Genetics

What is the inbreeding coefficient for VI-1? 0 0.0020 0.0039 0. 0078

0.0039

Describe the physical composition of a DNA microarray? List 4 specific uses of microarrays that are not described in your textbook.

A DNA microarray is made by dotting a slide with hundreds-thousands of single stranded DNA sequences from specific genes and then layering fluorescently labeled cDNA onto the slide, if the cDNA binds to an area it will have fluorescence indicating expression. 1. SNP detection - A type of DNA microarray used to detect polymorphisms in a population. 2. Multi-stranded DNA microarrays - DNA microarrays used to identify new drugs that bind to multi-stranded nucleic acid sequences. 3. Fusion genes microarray - detect fusion transcripts, for example, from cancer specimen 4. Tiling array - Overlapping probes designed to densely represent a genomic region of interest, and this can be as large as an entire human chromosome. This can detect expression of transcripts or alternatively spliced forms that may not have been known previously

The disorder known as cystic fibrosis (CF) is inherited in a recessive autosomal manner. In populations of Northern European descent, the frequency of people with CF is about 1 in 2,500. Assuming Hardy-Weinberg equilibrium for this disorder, answer the following: A. Assuming random mating, what is the probability that two phenotypically unaffected heterozygous carriers will choose each other as mates? B. What are the frequencies for the normal and CF alleles? C. What are the genotype frequencies of homozygous unaffected, heterozygous, and homozygous affected individuals?

A. q^2=1/2500=0.0004 q=sqrt(0.004)=0.02 p+q=1 p=0.98 2(0.98)(0.02)=0.0392 0.0392^2=0.154% B. q^2=1/2500=0.0004 q=sqrt(0.004)=0.02 (CF Alleles) p+q=1 p=0.98 (Normal Alleles) C. p^2=(0.98)^2=0.9604 2pq=2(0.98)(0.02)=0.0392 q^2=(0.02)^2=0.0004

One method of molecular pharming involves the synthesis of human proteins into the milk of cows. As an example, let's consider the production of human antibodies in cow's milk. Which of the following components is not needed to achieve this goal? Coding sequence of a human antibody gene Cow oocyte Beta-lactoglobulin promoter from a cow All of the above are needed.

All of the above are needed.

Which of the following statements regarding genetic drift is/are false? Genetic drift is more prominent in smaller populations. Genetic drift may lead to allele fixation. Genetic drift may lead to allele loss. All of the above statements are true.

All of the above statements are true.

The statistic called a covariance describes the degree of variation between two variables in a group. is needed to calculate the correlation coefficient. is calculated as the square of the variance. Answer Both A and B are correct.

Answer Both A and B are correct.

Which of the following is a cause of spontaneous mutations? A tautomeric shift An error in DNA replication UV light Both A and B

Both A and B

Which component of the CRISP-Cas system is not needed if the goal is to create a small deletion in a target gene? Cas9 protein Spacer region of the sgRNA TracrRNA region of the sgRNA Donor DNA

Donor DNA

Which of the following statements regarding heritability is/are true? Heritability applies to a specific population raised in a particular environment. Heritability in the narrow sense takes into account all types of genetic variance. Heritability is a measure of the amount that genetics contributes to the outcome of a trait. Both A and C are true.

Heritability applies to a specific population raised in a particular environment.

Which of the following would be consistent with the idea that a human disease has a genetic component? Identical twins share the disease symptoms more often than fraternal twins. The disease is less likely to occur in relatives living apart than in relatives living together. A correlation cannot be found between the presence of a mutation in a particular gene and the occurrence of the disease. All of the above are consistent.

Identical twins share the disease symptoms more often than fraternal twins.

Which of the following is a factor that, by itself, does not promote widespread changes in allele or genotype frequencies? New Mutation Migration Genetic drift Natural selection

New Mutation

Which of the following statements regarding polymorphism is false? Polymorphism is rare in natural populations. Polymorphisms can occur at the level of phenotype. When a gene exists in two or more alleles in a population, it is said to polymorphic. All of the above are true.

Polymorphism is rare in natural populations.

The Rb gene is a tumor suppressor gene that encodes the Rb protein that negatively regulates E2F. Which of the following genotypes would promote cancer at the cellular level? Note: A superscript with a minus sign denotes a loss-of-function allele. Rb-Rb- E2F- E2F- Rb+Rb- E2F- E2F- Rb+Rb- E2F+ E2F- Rb-Rb- E2F+ E2F-

Rb-Rb- E2F+ E2F-

For the examples shown below, which of them are probably caused by a somatic mutation? One apple tree, in a very large orchard, produces its apples two weeks earlier than any of the other trees. 60-year-old smoker develops lung cancer. A purple flower has a small patch of white tissue. One child, in a family of seven, is an albino.

Somatic mutations are seen in somatic cells, these cells can consist of muscle, nerve, and skin cells. Somatic cells are cells that are not germline cells or gametes. This type of mutation only affects a small portion of the organism. In examples B and C this type of mutation occurs in cells that are not germline cells and only affect a portion of the organism. In A and D, the entire organism is affected by the mutation rather than a portion that is seen with somatic mutations.

Pairs of genetically related people were examined with regard to the rate at which they metabolize glucose. For 25 pairs of individuals that had each of the following relationships, correlation coefficients for glucose metabolism were determined: 1. Mother-daughter: 0.36 2. Mother-granddaughter: 0.17 3. Sister-sister: 0.39 4. Sister-sister (fraternal twins): 0.40 5. Sister-sister (identical twins): 0.77 What is the average heritability for the rate of glucose metabolism in this group of females?

Using the narrow sense heritability equation: 1. 0.36/0.5=0.72 2. 0.17/0.25=0.68 3. 0.39/0.5=0.78 4. 0.40/0.5=0.8 5. 0.77/1.0=0.77 (0.72+0.68+0.78+0.8+0.77)/5=0.75 is the average heritability

The alpha-globin gene in humans and beta-globin gene in humans can be described as homologs. paralogs. members of a gene family. all of the above.

all of the above.

