End of Chapter Review Questions

The graph indicates the stem lengths observed in a plant population. (HW #16 chapter 17) What is the mean stem length in the plant population indicated in the graph?

Answer: -20 cm (it is in the center of the normal distribution) Explanation: -The mean is the average value of a quantifiable trait.

In plants, an increase in ploidy is one common mechanism of evolution. How does an increase in ploidy contribute to evolution? Which statements address why polyploidy is more common in plants than in animals?

Answer: -One copy of a gene retains its function, whereas the other gene copy develops a new function. -Self‑fertilization in plants allows new polyploids to mate with themselves. -Behavioral isolation mechanisms prevent different animal species from interbreeding.

James Noonan and his colleagues (2005. Science 309:597-599) set out to study genome sequences of an extinct species of cave bear. They extracted DNA from 40000‑year‑old bones from a cave bear and used a metagenomic approach to isolate, identify, and sequence the cave bear DNA. Why did they use a metagenomic approach when their objective was to sequence the genome of one species, the cave bear?

Answer: -The cave bear samples could be heavily contaminated so scientists used metagenomics to sequence all of the DNA. They then compared the sequences obtained with modern bear DNA to idenitfy which DNA frafments were from the cave bear Explanation: -Metagenomics sequences all the DNA in an enivornmental sample without purifying the sample of interest.

Order the steps required to analyze gene expression from a particular cell type using a DNA microarray.

Answer: -extract mRNA from cells -reverse transcribe mRNA to cDNA -label cDNA with a fluorescent molecule -add cDNA to microarray and incubate -wash away unbound cDNA -visualize microarray and analyze resulting data Explanation: -In order to analyze gene expression from a particular cell type, a pool of mRNA must first be isolated from a sample. Is mRNA typically added directly to a microarray? Consider what other steps must occur first.

You discovered a mouse gene with an unknown function. You do not know the location or sequence of this gene in the mice genome, but a similar gene has been isolated and sequenced in yeast. How might you determine whether this gene is essential for development in mice?

Answer: -Create a targeting vector using a cloned copy of the gene to develop a knockout mouse. If the gene is essential for development, no offspring homozygous for knockout will be born. Explanation: -You can suppress a gene and determine how the phenotype changes to determine a gene's function in vivo.

How are different mouse strains valuable in studies of human disease?

Answer: -Distinct strains of mice have bee created as models of specific human diseases Explanation: -Laboratory mice have been carefully bred to create groups of mice with specific traits that are useful in studies of human disease.

Which of the statements describe an aspect of a distribution?

Answer: -the variance of the number of vertebrae in the eelpout fish -the mean weight of adult grizzly bears in Canada Explanation: -Keep in mind that distributions describe quantifiable traits in populations. For each statement, consider how you would define the population.

Classify each statement about the heritability of a phenotype according to the narrow‑sense heritability value it describes.

Answer: -0.0 -the phenotypic variance is determined by environmental variance -The phenotype of offspring is not related to the phenotypes of parents -0.5 -half of the observed variation is due to the additive genetic effects of alleles -1.0 -a species of birds passes down the long-beaked trait to their offspring in the absence of environmental factors -This heritability value exists in the absence of environmental changes Explanation: -Narrow‑sense heritability takes into account both genetic variance and environmental variance.

Describe and list examples of next-generation sequencing

Description: -all produces similar fragment lengths -able to determine the sequence of the whole, or large part of the genome -produce large amounts of data -sequences many fragments at a time -Computers are used to read the frequences (also in Dideoxy sequencing) -The massive amounts of data are processed through a computer -Based on DNA polymerase (also in Dideoxy sequencing) Examples: -suitable for personal genomics applications - Many simultaneous reactions

Polymerase chain reaction (PCR) is a technique used to amplify (copy) DNA. Suppose a single, linear molecule of double‑stranded DNA (dsDNA) is amplified by PCR. After 30 PCR cycles (a typical number of cycles), how many molecules of dsDNA will there be? Consider that a typical PCR does not start with a single molecule of template DNA, but rather something in the range of 25 nmol of template DNA. What does this tell you about the potential of PCR to amplify DNA?

Answer: - ~60 billion -PCR is an efficient technique with the potential to produce a large amount of DNA

How is anagenesis different from cladogneesis?

Answer: -Anagenesis -occurs within a single lineage -results in evolution of a species -cladogenesis -occurs when a lineage diverges -results in speciation Explanation: -when a lineage diverges, it forms a clade of closely related species

Suppose the given data was collected on five populations of sheep. Breed DD. Dd. dd Adal 0.36~ 0.48 ~0.16 Chios 0.09~ 0.42~ 0.49 Dala 0.16~ 0.53~ 0.31 Columbia 0.23~ 0.63 ~0.14 Dorset 0.46~ 0.50~ 0.04 Which of the populations are in Hardy-Weinberg equilibrium?Columbia

Answer: -Chios -Adal Explanation: -The equation for Hardy-Weinberg equilibrium is 𝑝2+2𝑝𝑞+𝑞2=1,p2+2pq+q2=1, where 𝑝p is the allele frequency of the dominant trait and 𝑞q is the allele frequency of the recessive trait. (HW #5 chapter 28)

Which statements identify differences between proteomics and genomics?

Answer: -Genomics evaluates a relatively static molecule, whereas proteomics evaluates variable molecules. -Mass spectrometry is a technique that is used in proteomics, but not in genomics. Explanation: -Proteomics is the investigation of all of the proteins in an individual, whereas genomics is the investigation of all the genes in an individual.

Describe and list examples of Dideoxy sequencing

Description: -creates set of DNA strands of different lengths -color coded -more for small laboratory experiments -sequences one fragment at a time -Computers are used to read the frequences (also in next-generation sequencing) -The computer detects the peaks to determine the sequencing -Based on DNA polymerase (also in next-generation sequencing) Examples: -reaction produces set of DNA strands of different lengths -Used to determine sequence of single DNA fragment

How does comparative genomics aid in the search for factors that contribute to complex human conditions?

