haploid gametes occurs after meiosis.

Dominant mutations can be categorized according to whether they increase or decrease the overall activity of a gene or gene product. Although a loss-of-function mutation (a mutation that inactivates the gene product) is usually recessive, for some genes, one dose of the normal gene product, encoded by the normal allele, is not sufficient to produce a normal phenotype. In this case, a loss-of-function mutation in the gene will be dominant, and the gene is said to be haploinsufficient. A second category of dominant mutation is the gain-of-function mutation, which results in a new activity or an increased activity or expression of a gene or gene product. The gene therapy technique currently used in clinical trials involves the "addition" to somatic cells of a normal copy of a gene. In other words, a normal copy of the gene is inserted into the genome of the mutant somatic cell, but the mutated copy of the gene is not removed or replaced. Will this strategy work for either of the two aforementioned types of dominant mutations?

yes, but only for loss-of-function mutations.