A mutation changes a codon to another codon that does not change the original amino acid to a different amino acid. This is a silent mutation. missense mutation. nonsense mutation. frameshift mutation.

silent mutation.

In a population in Hardy-Weinberg equilibrium, the percentage of individuals exhibiting a recessive disorder is 4%, which is the same as 0.04. What is the frequency of heterozygotes? 0.96 0.48 0.32 0.16

0.32

In a population of humans, the correlation between height for fathers and their adult sons is 0.24. What is the narrow sense heritability for weight in this population? Note: hN2 = robs/rexp 0.12 0.34 0.48 0.96

0.48

Let's suppose the frequencies of two alleles in a particular population are B = 0.67 and b = 0.33. The genotype frequencies are BB = 0.50, Bb = 0.37, and bb = 0.13. Do these numbers indicate inbreeding? Explain why or why not.

BB = (0.67)^2 = 0.45 Bb = 2(0.67)(0.33) = 0.44 bb = (0.33)^2 = 0.11 When comparing the values found through the equation and the actual data values we find that the actual data has higher percentages of homozygotes and lower percentages of heterozygotes. These findings are consistent with inbreeding as inbreeding increases the percentage of homozygotes.

Which of the following is not an example of personalized medicine? Using a certain type of chemotherapy based on the genetic characteristics of a tumor Giving a diabetic patient insulin that was made by genetically modified bacteria Giving a patient a certain dosage of a drug based on their genotype Prescribing medicines or activities to prevent a disease (such as heart disease) based on a person's genotype

Giving a diabetic patient insulin that was made by genetically modified bacteria

For a quantitative trait that is polygenic, which of the following would tend to promote a continuum of phenotypes? Increasing the number of genes that affect the trait Decreasing the effects of environmental variation Increasing the mutation rate Both A and B

Increasing the number of genes that affect the trait

Which of the following steps is not needed to make a cDNA (complementary DNA) library? Attach linkers to the cDNA. Cut cDNA and vector with a restriction enzyme, and ligate the cDNA into the vector. Isolate chromosomal DNA and digest with a restriction enzyme. Isolate mRNA and make cDNA using reverse transcriptase and a poly-dT primer.

Isolate chromosomal DNA and digest with a restriction enzyme.

Let's consider the trait of height in people Researchers determined the correlation coefficient (r) for height between fathers and sons in two different populations. In both cases, r = 0.53. One population had an N value of 29 and the other had an N value of 5. Which of the following statements is true. Only the r value in which N = 5 is statistically significant. Only the r value in which N = 29 is statistically significant. Both r values are statistically significant. Neither r value is statistically significant.

Only the r value in which N = 29 is statistically significant.

During real-time PCR, why does the level of fluorescence given off by the TaqMan probe increase over time? Reverse transcriptase cleaves the TaqMan probe, which separates the quencher and the reporter. Taq polymerase cleaves the TaqMan probe, which separates the quencher and the reporter. Taq polymerase binds to the TaqMan probe and causes its fluorescence to increase. As the PCR products accumulate, the Taqman probe binds to more template strands and thereby gives off more fluorescence.

Taq polymerase cleaves the TaqMan probe, which separates the quencher and the reporter.

Table 20.3 shows the recognition sites for five different restriction enzymes. After these restriction enzymes have cleaved the DNA, four of them produce sticky ends that can hydrogen bond with complementary sticky ends. The efficiency of sticky ends binding together depends on the number of hydrogen bonds; more hydrogen bonds makes the ends "stickier" and more likely to stay attached. Rank these five restriction enzymes in Table 20.3 (from best to worst) with regard to the efficiency of their sticky ends binding to each other.

1. BamHI, 10 hydrogen bonds 2. Sau3AI, 10 hydrogen bonds 3. PstI, 10 hydrogen bonds 4. EcoRI, 8 hydrogen bonds 5. NaeI, 0 hydrogen bonds Note: 1, 2, and 3 are all equal in their "stickiness" because they have the same number of hydrogen bonds in their sticky ends.

For the descriptions that follow, is each one an allele, genotype, and/or phenotype frequency: A. The percentage of carriers of the cystic fibrosis allele in the U.S. is approximately 2%. B. Approximately 1 in 2,500 in the U.S. is born with cystic fibrosis. C. The number of new mutations for achondroplasia, a dominant genetic disorder, is approximately 5 × 10-5.

A. Genotype frequency B. Phenotype and Genotype frequency C. Allele frequency

In a strain of tomatoes, the variance for fruit weight is 441 g2, and the mean weight is 412 g. How heavy would a tomato have to be to rank in the top 2.3%? Equal to or greater than 433 g Equal to or greater than 454 g Equal to or greater than 475 g Equal to or greater than 824 g

Equal to or greater than 454 g

Researchers conducted Northern blotting on a sample of cells from mouse kidney, brain, and liver using a probe that is complementary to a gene that encodes a glucose transporter. All of the samples were taken from the same mouse. The following results were obtained: If we assume that equal amounts of total cellular RNA was loaded into each lane, which of the following statements is false? The pre-mRNA is alternatively spliced. The glucose transporter is highly expressed in brain cells. The glucose transporter is highly expressed in kidney cells. The glucose transporter is highly expressed in liver cells.

The glucose transporter is highly expressed in brain cells.

When certain genotypes are preferentially found in a particular environment, this is termed a genotype-environment association. a genotype-environment interaction. the additive effects of alleles. both A and B.

a genotype-environment association.

A gene exists in two alleles and the heterozygote has the highest fitness. This scenario is likely to result in directional selection. stabilizing selection. disruptive selection (also known as diversifying selection). balancing selection.

balancing selection.

Below is a pedigree involving a human disease that is inherited in a simple Mendelian manner. Which of the following patterns is this pedigree consistent with? autosomal recessive only X-linked recessive only autosomal dominant only both a and c

both a and c

Which of the following would not promote polymorphism? directional selection diversifying (disruptive) selection heterozygote advantage negative frequency-dependent selection

directional selection

Glofish are zebrafish that contain a gene from either a coral or jellyfish. This is an example of gene replacement. gene addition. a gene knockout. both A and B.

gene addition.