Answer: -Conserved genetic sequences likely influence similar phenotypic traits across species. Explanation: -Comparative genomics involves comparing genomics features of different organisms. For example, a locus may be compared for differences between mouse and human.

The table shows where different restriction enzymes cleave DNA. The abbreviation R represents the purines (adenine and guanine). The pyramids (cytosine, thymine, and uracil) are abbreviated as Y. The abbreviation W represents adenine or thymine. (table here) Which restriction endonucleases would cleave a DNA molecule at the given sequences? The complementary DNA substrate strand is omitted for clarity.

Answer: -EcoRV Explanation: - Sequence in question to be cut: 5' GAAGATATCCTCGC 3' -EcoRV will cut (5' GATATC 3') -The cleavage column did not determine whether or not an enzyme would cut the sequence

As depicted in the table, certain plant genomes are significantly larger than mammal genomes, including the human genome. Why do these diploid plants have more DNA, but less organismal complexity compared with humans?

Answer: -Repetitive DNA from transposons accounts for the majority of these plant genomes Explanation: -Most eukaryotes have about the same number of genes -some plants have 25 or 30 thousand genes and human had around 20 thousand genes -restate the question: (how can the genome be larger without many more genes?) -transposons can make up a large portion of genomes. They make up over 40% of the human genome. - transposons are not important in gene expression

Which of the techniques are examples of biotechnology?

Answer: -Using DNA fragments to harvest insulin for the treatment of diabetes -breeding different dog breeds for hinting or racing -using yeast to ferment fruits or grains to make beer, wine, whiskey, or vodka -developing transgenic crops to use as biofuels and reduce carbon emissions Explanation: -Biotechnology: any innovation in the use of biological resources like plants, animals, or microbes to make products of use to humans. -Biotechnology includes any use of biological resources to make a product of use for humans -Biotechnological techniques include both modern ones and ancient ones

How does the concept of molecular clocks contribute to addressing the question of genetic identity between different species?

helps estimate time since species had a common ancestor

Suppose that the narrow‑sense heritability of ear length in Reno rabbits is 0.4.0.4. The phenotypic variance (𝑉𝑃)(VP) is 0.6,0.6, and the environmental variance (𝑉𝐸)(VE) is 0.1.0.1. Calculate the additive genetic variance (𝑉𝐴)(VA) for ear length in these rabbits.

Answer: -0.2 Explanation: -Consider that narrow‑sense heritability accounts for only additive genetic variance. narrow-sense heritability: h^2=VA / VP -VA= h^2 x VP =0.4 x 0.6 =0.2

Which of the statements defines somatic gene therapy?

Answer: -The transfer of recombinant DNA in people with disease to cure or decrease the symptoms of the disease Explanation: -Think about the possible therapeutic solutions for individuals affected with a genetic disease. Consider that integrating new DNA into an individual's cells can counteract mutations and change their phenotype.

Select the ways in which mass spectrometry is used to identify proteins in a cell.

Answer: -to identify the amount of each protein present in the proteome of a cell -to identify individual proteins from a database of known protein profiles Explanation: -Mass spectrometry is a technique where a molecule's precise mass and charge is measured and a profile is generated.

According to the biological species concept, what is the definition of a species?

organisms that can interbreed in nature to produce healthy offspring

Classify each statement about the heritability of a phenotype according to the narrow-sense heritability value it describes

Answer: -0.02 (va is small): -Genes do not likely contribute to the phenotypic variability in the population -A relationship between the phenotypes of parents and their offspring is mostly determined by environmental factors -0.5 (va is about half): -half of the observed variation is due to the additive genetic effects of alleles -0.98 (va is almost all): - Parents and their offspring share heritable phenotypes -Phenotypic differences result from genetic variation Explanation: -(va/vp)

In humans, red blood cells have a number of proteins embedded in the cell membrane. One type of protein, the Rh factor, is controlled by a single gene and is either present or missing from the red blood cells. If present, the individual has the Rh+ phenotype. If missing, the individual has the Rh− phenotype. Rh+ is the dominant to Rh−. Suppose that, in the Welsh population, the frequency of the Rh− phenotype is 0.04 Using the Hardy-Weinberg equations, calculate the frequency of the Rh+ allele to at least two decimal places.

Answer: -0.8 Explanation: -For recessive phenotypes, the phenotype frequency is equal to the recessive homozygote genotype frequency. p^2 + 2pq + q^2 = 1 p + q = 1 √0.04 = 0.2 1-p=q 1-0.2=0.8

With current sequencing technology, contigs are integral to understanding the entire genome of an organism. What is a contig?

Answer: -a continuous DNA sequence constructed from overlapping DNA fragments Explanation: -The term contig comes from the word contiguous which means connected without a break.

green= intestine; gene D red= stomach; gene C Genes A and B are responsible for basic cellular functions and are equally expressed in stomach and intestinal tissues.

Answer: Gene C=red Gene D=green Gene A and B = yellow

Different types of environmental studies are listed. Select the metagenomic studies.

Answer: -DNA of primate gut contents is sequenced to determine microbial diversity in their food source. -Genetic analyses of deep sea sediment samples characterize regional microbial composition. Explanation: -Metagenomics is also sometimes referred to as environmental genomics.

Use the artificial selection interactive to answer the questions. Cross two plants with narrow flower diameters in the artificial selection interactive and observe the results. Complete the statement about the experimental data. Next, select the correct conclusion regarding the relationship between parental and offspring flower diameters. Parents with narrow flower diameters produce offspring with ________ flowers on average than the parental generation. What kind of relationship is there between parental and offspring flower diameter?

Answer: -Narrower -Direct

Describe the construction of a recombinant plasmid containing the gene for the green fluorescent protein (GFP) and the insertion of the plasmid into a bacterial cell by placing the steps in order.