If a computer program is designed to recognize a specific regulatory element such as a GRE (glucocorticoid response element), this is an example of pattern recognition. search by signal, also known as sequence recognition. search by content (e.g., codon bias). both A and B.

search by signal, also known as sequence recognition.

Let's consider the trait of weight is a herd of cattle. The variance for weight in this herd is 484 lb2. and the mean weight is 562 lb. How heavy would an animal have to be to rank in the top 2.3% of the herd or in the bottom 0.15%?

sqrt484= 22 Top 2.3% → 562 + 22 (2 standard deviations) = 606 lbs Bottom 0.15% → 562 - 22 (3 standard deviations) = 496 lbs

According to the double-strand break model for homologous recombination, what happens right after DNA strand degradation at the double-stranded break site? the breakage of DNA strands in the homologous chromosome. the formation of a D loop. the migration of a Holliday junction. the resolution of the Holliday junction into two separate DNA molecules.

the formation of a D loop.

A key event that promotes trinucleotide repeat expansion is chromosome breakage. the presence of an AT-rich region. the presence of a lesion in the DNA. the formation of a hairpin.

the formation of a hairpin.

Describe the steps of nucleotide excision repair.

1. A protein complex that has two UvrA molecules and one UvrB molecule goes along the DNA and tracks down the damaged DNA. The complex detects it by its distorted double helix. 2. When the damage is identified, the UvrA proteins release and UvrC binds 3. UvrC makes a cut on both sides of the damaged site 4. UvrD, a helicase, recognizes the region and separates the two strands of DNA, which releases a DNA segment that contains the damaged region 5. UvrB and UvrC are released 6. DNA polymerase fills in the gap using the undamaged strand as a template 7. DNA ligase connects the newly made DNA to the original strand

The narrow-sense heritability (hN2) for potato weight in a starting population of potato plants is 0.42, and the mean weight is 1.4 pounds. If a breeder crossed plants with average potato weights of 1.9 and 2.1 pounds, respectively, what is the predicted average weight of potatoes from the offspring? Note: hN2 = (Xo − X)/(Xp − X) Where X is the mean weight of the starting population Xo is the mean weight of the offspring Xp is the mean weight of the selected parents 1.4 pounds 1.65 pounds 2.0 pounds 2.35 pounds

1.65 pounds

A primer used in dideoxy DNA sequencing is 20 nucleotides long and has the following sequence: 5'-GGATCCATGACTAGTCCGAC-3'. A segment of DNA is cloned into a vector and then the vector DNA is denatured and subjected to the dideoxy DNA sequencing method. Below is the DNA sequence from a region of the vector. The primer-annealing site is shown bold and underlined. 3-CCCGATCGGCCTAGGTACTGATCAGGCTGAATGACTCTTCAGA-5' Based on the sequence above, what would be the size(s) of the band(s) (i.e., the number of nucleotides in each band) in which dideoxyG had been added to the sequencing reaction? 6, 8, and 11 4 and 13 26, 28, and 31 24 and 33

26, 28, and 31

What is a QTL (quantitative trait locus)? A QTL is a site in a cell where one on or more chromosomes affecting quantitative traits is/are located. A QTL is a site in a cell where one on or more chromosomes affecting quantitative traits is/are expressed. A QTL is a site on a chromosome where one on or more genes affecting quantitative traits is/are located. A QTL is a site on a chromosome where one on or more genes affecting quantitative traits is/are expressed.

A QTL is a site on a chromosome where one on or more genes affecting quantitative traits is/are located.

The BRCA-1 gene in humans is a tumor-suppressor gene that can play a role in certain types of inherited breast cancers. Which of the following statements regarding this gene is false? A pre-disposition to develop breast cancer due to a BRCA-1 mutation displays a dominant inheritance pattern in a pedigree. The mutation in BRCA-1 that promotes cancer is a loss-of-function mutation. At the cellular level, the BRCA-1 mutation is recessive. A deletion of the BRCA-1 gene would not promote breast cancer.

A deletion of the BRCA-1 gene would not promote breast cancer.

What is the key feature of a quantitative trait? Give 3 examples. How are these traits described within groups of individuals?

A key feature of quantitative traits is that they are able to be described numerically. Examples include: height, blood pressure, blood oxygen content, etc. These groups are described in numerical terms and in a frequency distribution, these are also typically continuous traits.

Which of the following describes how Dolly (a cloned sheep) was created? A mammary cell was made totipotent by treating it with hormones. A mammary cell had its nucleus removed and the nucleus from an oocyte was inserted into it. A mammary cell was fused with an egg cell that had its nucleus removed. A sperm and egg cell were fused with a mammary cell that had its nucleus removed.

A mammary cell was fused with an egg cell that had its nucleus removed.

Do you think that a random mutation would be more likely to be beneficial or harmful? Explain your reasoning.

A random mutation would be more likely to be harmful because genes within each species have evolved to function properly. They have various features, such as promoters, coding sequences, and terminators, that allow them to be expressed. A random mutation within a functional gene is more likely to disrupt the gene rather than improve its function.

A point mutation can be a transition, transversion, or a frameshift mutation. What would you expect each of the following mutagens to cause? A. 5-bromouracil B. Nitrous acid C. Proflavin

A. 5-bromouracil (5BU) causes a transition because it undergoes a tautomeric shift and base pairs with guanine at a relatively high rate. When this occurs in DNA replication, 5BU causes an A-T base pair to change to a G-5BU base pair. The adenine changes to guanine; both are purines. B. Nitrous acid causes a transition because it replaces amino groups with keto groups through deamination. This causes cytosine to change to uracil and adenine to change to hypoxanthine. C. Proflavin causes a frameshift mutation because when DNA containing this mutagen is replicated, single nucleotide additions and/or deletions occur in the newly made daughter strands.

The rb gene encodes a protein that inhibits E2F, a transcription factor that activates several genes involved in cell division (see Figure 25.14). Mutations in rb are associated with certain forms of cancer, such as retinoblastoma. Under each of the following conditions, would you expect cancer to occur? A. One copy of rb is defective; both copies of E2F are functional. B. Both copies of rb are defective; both copies of E2F are functional. C. Both copies of rb are defective; one copy of E2F is defective. D. Both copies of rb and E2F are defective.

A. Cancer would not occur because in this condition E2F would be inhibited by the rb protein, preventing the cell from advancing through the cell cycle. B. Cancer would occur because E2F would not be inhibited by the rb protein since both copies are defective. The cell would be able to advance through the cell cycle since E2F is not inhibited. C. Cancer would occur because E2F would not be inhibited by the rb protein since this protein is defective and there is one E2F that is functional. D. Cancer would not occur because the E2F would be inhibited since the rb and E2F copies are both defective.

The heritability for the number of apples per tree in an orchard of trees in Minnesota is 0.88. Consider the following statements and indicate whether you think they are true or false. For those that are false, explain why. A. The environment in Minnesota has very little impact on the outcome of this trait. B. Most of the phenotypic variation for this trait in this group of apple trees is due to genetic variation. C. Based on the observation of the heritability in Minnesota, it is reasonable to conclude that heritability for apple number in an Oregon orchard is also high.

A. False - The variance in the number of apples per tree in the orchard is mostly due to genetic variation. Even though the variance in the number of apples per tree in the orchard is very little impacted by the environment, the environment still affects the outcome of the trait. B. True C. False - It is not possible to conclude that the heritability for apple number per tree in an Oregon orchard is also high because there could be differences in genetic variation among the Oregon orchard population and the environment in which the orchard was raised in.

Inherited diseases such as phenylketonuria (PKU) have a genetic basis. Are the following statements accurate with regard to the genetic basis of any human disease (not just PKU). A. An individual must always inherit two copies of a mutant allele to have disease symptoms. B. A genetic predisposition means that an individual has inherited one or more alleles that make it more likely for them to develop disease symptoms compared to individuals who have not inherited such alleles. C. A genetic predisposition to develop a disease may be passed from parents to offspring. D. The genetic basis for a disease is always more important than the environment.

A. False. They could develop a mutation in the second copy or one mutant may be enough for disease to occur. (haploinsufficiency) B. True. C. True. D. False. The environment can have carcinogens that cause mutations to occur at higher rates resulting in the disease occurring. For example, at least 80% of all human cancer is related to exposure to environmental agents.

A family pedigree is shown below: A. Is individual IV-4 inbred? If so, who is/are the common ancestor(s) and calculate the inbreeding coefficient. B. Is individual IV-1 inbred? If so, who is/are the common ancestor(s) and calculate the inbreeding coefficient.

A. No. B. Yes. Common Ancestors: I-2 Inbreeding Coefficient: F=Sum((½)^5(1+0))=1/32=3.125%

Let's suppose that four birds of the same species fly to a new location and form a new colony. Three are homozygous AA, and one is heterozygous Aa. A. Calculate the probability that the a allele will become fixed in the population via genetic drift? B. If fixation of the a allele occurs, how long will it take? C. How will the growth of the population, from generation to generation, affect the answers to parts A and B? Explain.

A. Probability of fixation=1/2N=1/(2*4)=0.125=12.5% B. Bar t=4N=4*4=16 generations C. As the population becomes larger the probability of fixation will decrease as it becomes more likely that the new mutation will be eliminated due to genetic drift. The amount of generations it takes to fix the new mutation will increase as the population does as well.

A sheep breeder had a herd with a mean weight of 254 lb at 3 years of age. He chose animals with mean weights of 281 lb as parents for the next generation. When these offspring reached 3 years of age, their mean weights were 269 lb. A. Calculate the narrow-sense heritability for weight in this herd. B. Using the heritability value that you calculated in part A, what mean parental weight would you have to choose to get offspring that weigh 275 pounds on average (at 3 years of age)?

A. R = 269-254 = 15 S = 281-254 = 27 hN2 = 15/27 = 0.556 B. R = 275-254 = 21 0.556 = 21/S; S = 37.77 37.77= Xp- 254; Xp = 291.77 Ibs

To identify the following types of genetic occurrences, would a program use sequence recognition, pattern recognition, or both? A. Whether a segment of Drosophila DNA contains a P element (which is a specific type of transposable element) B. Whether a segment of DNA contains a stop codon C. In a comparison of two DNA segments, whether there is an inversion in one segment compared to the other segment Pattern recognition would be used D. Whether a long segment of bacterial DNA contains one or more genes

A. Sequence recognition would be used B. Sequence recognition would be used C. Pattern recognition would be used D. Both can be used

Consider the following examples, and decide if each of them would result in directional, disruptive, balancing, or stabilizing selection? A. Birth weight in humans B. Sturdy stems and leaves among plants exposed to windy climates C. Polymorphisms in snail color and banding pattern as described in Figure 27.12 D. Thick fur among mammals exposed to cold climates

A. Stabilizing selection - Stabilizing selection is typically directed at quantitative traits, such as body weight and offspring number. B. Directional selection - this will favor this single phenotype which will make the plants more likely to survive and reproduce in the windy climate C. Disruptive selection - There are specific environments where the snails are located, and depending on the environment, a certain phenotype allows them to blend in better, so two different phenotypes will be favored. D. Directional selection - in cold climates, this will favor the phenotype of thick fur, which will make the mammals more likely to live and reproduce in the cold.

Western blotting, which is described in Chapter 21, is a method to detect the amount of a given polypeptide. In this method, a particular polypeptide or protein is detected by an antibody that specifically recognizes a segment of its amino acid sequence. After the antibody binds to the polypeptide within a gel, a secondary antibody (which is labeled) is used to visualize the polypeptide as a dark band. For example, an antibody that recognizes α-galactosidase A could be used to specifically detect the amount of α-galactosidase A protein on a gel. The enzyme α-galactosidase A is defective in people who have Fabry disease, which follows an X-linked recessive pattern of inheritance. Amy, Nan, and Pete are siblings, and Pete has Fabry disease. Aileen, Jason, and Jerry are brothers and sister, and Jerry has Fabry disease. Amy, Nan, and Pete are not related to Aileen, Jason, and Jerry. Amy, Nan, and Aileen are concerned that they could be carriers of a defective α-galactosidase A gene. A sample of cells was obtained from each of these six individuals and subjected to Western blotting, using an antibody against α-galactosidase A. Samples were also obtained from two unrelated unaffected females (lanes 7 and 8). The results are shown next. (Note: Due to X-chromosome inactivation in females, the amount of expression of genes on the single X chromosome in males is equal to the amount of expression from genes on both X chromosomes in females.) A. Explain the type of mutation (i.e., missense, nonsense, promoter, etc.) that causes Fabry disease in Pete and Jerry. B. What would you tell Amy, Nan, and Aileen regarding the likelihood that they are carriers of the mutant allele and the probability of having affected offspring?