Answer: -Use PCR to amplify the gene for GFP -Perform a restriction digestion of the GFP gene and the plasmid -Ligate together the GFP gene and the plasmid to generate a recombinant plasmid -Transform bacteria with the recombinant plasmid using electroporation -Plate the bacterial cells, and screen for positive transformants Explanation: -The first step to construct the recombinant plasmid involves making many copies of the green fluorescent protein (GFP) gene. A recombinant plasmid is a circular piece of DNA that includes the plasmid backbone and the gene of interest.

Suppose a researcher performed an experiment to determine the number of genes affecting fruit size in tomatoes. She crossed the domestic tomato, Lycopersicon esculentum, and the wild species, Lycopersicon cheesmanii, for two generations to obtain F2 individuals. Consider the table, which reports the fruit diameter and number of offspring obtained by the researcher in the F2 generation. In the parental lines, L. cheesmanii had a fruit diameter of 7 cm7 cm and L. esculentum had a fruit diameter of 17 cm17 cm . Fruit diameter. Number of offspring 7 cm. 1 8 cm. 10 9 cm. 45 10 cm. 120 11 cm. 210 12 cm. 252 13 cm 210 14 cm. 120 15 cm. 45 16 cm. 10 17 cm. 1 Calculate the total number of genes involved in fruit diameter in tomatoes.

Answer: -(n) number of genes: 5 Explanation: -The ratio of F2 individuals showing extreme phenotypes is equal to (1/4)^𝑛 -n=(ln(1)-ln(1024)) / (ln(1)-ln(4))

A geneticist uses a cloning plasmid that contains the lacZ gene and a gene that confers resistance to penicillin. She inserts a piece of foreign DNA into a restriction site located within the lacZ gene and uses the plasmid to transform bacteria. She then grows the bacteria on selective media containing penicillin and X-gal. Explain how the geneticist can identify bacteria that contain a plasmid with the foreign DNA.

Answer: -Bacteria with the desired plasmid produce white colonies Explanation: -Selective media allows a scientist to screen for bacterial cells containing a desired plasmid. Penicillin ensures that only cells containing the plasmid that confers antibiotic resistance will form colonies. X‑gal discriminates between cells with or without functioning lacZ genes. -Blue = unsuccessful plasmid/gene transfer -White = successful plasmid/gene transfer

As depicted in the table, certain plant genomes are significantly larger than mammal genomes, including the human genome. Genome. Approximate Genome Size Human. 3,200,000,000 Rat. 2,750,000,000 Norway Spruce. 19,600,000,000 Whisk fern. 250,000,000,000 Why do these diploid plants have more DNA, but less organismal complexity compared with humans?

Answer: -Repetitive DNA from transposons accounts for the majority of these plant genomes.

Use the Artificial Selection Interactive to increase the average flowering head size of sunflowers. Select the statements that describe the change in distribution as the diameter of the flowering heads increases.

Answer: -The phenotypic variance of each genotype decreases with each successive generation. -As the diameter of the sunflower head increases, the mean increases. Explanation: -The mean of the data is the average of the flowering head diameters. The variance is the average squared deviation of the flowering head diameters from their mean value.

Plasmids are small circular DNA molecules found in bacteria that replicate separately from chromosomes. Why are plasmids essential for recombinant DNA technology?

Answer: -DNA from a gene of interest can be inserted into a plasmid, then the modified plasmid can be inserted into a bacterial cell to replicate a gene of interest many times Explanation: -Plasmids serve as vectors for the generation of recombinant DNA. A vector is a small piece of DNA in which foreign DNA can be inserted and stably maintained when the combination is inserted into a cell or organism.

Suppose that archaeologists find a tomb in Egypt containing a mummy. The tomb has been looted and any identifying hieroglyphs on the walls have worn away. The mummy does not have any visible distinguishing features, but her tissues still contain intact DNA. What could scientists do to identify who this mummy was?

Answer: -Use genome analysis to compare her DNA with DNA from other identified mummies in the area Explanation: -DNA contains information about an individual's traits and is passed on from parents to offspring.

Match each vector type to the DNA fragment size that vector may clone.

Answer: -4 kb = plasmid -20 kb = bacteriophage -35 kb = cosmid -100 kb = bacterial artificial chromosome (BAC) Explanation: -Different types of vectors accommodate different sizes of DNA inserts. These vectors derive from natural sources. Scientists engineer plasmid vectors from natural bacterial plasmids, or small pieces of DNA that replicate independently of larger bacterial chromosomes. -Bacteriophage can only clone DNA fragment sizes up to 15 kb

Classify the examples as prezygotic or postzygotic barriers.

Answer: -prezygotic -one fly species mating in the morning, whereas another fly species mates in the afternoon -one bird prefers open areas, whereas another bird species prefers wooded areas -postzygotic -individuals from two different species mate, but the resulting embryo dies before birth -A horse and donkey mate to produce a sterile mule Explanation: -Prezygotic barriers prevent a zygote from being created, whereas postzygotic barriers prevent two species from generating viable offspring or generating offspring that are able to reproduce.

Suppose that three loci, each with two alleles (A and a, B and b, C and c), determine the differences in height between two homozygous strains of a plant. These genes are additive and equal in their effects on plant height. One strain (aa bb cc) is 10 cm in height. The other strain (AA BB CC) is 22 cm in height. The two strains are crossed, and the resulting F1 are interbred to produce F2 progeny. Match the F2 phenotypes and expected proportions of progeny.

Answer: -Phenotype (Height). Proportion of F2 10 cm. 1/64 12 cm. 6/64 14 cm. 15/64 16 cm. 20/64 18 cm. 15/64 20 cm. 6/64 22 cm. 1/64 Explanation: -Consider using a Punnett square to represent the F2 progeny.