A. The figure illustrates that the polypeptide from Pete is shorter than normal. This could be caused by either a deletion or a nonsense mutation. The figure illustrates that there is no polypeptide from Jerry, which could mean that there is no gene expression due to a deletion mutation or a promoter mutation that prevented the gene's expression. B. Amy has two normal copies of the gene, and therefore, has a 0% chance of passing on the mutant alleles or having affected offspring. Nan is heterozygous for the allele, and therefore, has a 50% chance of passing the mutant allele to offspring meaning half of her sons will be affected. Aileen also appears to be heterozygous as the band width is half that of the normal band width. This means she would have the same probability as Nan in passing on the mutant allele to offspring at 50% with half of her sons being affected.

An original DNA strand had the sequence, 5'-ATGGGACTAGATACC-3'. (Note: Only the coding strand is shown; the first codon specifies methionine.) Is each of the following mutations a silent, missense, nonsense, or frameshift mutation? A. 5'-ATGGGTCTAGATACC-3' B. 5'-ATGCGACTAGATACC-3' C. 5'-ATGGGACTAGTTACC-3' D. 5'-ATGGGACTAAGATACC-3,

A. silent mutation because both GGA and GGT code for the same amino acid B. Missense mutation because CGA encodes for a different amino acid than GGT C. Missense mutation because GTT and GAT encode for different amino acids D. Frameshift mutation because an extra base, A, was added into the sequence.

Let's suppose that the frequency of individuals with a recessive autosomal disorder is 0.01, which equals 1%. What percentage of people in the population would be expected to be heterozygous carriers?

Assume that this population is in Hardy-Weinberg equilibrium. Let q^2=0.01 Observe q=0.1 p=1-q=1-0.1=0.9 2pq=2(0.9)(0.1)=0.18=18%of people in the population would be expected to be heterozygous carriers.

For a rare genetic disorder that is equally probable in males and females, the concordance for identical twins is 91%. Which of the following is/are a possible reason(s) why the concordance value between identical twins is not 100%? The disorder may not be completely penetrant. In some cases, the disorder may be due to a new mutation in one of the twins. Identical twins may have inherited different types of mitochondria. Both A and B are possible.

Both A and B are possible.

With regard to heterosis, which of the following scenarios is/are not consistent with the overdominance hypothesis? AA and Aa have the same fitness, and aa has a lower fitness. AA and aa have the same fitness, and Aa has a lower fitness. A1A2 has a higher fitness compared to A1A1 and A2A2. Both A and B.

Both A and B.

Which of the following is an example of a somatic mutation? A mutation in an embryonic muscle cell A mutation in a sperm cell A mutation in an adult nerve cell Both A and C

Both A and C

According to the Holliday model for homologous recombination, what is/are the possible end result(s) of the resolution step? Recombinant chromosomes without a heteroduplex region Recombinant chromosomes with a heteroduplex region Nonrecombinant chromosomes with a heteroduplex region Both B and C are possible.

Both B and C are possible.

Below are 4 steps that occur in the nucleotide excision repair system. What is their correct order? A. Removal of the damaged region B. Endonuclease cleavage on either side of the lesion C. Sensing of a DNA lesion D. Use of the opposite strand (without a lesion) as a template to make new DNA A, B, C, D C, B, A, D C, A, B, D B, C, A, D

C, B, A, D

In the method of real-time PCR, what is the cycle threshold? How is this value used to quantitate the level of a specific RNA in a sample?

Cycle Threshold: when the accumulation of fluorescence is significantly greater than the background level. Standards can be used and compared to the sample to quantitate the unknown amount of RNA. The standard is often a known concentration and the relative concentration compared to the sample can be determined. When the concentration is higher, the cycle threshold is reached at a shorter amount of time which helps determine the sample concentration.

Why is it useful to search a database to identify sequences that are homologous to a newly determined sequence?

Databases store so much information in files that are easily accessible by computers. Computers are much faster and accurate than humans in detecting patterns and sequences located within a multi thousand base pair sequence. Because it will be faster and more accurate, it will save the researcher much more time. Since homologous sequences carry out identical or similar functions most of the time, the annotated sequences that a newly determined sequence is homologous to can provide possible functions of that sequence.

During PCR, what is the order of events in a single cycle? Annealing of the primers, denaturation of the DNA, synthesis of complementary strands Annealing of the primers, synthesis of complementary strands, denaturation of the DNA Denaturation of the DNA, synthesis of complementary strands, annealing of the primers Denaturation of the DNA, annealing of the primers, synthesis of complementary strands

Denaturation of the DNA, annealing of the primers, synthesis of complementary strands

A gene exists in two alleles, which we will call B and b. The gene is 1123 bp in length, and the B and b alleles exhibit single base pair differences at six different sites. If gene conversion changed the b allele into the B allele, which mechanism would you favor to explain the conversion? Mismatch repair Gap repair synthesis Nucleotide excision repair Direct repair

Gap repair synthesis

Search the web and describe three applications of gene cloning that are not described in your textbook.

Gene cloning is commonly used in research to produce a virtually limitless supply of a gene to analyse and test. Gene cloning in the medical industry is used to produce important vitamins, hormones, and antibiotics. Specifically, in agriculture, gene cloning is commonly utilized to increase nitrogen fixation in plants which allows for more hardy crops. More specific applications include the following: 1. Scientists have Inserted recombinant plasmids into bacteria to synthesize biofuel. The cloned gene of interest stimulates the conversion of glucose to triacylglycerol, which can be converted into hydrocarbons used as fuel. 2. To develop a vaccine for Hepatitis B, scientists use a recombinant subunit to immunize patients against the disease without exposing the patients to the Hepatitis B virus itself. 3. Gene cloning has been used to rapidly get disease resistant genes from wild plants and transfer them to domestic crops using a technique called AgRenSeq, also known as speed cloning

The restriction enzymes shown below cut DNA in the following manner. The slashes indicate where cutting occurs. BamHI G\GATCC CCTAG\G EcoRI G\AATTC CTTAA\G HaeII GGCGC\C C\CGCGG PstI CTGCA\G G\ACGTC These enzymes will generate sticky ends that will bind to complementary sticky ends? Rank them from those that would produce the most stable sticky ends to the least stable. EcoRI > PstI = BamH1 > HaeII HaeII > PstI = BamH1 > EcoRI EcoRI > HaeII > PstI = BamH1 HaeII > EcoRI > PstI = BamH1

HaeII > PstI = BamH1 > EcoRI

Gene conversion may occur via mismatch repair or gap repair synthesis. Let's suppose a gene exists in two alleles, F and f. The gene is 2066 bp in length, and the F and f alleles exhibit single base pair differences at 8 different sites that are very close together. If gene conversion changed the F allele into the f allele, which mechanism would you favor to explain the conversion—mismatch repair or gap repair synthesis?. Explain your choice.

If gene conversion changed the F allele into the f allele, the mechanism that would explain this conversion is gap repair synthesis. This is because in gap repair synthesis there is a double-stranded break in the DNA that, in this case, degrades the F allele. Gap repair synthesis would be implemented to use DNA of the f allele to repair the F allele. Through strand invasion that causes a D-loop formation, the gaps are filled in with the f allele. Additionally, gap repair synthesis would be favored because there are multiple differences that are close together, making it more efficient. It is worth noting that in mismatch repair, there is a (¼)8 chance of changing the F allele into the f allele, whereas this is increased to ½ in gap repair synthesis as the eight different sites are very close together.

Which of the following is not expected to occur as a result of bidirectional migration? It tends to reduce allele frequency differences between two populations. It tends to increase the rate of genetic drift. It can enhance genetic diversity within a population because new mutations in one population can be introduced to neighboring population. All of the above are expected to occur as a result of bidirectional migration.

It tends to increase the rate of genetic drift.

Compare and contrast the advantages and disadvantages of mutations.

Mutations allow organisms to adapt to their environments, but they can also cause disruptions in normal gene function. For example, a beneficial mutation causing a change of color of peppered moths allowed them to adapt to changing environments during the industrial revolution and blend into their habitat better, making it harder to be spotted by predators. Additionally, mutations facilitate genetic diversity. Conversely in humans, a single point mutation is responsible for sickle cell anemia, an incurable disease that can be fatal if left untreated.

Most inherited forms of cancer show a dominant pattern of inheritance in a pedigree. An example is a mutation in the BRCA-1 gene. However, the formation of cancer is recessive at the cellular level. Explain how this is possible.

Mutations in DNA repair genes, such as BRCA-1, can contribute to a cell becoming cancerous. There are two copies of DNA repair genes in each cell. One copy of DNA repair genes is sufficient to repair the DNA in a cell and prevent a cell from becoming cancerous. Therefore, DNA repair genes require mutations on both copies to contribute to a cell becoming cancerous. This is why the formation of cancer is recessive at the cellular level.

Which of the following types of genetic drift involves a dramatic shrinkage in population size followed by a rebound in the population size? The bottleneck effect The founder effect Negative-frequency dependent selection Both A and B

The bottleneck effect

What is the difference between gene addition versus and gene modification. Are the following examples of gene addition or gene modification? A. A mouse model to study sickle cell disease B. Introduction of a pesticide-resistance gene into corn using the Ti plasmid of A. tumefaciens

The difference between gene addition and gene modification is gene addition involves the insertion of a cloned gene into a genome, whereas a gene modification focuses on altering the gene sequence that is already present in the organism. A. Since mice don't have the genes that could be adequately used to study human sickle cell disease, a gene addition would have to occur, making the mice transgenic models. The normal human α-globin gene knockin and human β-globin gene knockin are used. Now that the genes were introduced and have the ability to adequately express sickle cell, a gene modification would have to occur to knock out the mouse α-globin gene and mouse β-globin gene. This could be done with the use of CRISPR, which would alter a gene sequence by introducing a specific mutation. B. This would be gene addition because researchers are introducing and inserting a pesticide-resistance gene into the corn. Researchers are introducing a gene that is not already present in the corn's genome.

The method of Northern blotting is used to determine the amount of a particular RNA transcribed in a given cell type and the size of the mRNA. The Northern blot shown below was produced using a DNA probe that is complementary to the mRNA encoded by a particular gene. The mRNA in Lanes 1-4 was isolated from different cell types, and equal amounts of total cellular mRNA were added to each lane.

The different bands indicate that the mRNA of each has been alternatively spliced to produce mRNAs with different molecular weights (MW) so that the lower MW is furthest down the gel. The thicker and darker the bands indicate more mRNA present in the respective cell type. For example, since lane 1 has the thickest band, this shows that nerve cells contain a large amount of this large-sized mRNA. Spleen cells produce a moderate amount of smaller mRNA. Since there is no band in the second lane, this indicates that the kidney cells do not transcribe the gene. Since the muscle cells have two bands, this indicates that there are two types of alternatively spliced products produced in small amounts.The muscle cells transcribe both the larger form of the mRNA as well as the smaller form of the mRNA.

Cultivated roses have been the subject of intense selective breeding. Many newer varieties have particularly large and brightly colored flowers. Unfortunately (perhaps), most of these newer varieties do not have very fragrant flowers compared to the wild-type species and older varieties. Propose two explanations why some of these newer varieties with showy flowers are not as fragrant.