Recent advances in genome sequencing technology facilitate whole‑genome sequencing of non‑model organisms. Select the scientific questions that a geneticist can answer with the whole‑genome sequence of a single individual of a non‑model organism.

Answer: -What are adaptive and neutral loci in an organism? -How are organisms related?

Match each term to the most appropriate definition or example.

Answer: -A trait that has a range of phenotypic variation= QUANTITATIVE TRAIT -The number of puppies in a dog's litter= MERISTIC TRAIT -The incidence of high blood pressure in a person= THRESHOLD TRAIT -A trait that is influenced by genes at multiple loci= POLYGENIC TRAIT -The amount of milk a cow produces in a day= CONTINUOUS TRAIT Explanation: -Some of the terms are subsets of other terms. Match specific terms and definitions or examples before matching the general ones.

Most scientists consider the Human Genome Project (HGP) to be the most significant scientific project of the 21st century. Choose the statements that describe the key findings of the Human Genome Project.

Answer: -The human genome contains approximately 25000 genes. -There are approximately three billion base pairs in the human genome. Explanation: -Start by eliminating discoveries made prior to the completion of the first draft of the human genome sequence.

The genome of Drosophila melanogaster, a fruit fly, was sequenced in 2000. However, this sequence consisted mainly of euchromatin. The heterochromatin was not sequenced until 2007. Most completed genome sequences do not include heterochromatin. Why is heterochromatin usually not sequenced in genomic projects?

Answer: -Heterochromatin contains many short repeated sequences, which makes it difficult to assemble heterochromatin sequences into large contigs. Explanation: -Next‑generation sequencing technology uses a shotgun approach that breaks large genomic DNA into short segments for easier sequencing. Unique overlapping regions of these short segments are used to computationally reassemble the pieces into long stretches of DNA. -Nuclear DNA comes in two forms: euchromatin and heterochromatin -Euchromatin is lightly packed and contains actively transcribed genes. -Heterochromatin is densely packed, and genes located in heterochromatic regions are largely silenced

Simulations were run on a population with an effective population (Ne) size of 1000 and15 , respectively, to simulate the change in allele frequencies over time. Ten runs were simulated across 100 generations for each population. The beginning frequencies for each allele were set at 0.50. The graphs show only the wildtype allele. (graphs similar to this one) 1000 Ne population 15 Ne population Compare the graphs and determine which evolutionary force is occurring in the population of 15 Ne individuals and not in the population of 1000 Ne indviduals.

Answer: -genetic drift Explanation: -Compare the allele frequencies in each of the graphs to determine how the wildtype allele may fluctuate over time. Each line represents the theoretical change in allele frequency in the population.

For both questions, assume that the four types of nucleotides are equally likely to be found in DNA. Calculate how often, on average, a type II restriction endonuclease is expected to cut a DNA molecule if the recognition sequence for the enzyme has 5 bp. How bout 8bp?

Answer: -5: Once in every 1024 bp -> (4^5) -8: Once in every 65536 bp -> (4^8) Explanation: -Consider that each position in a DNA sequence is an independent variable when compared to each other position. Accordingly, the entire sequence will be multiplicative. -A restriction endonuclease that recognizes a specific sequence that is 5 bp long will cut once whenever the sequence appears. Consider that there are 4 possible nucleotides in a DNA sequence, and each base is independent of the other bases. -Expressed mathematically, a sequence of n bases will appear 1/(4^n) of the time. In other words, the sequence will appear once in every 4^n bases.

Suppose a strawberry farmer calculates that the average strawberry produced from his garden weighs 2 g.2 g. The farmer's 10 most productive plants grow strawberries with a mean weight of 6 g.6 g. The farmer interbreeds these 10 plants, and the progeny produce strawberries with an average weight of 5 g.5 g. If the farmer interbreeds plants that produce strawberries with an average weight of 4 g,4 g, how much would strawberries produced by the progeny of these selected plants weigh?

Answer: -3.5 Explanation: -Use the equation to estimate the response to selection, 𝑅,R, in the selected plants. 𝑅=ℎ2×𝑆 Estimate narrow‑sense heritability, ℎ2,h2, using the example trait data from the 10 most productive plants. 𝑆S represents the selection differential, or the difference in trait values between the individuals selected as parents and the starting population. R= 5g - 3g = 3g S= 6g - 2g = 4g h^2=R/S =3/4 .75 S = 4g - 2g = 2g 𝑅=ℎ2×𝑆 =.75 x 2 =1.5 Progeny weight = mean wieght of the orignial population + R Progeny weight = 2g + 1.5g = 3.5g

A linear piece of DNA was broken into random, overlapping fragments and each fragment was then sequenced. The sequence of each fragment is shown here, listed in random order. 5′−ATCAAAAGC−3′ 5′−TCAAAAGCATAGAGGTACC−3′ 5′−GAGGTACCCTTC−3′ 5′−CTATCAAAAGCAT−3′ 5′−AAAAGCATAGAGGTACC−3′ 5′−AAGCATAGAGGTACC−3′ Using the overlapping regions shared between sequences, create a contig sequence of the original fragment by placing the sequence that is closest to the 5' end first and moving towards the 3' end.

Answer: -5′−CTATCAAAAGCAT−3′ -5′−ATCAAAAGC−3′ -5′−TCAAAAGCATAGAGGTACC−3′ -5′−AAAAGCATAGAGGTACC−3′ -5′−AAGCATAGAGGTACC−3′ -5′−GAGGTACCCTTC−3′ Explanation: -Begin by looking for a region that is shared between as many fragments as possible and place that fragment with the longest 5' region first. Then, add on to the fragment from that starting point by looking for overlap in either direction.