The flower color and fragrance could be affected by additive alleles. Perhaps the traits are influenced in a way that more robust flowers include less fragrance. Otherwise, there could be inbreeding depression as the desired roses are repetitively selected to breed with each other. Another reason could be environmental factors, such as the type of soil or fertilizer this specific plant uses. This is also related to the genes of the flower and its ability to take up certain types of nutrients.

For a genetic disease that is equally probable in females and males, the concordance value for identical twins is 91% and for fraternal twins it is 45%. Do these concordance values suggest a recessive or dominant pattern of inheritance? Provide two reasons why the concordance value between identical twins is not 100%.

The genetic disease is likely autosomal as it is equally probable in males and females. For fraternal twins, it is expected that the concordance value to be 50% for dominant disorders when assuming only one parent is heterozygous for the disease. The concordance value for fraternal twins being 45% suggests that this disease follows a dominant pattern of inheritance. Theoretically, identical twins would be expected to have a concordance value of 100%. However, possible reasons for the concordance value being 91% for the identical twins include the genetic disease not being fully penetrant or one twin developing the disorder due to a new mutation that occurred after fertilization.

Let's suppose a sheep that provided the oocyte for reproductive cloning was homozygous for a gene that causes a small head. This gene follows a maternal effect pattern of inheritance. The oocyte had its nucleus removed and then was fused with a mammary cell that was homozygous for the nonmutant allele, which results in a normal-sized head. Would you expect the resultant lamb to have a small or normal-sized head? Explain your choice.

The lamb would have a small head because the maternal effect of the oocyte would cause the formation of a small head before the donor genes are able to be effective. The donor nucleus from the mammary cell and the maternal proteins within the enucleated egg initiate the development of the egg into an embryo. Since the maternal effect pattern of inheritance is observed, the mother's genotype directly determines the phenotype of the offspring. Therefore, even though the nucleus was removed and replaced with a donor nucleus, the offspring would still not be affected by the genes in the donor nucleus. It would not express the same phenotype as the donor. Instead, it is affected by the proteins that facilitate the maternal effect in the enucleated egg. Since the sheep that provided the enucleated egg was homozygous for the small head, the resulting sheep would also have a small head. This phenomenon is similar to the shell spiral direction in snails.

Which of the following is/are a possible reason why a mutation may have a position effect? The mutation is located next to the gene's promoter. The mutation alters the start codon. The mutation is an inversion that moves a gene's promoter next to the regulatory sequences of a different gene. All of the above are mutations that can have a position effect.

The mutation is an inversion that moves a gene's promoter next to the regulatory sequences of a different gene.

A primer used in dideoxy DNA sequencing is 5'-GGATCCATGACTAGTCCGAC-3'. A segment of DNA is cloned into a vector and then the vector DNA is denatured and subjected to dideoxy DNA sequencing method. Below is the DNA sequence from a region of the vector. 1 20 24 2829 32 35374041 | | | || | | | || 5'-GGGCTAGCCGGATCCATGACTAGTCCGACTTACTGACCATCGACTCATCC-3' 3'-CCCGATCGGCCTAGGTACTGATCAGGCTGAATGACTGGTAGCTGAGTAGG-5' To which DNA strand does the primer bind, the top or bottom one? Based on the sequence above, what would be the sizes of the bands (i.e., the number of nucleotides in each band) in which dideoxyC had been added to the growing strand?

The primer binds to the bottom strand because it contains the region that is complementary to the primer (region that the primer binds to is highlighted). Based on the sequence above, the sizes of the bands in which dideoxyC had been added to the growing strand are as follows: 24, 28, 29, 32, 35, 37, 40, and 41. The nucleotides are added next to the primer site. Addition of nucleotides stops when dideoxyC is added.

What are the two molecular mechanisms that can result in gene conversion? Can both of them occur in the Holliday model and in the double-strand break model?

The two mechanisms that can result in this are DNA mismatch repair and DNA gap repair synthesis. Gap repair synthesis occurs in the double-stranded break model and DNA mismatch repair occurs in the Holliday model. The double-stranded break model facilitates DNA gap synthesis repair because there is a double-stranded break on one chromosome. The Holliday model facilitates DNA mismatch repair because of the heteroduplexes formed.

Take a look at Figure 20.6. After 10 cycles of PCR, which type of PCR product predominates? Explain why?

The two parent DNA strands are denatured at the start of PCR. A primer is attached to the 3'-end of each DNA strand. Then, Taq polymerase fills in the complementary nucleotide sequence after the primers. After that, the strands of DNA split, rendering four strands of DNA. Two of these strands are the original parent strands while the other two are the newly synthesized daughter strands. When the primer binds to the daughter strands, Taq polymerase synthesizes the complementary strands and will be going towards the shorter end of the sequence. This results in strands that are short, and when the primers anneal on these strands, Taq polymerase makes the complementary strand, which results in the generation of strands that have the desired length. This means that in every generation, the two parent strands synthesize two long strands. However, as the number of cycles increases, the number of short strands grows by 2^n while the number of long strands grows by 2n. The difference in exponential vs. linear growth explains why the desired short nucleotide sequence will be quite prominent in the system. After 10 cycles, the majority of the PCR product will be of DNA that contains only the region that is flanked by the two primers.

An allele in corn (FR) provides resistance to infection by a particular fungal pathogen. The other allele (FS) results in sensitivity to the pathogen. However, homozygotes carrying the resistance allele (FRFR) have a lower fitness because they are deficient in taking up iron from the soil, but heterozygotes (FRFS) do not have this deficiency. The heterozygotes are still resistant to the fungal pathogen. In an area where the fungal pathogen is present, a heterozygote has a survival advantage. If the relative fitness values for FRFS, FRFR, and FSFS individuals are 1.0, 0.37, and 0.19, respectively, in areas where the fungal pathogen is present, calculate the allele frequencies at equilibrium. How would this equilibrium be affected if the fungal pathogen was eradicated?