The domestic cat, Felis catus, is a domesticated furry creature often kept as a pet in many countries. Suppose a volunteer at a local animal shelter wanted to find out a little more about the cats in her care. The volunteer carefully measured the height for the 10 fully grown adult cats at her shelter. The height measurements were taken from the base of the foreleg to the shoulder. The table shows the results of the measurements. Sample Height (in) 1 9 2 12.8 3 7.4 4. 8 5. 8.6 6. 12.3 7 9.2 8 11.1 9. 7.8 10 6. Calculate the mean, variance, and standard deviations for cat height for the sample of shelter cats. Report all your answers to two decimal points.

Answer: -Mean: 9.27 in -Variance: 4.50 in^2 Explanation: -Variance is the discrepancy between the observed value and the mean, and the standard deviation measures how spread out the numbers are from the mean. -Mean: (calculate like normal) -Variance: 1) begin by subtracting the mean value from each cat's height measurement. 9-9.27= -.027 12.8-9.27= 3.53 7.4-9.27= -1.87 ..... 2) Next square each difference and divide by (n-1) ((-.027)^2 + (3.53)^2 + (-1.87)^2 + ....) / (n-1) = 4.50

The color distribution for a specific population of geckos is 150 red, 60 orange, and 40 yellow. The allele for the color red is represented by RR, whereas the allele for the color yellow is represented by RY. Both alleles demonstrate incomplete dominance. What are the genotype frequencies in the population? Calculate to at least two decimal places. What is the allele frequency of RR in this population? Calculate to at least two decimal places.

Answer: -RRRR: 0.60 -RRRY: 0.24 -RYRY: 0.16 -allele frequency of RR: 0.72 Explanation: -The genotype frequency is the proportion of total individuals in a population that have the same genotype. The allele frequency is the proportion of individuals in a population that have the same allele. (AA)= AA / N

Which of the following is an example of gene therapy?

Answer: -Replacing a disease-causing allele for a gene with a wild-type allele Explanation: -Mutations in a particular gene can result in the absence of a functional protein and cause disease in affected individuals. Gene therapy can be used to restore the function of a mutated gene, so that the affected individual can produce the correct protein, and therefore cure, or lessen the severity of the disease.

Complementary DNA (cDNA) is a double‑stranded molecule. In the laboratory, how is cDNA generated from a eukaryotic messenger RNA (mRNA)?

Answer: -Reverse transcriptase generates a single-stranded cDNA and then DNA polymerase synthesizes the complementary strand Explanation: -Reverse transcriptase can synthesize a single‑stranded DNA molecule from a single‑stranded eukaryotic mRNA. What enzyme that is used within the cell to synthesize new DNA from a DNA template can be used to generate the complementary strand? -cDNA: double-stranded DNA molecule that is a copy of a messenger RNA molecule -cDNA can be generated from mRNA using an enzyme called reverse transcriptase

How do the goals of the HapMap project differ from the goals of the Human Genome Project?

Answer: -The HapMap Project identifies common genetic variations among individuals, whereas the Human Genome Project sequenced the entire human genome Explanation: -The Human Genome Project was completed before the HapMap Project. The information gained from the Human Genome Project is being used to identify potential targets in the HapMap Project.

Suppose that seed size in a particular plant species is a polygenic trait. A grower crosses two different pure-breeding varieties of plant, and measures seed size in the F1 progeny. She then backcrosses the F1 plants to one of the parental varieties and measures seed size in the backcross progeny. The grower finds that seed size in the backcross progeny has a higher variance than seed size in the F1 progeny. What is the most likely explanation for the higher seed size variability in the backcross progeny?

Answer: -The backcross progeny are not uniformly heterozygous at the seed size loci Explanation: -Polygenic: multiple genes involved -Any variance that occurs in pure-breeding is due to environmental factors -If you have an F1 that is heterozygous for all those genes, many different combinations will segregate in the backcross -There are more genotypes with the backcross of the heterozygous -the backcross progeny are not uniformly heterozygous at all seed size loci (this is actually the F1 progeny) -cell divisions and mutations were not discussed in the heritability section

Researchers used microarrays to examine the expression pattern of 25,000 genes from the primary tumor cells in women who had breast cancer. They determined that the gene expression patterns found in these tumor cells accurately predict the recurrence of cancer within five years after treatment. In this experiment, mRNA from cancer cells and noncancer cells was converted into cDNA and labeled with red fluorescent and green fluorescent markers, respectively. (tables here) Determine whether most of the genes represented on the microarray are overexpressed or underexpressed in cancer cells from patients who remained cancer free for five years after treatment, and select the reason why.

Answer: -Underexpressed, because green indicates genes that are highly expressed in noncancer cells compared to cancer cells. Explanation: -A green dot indicates that the gene was more highly expressed in noncancer cells, whereas a red dot indicates that the gene was more highly expressed in cancer cells.

The Northern spotted owl and Mexican spotted owl arose from geographically isolated populations of the same ancestral owl species. The geographic ranges of the Northern spotted owl and the Mexican spotted owl are shown in the image. The evolution of the Northern spotted owl and Mexican spotted owl is an example of which type of speciation?

Answer: -allopatric speciation Explanation: -The ancestral owl populations that gave rise to the Northern spotted owl and Mexican spotted owl species were separated geographically into the Northern and Southern populations.

Whole‑genome shotgun sequencing is a method used to sequence genomic DNA. Arrange the steps of whole‑genome shotgun sequencing in order from first to last. First step -> Last step

Answer: 1) Isolate genomic DNA 2) Break DNA into large fragments that are approximately 150,000 base pairs in length 3) Insert DNA into BACs 4) Break DNA into fragments that are approximately 1000 base pairs in length 5) Insert DNA into plasmids 6) Sequence DNA 7) Computationally assemble sequences from each BAC 8) Computationally assemble the genome from overlapping BACs Explanation: -Whole‑genome shotgun sequencing requires generating bacterial artificial chromosomes (BACs), which allow bacteria to maintain and replicate very large fragments of foreign DNA. After BACs are generated, plasmids are used to amplify smaller fragments of foreign DNA. Whole‑genome shotgun sequencing works by reconstructing the genome from small, overlapping pieces.