There appears to be Heterozygote Advantage in this scenario. Let wFRFS=1.0 wFRFR=0.37 wFSFS=0.19 sFRFS=1-1.0=0 sFRFR=1-0.37=0.63 sFSFS=1-0.19=0.81 The population reaches equilibrium when sFRFR*p=sFSFS*q Observe Case 1: q=1-p FR = p=sFSFSsFSFS+sFRFR=0.810.81+0.63=0.5625 Case 2: p=1-q FS=q=sFRFRsFSFS+sFRFR=0.630.81+0.63=0.4375 If the fungal pathogen was eradicated, then the corn FS allele would have no pathogen to be sensitive to. Therefore, we would observe a rise in the FS allele frequency because it confers the advantage of being able to take up iron from the soil. We would also observe a decrease in the FR allele frequency because it would not be useful as there is no pathogen for the corn to be resistant to. Additionally, this allele confers a disadvantage because it does not facilitate the corn taking up iron from the soil.

Certain cells of the body, such as skin cells, continue to divide throughout life. Why don't we run out of stem cells that are needed to replenish worn out skin cells?

There are two characteristics that help explain why we do not run out of stem cells that are needed to replenish worn out skin cells. The first characteristic is that stem cells have the ability to asymmetrically divide. The second characteristic is that stem cells have the ability to differentiate. The result of these characteristics is that upon cell division, one of the daughter stem cells can differentiate into a skin cell while the other daughter stem cell remains an undifferentiated stem cell. This type of asymmetrical division/differentiation pattern facilitates a relative constant population of stem cells while providing a population of skin cells, which prevents us from "running out" of them. Below is a diagram from the Brooker textbook that substitutes red blood cells for skin cells.

Below is a pedigree involving a human disease that is inherited in a simple Mendelian manner. The filled (black) symbols are affected individuals. Is this pedigree consistent with autosomal recessive, autosomal dominant, and/or X-linked recessive inheritance? If the pedigree is not consistent with one or more of these patterns, explain why.

This pedigree is consistent with autosomal dominant ONLY. Can't be autosomal recessive - as seen in III-8 and III-9, if this were autosomal recessive, two affected parents would not be able to have unaffected children. Can't be X-linked recessive - as seen in II-1 and II-2, if the male is unaffected and the female is affected, all the males would be affected, but III-3 is unaffected.

You suspect that a chemical, which we will call mutagen A, might be a mutagen. To determine if it is, you carry out an Ames test. For both the control and experimental plates, one million bacteria were plated on media that lacked histidine. The following numbers of colonies were observed to grow: Trial Control With mutagen A. 1 5 43 2 9 20 3 4 27 4 5 33 5 7 44 Conduct a t-test on these data. What are the results of the t-test. Would you conclude that mutagen A is really a mutagen?

When a t-test is conducted on these data, p < 0.05. Therefore, the null hypothesis that the control and experimental data are not different from each other can be rejected. So, we can accept the hypothesis that mutagen A is causing a higher mutation rate and is really a mutagen.

Restriction fragment length polymorphisms are a type of molecular marker that can be used in QTL mapping studies (see Table 23.1). Two breeds of sheep differ in 474 RFLPs. One breed has much thicker wool than the other. The two breeds are crossed to each other, and the members of the F1 generation are backcrossed to the original breeds to produce an F2 generation. Five different RFLPs, which are not close to each other, are associated with F2 sheep with thicker wool. How would you interpret these results?

When an RFLP is associated with a trait, this means that a gene that influences the trait is closely linked to that RFLP. At the chromosomal level, the gene of interest is closely linked to the RFLP that a crossover virtually never occurs. In this passage, there are five different RFLPs. This could provide evidence that there are five different genes that are associated with sheep with thicker wool. A possible explanation of this phenomenon could be that this trait is subject to each gene contributing to the trait in an additive way.

Two nearby populations of the same species occasionally intermix due to migration. Describe the consequences of intermixing with regard to allele frequencies and genetic variation?

When two nearby populations of the same species occasionally intermix due to migration the genetic variation would increase since the migrant population would alter allele frequencies in the recipient population. The genetic variation may also increase due to the possibility of a random mutation occurring in one population, this would create a new allele that could be introduced to the other population when migration occurs. Over time the occasional migration would reduce differences in allele frequencies due to the populations having similar allele frequencies.

Within a particular population of rodents in a cold climate, those with thick fur are more likely to survive than those with sparser fur. This scenario is likely to result in balancing selection. directional selection. disruptive selection (also called diversifying selection). stabilizing selection.

directional selection.

Haploinsufficiency follows a __________ pattern of inheritance. In this case, the protein encoded by the disease-causing allele is ___________. recessive, inactive recessive, overactive dominant, inactive dominant, overactive

dominant, inactive

Selective breeding can be conducted to alter the abdomen length in a species of fruit flies. Let's suppose you want to practice selective breeding to make longer flies. In a starting population, the average abdomen length is 1.01 mm. You choose 10 parents (five males and five females) with the following abdomen lengths: 0.97, 0.99, 1.05, 1.06, 1.03, 1.21, 1.22, 1.17, 1.19, and 1.20. You mate them and then analyze the abdomen lengths of 30 offspring (half male and half female): 0.99, 1.15, 1.20, 1.33, 1.07, 1.11, 1.21, 0.94, 1.07, 1.11, 1.20, 1.01, 1.02, 1.05, 1.21, 1.22, 1.03, 0.99, 1.20, 1.10, 0.91, 0.94, 1.13, 1.14, 1.20, 0.89, 1.10, 1.04, 1.01, 1.26. Calculate the realized heritability of abdomen length in this group of flies.

hN2=R/S, R=X0-X, S=Xp-X X= 1.01 X0= 1.09 Xp= 1.11 hN2=(1.09 - 1.01)/(1.11 - 1.01) = 0.8

A DNA microarray is a slide that is dotted with mRNA from a sample of cells. fluorescently labeled cDNA. known sequences of DNA. known cellular proteins.

known sequences of DNA.

Let's suppose that a chemical is a mutagen. If you tested this chemical using the Ames test, there will be_________ colonies on the plates in which the cells had been exposed to the mutagen, because the mutagen converts some of the cells from ____________. less, his+ to his- more, his+ to his- less, his- to his+ more, his- to his+

more, his- to his+