Some traits are considered to be quantitative even though phenotypes are classified only as present or absent. Genetic diseases often fit this description. What creates the underlying quantitative distribution for this type of trait?

Anwer: -Genetic and environmental factors determine an individual's susceptibility to the trait. Susceptibility exhibits continuous variation, and individuals above a certain threshold express the trait Explanation: -Why doesn't everyone with a certain allele show the certain phenotype? -You need a certain collection of alleles that increases the susceptibility above a certain level -quantitative traits are in many loci, not a single locus

A breeder is breeding dogs to select for weight. The weight, in pounds, for the eight dogs in the population is shown in the chart. The mean weight of the population of dogs is 60 pounds. Classify each mating pair according to the effect of its offspring has on the mean weight of the population. A 40 B 60 C 60 D 80 E 40 F 60 G 80 H 60

Anwer: -Increase: C and D -Stays the same: B and H, C and F -Decrease: A and E, E and a 30 pound dog Explanation: -anything above 60 will increase, and anything below 60 will decrease the mean weight of the population

Bipolar disorder is a psychiatric disorder with a strong hereditary basis, but the exact mode of inheritance is not known. Research has shown that siblings of patients with bipolar illness are more likely to develop the disorder than are siblings of unaffected persons. Findings from one study demonstrated that the ratio of bipolar brothers to bipolar sisters is higher when the patient is male than when the patient is female. In other words, relatively more brothers of bipolar patients also have the disease when the patient is male than when the patient is female. Determine what the evidence suggests about the inheritance of bipolar disorder.

Answer: -An X‑linked locus may contribute to this disorder. Explanation: -Consider which mode of inheritance most often results in a skewed gender ratio of affected progeny.

Some traits are considered to be quantitative even though phenotypes are classified only as present or absent. Genetic diseases often fit this description. What is the term used to describe this type of trait? What creates the underlying quantitative distribution for this type of trait?

Answer: -Threshold trait -Genetic and environmental factors determine an individual's susceptibility to the trait. Susceptibility exhibits continuous variation, and individuals above a certain threshold express the trait. Explanation: -Consider that both genes and the environment influence the magnitude of a quantitative trait. One example of this is the disorder phenylketonuria (PKU). Brain damage results in individuals with high levels of blood phenylalanine, which result from a combination of genetic effects and diet. Consider what kind of mechanisms trigger brain damage in a disease such as PKU.

Some traits are considered to be quantitative even though phenotypes are classified only as present or absent. Genetic diseases often fit this description. What is the term used to describe this type of trait?

Answer: -Threshold trait Explanation: -quantitative because there are many genes involved and they can be affected by the environment -Polygenic trait: multiple genes are underlying the phenotype -environmental trait: environment affects a trait, but this isn't actually a real term -meristic trait: only fixed values -epigenetic trait: controlled by the changes in genes other than changes in gene sequence -threshold trait: quantitative trait where the phenotype either shows are doesn't, but you need to accumulate enough factors (genetic, environmental)

After discovering a gene for the production of fur in mammals and examining the gene regulation in the genomes of a monkey, bear, zebrafish, wolf, and snake, it is evident that all of the genomes have an orthologous gene. Select the statements that show what analysis of the orthologous gene says about the function of the gene controlling the trait.

Answer: -The ancestral form of the gene was potentially involved in a different biological function. -The gene function is not limited to fur production. Explanation: -All of the animals have the orthologous gene, but the zebrafish and snake do not have fur. Consider the function of the gene as it relates to expression in the zebrafish and snake.

The graph illustrates frequency distributions for a trait influenced by a single quantitative trait locus (QTL). The distribution drawn in orange represents the entire population. (a graph similar to this one) Identify the true statements regarding the graph.

Answer: -The total population and the population of heterozygotes have the same mean value of the trait. -The distributions that represent the two homozygous populations have the same variance. -The distribution that represents the BB homozygous individuals has a greater mean value of the trait than the total population. Explanation: -Carefully consider the characteristics of a normal distribution that are represented by the height and width of each curve, as well as the relative measurement of the trait value over which it is centered.

A geneticist isolates a new restriction enzyme from the bacterium aeromonas ranidae. No other restriction enzymes have been isolated from this bacterial species. Use the standard convention for abbreviating restriction enzymes to name this new restriction enzyme.

Answer: -Ara1 Explanation: -The first three letters of a restriction enzyme name refer to the bacterial genus and species. A Roman numeral following the letters distinguishes multiple restriction enzymes isolated from the same species. -Example: "Eco" refers to E. Coli. A fourth letter may refer to the bacterial strain from which the enzyme was isolated. For example, the "R" in EcoRI indicates that this enzyme was isolated from E. coli strain RY13. -Roman numerals following the first three or four letters identify different enzymes isolated from the same species. The first enzyme to be isolated from a species would receive the Roman numeral 1.

Suppose a farmer is interested in developing a breeding program on his chicken farm. The farmer would like to artificially select for egg eight, egg shape, shell color, and shell thickness. In a large population of his chickens, the farmer has measured these four continuous egg traits of interest and calculated their variances, which are shown in the table. (Narrow-sense heritability) Which trait would best respond to artificial selection by the farmer?

Answer: -Shell color (59/112= .53) Explanation: -The table gives the calculated variances, which gives us the heritability -under conditions of selection the response to selection R is R=H^2 x S -Response = heritability x selection differential -additive genetic and phenotype columns are very important -The answer is the trait with the highest H^2 -Additive genetic variance / phenotypic variance= answer -Weight (12/503=.02) -Shape (100/745=.13) -Shell thickness (20/736=.02)

Polymerase chain reaction (PCR) is a technique used to amplify (copy) DNA. Suppose a single, linear molecule of double‑stranded DNA (dsDNA) is amplified by PCR. After one PCR cycle, how many molecules of dsDNA will there be? After 3?

Answer: - 2 molecules of dsDNA after one cycle -8 molecules of dsDNA after three cycles Explanation: -Template DNA is incubated with primers, nucleotides, and a thermostable polymerase in a buffer with Mg2+Mg2+ ions (required for polymerase activity). -PCR reaction: -There is an initial denaturation step at 95 °C and then the following steps are cycled 25-30 times. 1) 95 °C - Denaturation Time - to generate ssDNA 2) 55 °C - Annealing Time - primers bind to complementary regions of the template DNA 3) 68 °C - Extension Time - the polymerase extends the primers to form dsDNA. -The incubation times depend on the size of the DNA being copied. -After 25-30 cycles, there is a final extension step at 68 °C to ensure the polymerase completes the extension of all DNA strands. -After one PCR cycle, the dsDNA template will be denatured into two ssDNA molecules. Then, the polymerase will fill in the complementary strand to form two dsDNA copies. Using the same logic how many dsDNA copies are present after two PCR cycles?

Suppose a scientist is studying the platypus, one of the few venomous mammals. The male platypus produces venom in the crural glands located beneath the calcaneus spur on each of its hind limbs. The female platypus can develop calcaneous spurs, but the crural glands never develop. Which of the following libraries would be most useful to help the scientist identify the gene or genes responsible for venom production?

Answer: -cDNA library made from the mRNA isolated from the crural gland of a male platypus Explanation: -Remember that cDNA libraries are generated from isolated mRNAs, whereas genomic DNA libraries contain all the sequence from the organism's genome. -The genomic library will be the same regardless of the tissue from which it is isolated -Venom is comprised of a number of different proteins, so the scientist needs to look for genes that are expressed -scientist needs a library that can identify genes expressed in a specific tissue

Assume that the length of wheat leaves is controlled by three loci, each with two alleles: L and l, W and w, H and h. Determine the differences in leaf length between two homozygous strains of wheat. Assume each allele contributes equally to the length of the plant leaves. One homozygous strain, ll ww hh, has leaves that are 100 mm in length, and the other strain, LL WW HH, has 220 mm leaves. The two homozygous strains are crossed, and the resulting F1 are selfed to produce F2 progeny. What length will the leaves of the ll ww HH genotype plant be? What proportion of the F2 progeny will have the same phenotype as the ll ww HH genotype? Input answer as a decimal.

Answer: -140 -0.23 Explanation: -Calculate the length that each dominant allele adds to the leaf length phenotype by determining the difference in length between the homozygous genotypes. Divide the difference by the total number of dominant alleles in the homozygous dominant genotype. -To determine the length of the leaves for the llwwHH genotype, first, determine the total number of dominant alleles. Multiply the number of dominant alleles by the length that each dominant allele adds to the baseline leaf length. Add the result to the baseline leaf length/ -Length: 20mm x 2 + 100mm = 140mm -Proportion: (there are a total of 15 total genotypes with two dominant alleles) 15/64 = 0.23

Researchers construct different types of genomic maps to gain information about genes and their locations in the genome. Why would a researcher choose to construct a physical map of genes rather than a genetic map?

Answer: -To determine the locations of specific genes on a chromosome and the distance between them Explanation: -genetic maps indicate the relative position of genes as they are aligned along the chromosome, but does not give information about the space between the genes -physical maps are based on the sequence of the chromosome and give information on all the base pairs but may not give information on which base pairs are genes -don't indicate geographic locations of human populations -more than markers needed here -map does not give special information on their sequence such as height or weight

Suppose the human trait for dimples is controlled by a simple dominant and recessive relationship at one locus. Dimples, D, is the dominant allele, and a lack of dimples, d, is the recessive allele. In a college genetics class, the professor takes a tally of students who have dimples and of students with a lack of dimples. In this class of 117117 students, 3737 have dimples. Calculate the frequency of the dominant allele, D, and the heterozygous genotype, Dd. Express the frequencies in decimal form rounded to the nearest thousandth. Assume the class is in Hardy-Weinberg equilibrium for this trait.

Answer: -frequency of D: 0.173 -frequency of Dd: 0.286 Explanation: -Use the population data and the Hardy-Weinberg equations, 𝑝+𝑞=1p+q=1 and 𝑝2+2𝑝𝑞+𝑞2=1 to calculate the genotype and allele frequencies. -Students with recessive genotype = 117-37 = 80 Frequency of recessive genotype =80 / 117= 0.684 Frequency of recessive allele = √0.684 = 0.827 Frequency of dominant allele= 1-0.827= 0.173 frequency of heterozygotes= 2 x 0.173 x 0.827= 0.286

A researcher treats a DNA sample and a plasmid with the ampR and lacZ (for the enzyme β‑galactosidase) genes with the restriction enzyme EcoRI with the goal of making recombinant DNA molecules. The researcher places the lacZ gene in the cloning region so that it is disrupted by DNA inserts. Place the remaining steps for this process in the correct order. Not all steps will be used.

Answer: 1) All the fragments have identical sticky ends: 5' -AATT- 3' 2) Use DNA ligase to join the sticky ends of the DNA sample and the plasmid 3) Grow cells containing the recombinant plasmid in a medium with ampicillin and X-gal (a galactose derivative) 4) White colonies that survive on the culture medium contain the recombinant plasmid 5) Remove colonies containing the recombinant plasmid from the culture medium and continue growing Explanation: -The restriction enzyme EcoRI recognizes and cleaves the sequence G ∣∣ AATTCG | AATTC . Consider the sticky ends that result from this cleavage. -The ampR gene confers resistance to the antibiotic ampicillin. -The lacZ gene codes for β‑galactosidase, which hydrolyzes some galactose derivatives. Cells that contain an intact lacZ gene hydrolyze X‑gal, which yields a blue product. Inserted DNA can interrupt this gene. Consider the color of colonies that do not have an intact lacZ gene.

Samples of a plasmid containing a segment of unknown DNA are digested using the restriction enzymes EcoRI, BamHI, and a combination of EcoRI and BamHI. The digests are then run on an agarose gel in order to separate the resulting fragments by size. Use the results of the gel electrophoresis shown in the image to determine the sizes of the fragments and label the restriction map. Create a map of the DNA segment by dragging the enzyme name to the location it cuts (the vertical lines above the segment) and labeling the fragments with their lengths (numbers, in kilobases). You will not use all the numbers.

Answer: (map image)

Suppose that phenotypic variation in the tail length of mice is summarized in the table. Additive genetic variance (𝑉𝐴)(VA)= 0.4 Dominance genetic variance (𝑉𝐷)(VD)= 0.5 Genetic interaction variance (𝑉𝐼)(VI)= 0.4 Environmental variance (𝑉𝐸)(VE)= 0.2 Genetic‑environmental interaction variance (𝑉𝐺𝐸)(VGE)= 0.0 What is the narrow‑sense heritability (ℎ2) of tail length? What is the broad‑sense heritability (𝐻2) of tail length?

Answer: -Narrow-sense: 0.3 -Broad-sense: 0.87 Explanation: -Narrow‑sense heritability includes only additive genetic effects. However, broad‑sense heritability includes all sources of genetic variation, such as dominance and interaction effects. -Narrow: h2= VA / (VA+VD+VI+VE+VGE) h2= 0.4 / (0.4+0.5+0.4+0.2+0) -Broad: A similar strategy is used to calculate broad-sense heritability, but instead the variable VG is used in place of VA. VG is the total variance due to all genetic effects, so it includes VA, VD, VI, and VGE. H2=VG / VP H2= (VA+VD+VI+VGE) / (VA+VD+VI+VE+VGE) H2= (0.4+0.5+0.4+0) / (0.4+0.5+0.4+0.2+0)

The kernel color in wheat is a continuous trait determined by two additive genes, each with two alleles, that equally contribute to kernel color determination. The red kernels are determined by two genes and two dominant alleles (R1R1R2R2), and white kernels are determined by two recessive alleles at the same two genes (r1r1r2r2). The duplicate dominant alleles R1 and R2 contribute equally to kernel color and cumulatively control the intensity of the red kernel phenotype. Both these alleles are dominant over the recessive white alleles r1 and r2. A true breeding red plant and true breeding white plant are crossed, and the resulting F1 progeny are selfed. What is the expected phenotypic ratio of the kernel progeny of the F2 offspring? Determine which statement best describes the relationship between the phenotypes of the r1r1R2r2 and r1R1r2r2 genotypes.

Answer: -1:4:6:4:1 (1 red: 4 dark pink:6 pink: 4 light pink: 1 white) -The phenotypes of the r1r1R2r2 and r1R1r2r2 genotypes will be the same because the same number of dominant and recessive alleles are contributing in each genotype Explanation: -The color intensity of a kernel depends on the number of dominant alleles in the genotype.

Classify each characteristic as being commonly found in bacterial genomes, in eukaryotic genomes, or in both genomes.

Answer: -Bacterial genomes only: -generally have one chromosome -rarely contain introns -often contain plasmids -Eukaryotic genomes only: -have a diploid chromosome number -chromosomes are generally linear -nucleus contains chromosomes -Characteristics of both -have a DNA double-helix structure -genomes are necessary for survival -contain genes that code for proteins Explanation: -Bacteria and eukaryotes diverged from a common ancestor, so their genomes share some basic structural and functional characteristics common to all living organisms. However, bacteria are prokaryotes and therefore lack the membrane-bound organelles that eukaryotes possess.

Consider how the traits are affected by genetic and environmental factors. -Kernel color in a particular strain of wheat: Kernel color is either white, light red, or dark red in this strain of wheat. Two codominant alleles, which segregate at a single locus, determine these three phenotypes. -Presence or absence of leprosy: Family pedigrees indicate that an increased susceptibility to leprosy is heritable. However, the increased susceptibility does not follow a Mendelian inheritance pattern. Repeat exposure to the bacterium that causes leprosy also increases the probability of having leprosy. -Number of branches on a species of rose bush: Several loci control the formation of 4, 6, 8, or 10 branches in this species of rose bush. This rose bush species forms more than 8 branches per bush when the rainfall is high during a growing season. -Skin color in humans: The broad range of human skin color is due to variations in the type and presence of melanin. Melanin production and accumulation is controlled by at least six genes and is also influenced by environmental factors such as nutrition and exposure to sunlight. -Number of fingers in humans: One form of polydactyly, or having extra limb digits, is caused by an autosomal dominant allele. The presence of the dominant polydactyly allele results in 6, 7, or 8 fingers per hand, instead of the normal 5 fingers. -Presence or absence of cystic fibrosis: A recessive mutation in the CFTR gene, which encodes a membrane‑bound chloride channel, causes cystic fibrosis. Cystic fibrosis can be inherited from parents who do not have the disease but are both heterozygous for the recessive cystic fibrosis allele. Classify each trait to indicate whether it is a continuous trait, a meristic trait, a threshold trait, or a discontinuous trait.

Answer: -Continuous trait -skin color in humans -Meristic trait -number of branches on a rose bush -Threshold trait -presence or absence of leprosy -Discontinuous trait -kernel color in wheat -number of fingers in humans -presence or absence of cystic fibrosis Explanation: -Continuous, meristic, and threshold traits are different types of quantitative traits, which are determined by both genetic and environmental factors. Discontinuous traits are determined primarily by genetic factors and show Mendelian inheritance patterns.

Suppose a team of researchers sequences the genome and measures the proteome of a human skin cell and a human kidney cell. Will there be a greater difference between the genome or the proteome between these two cells?

Answer: -Differences in gene expression and post-transcriptional mRNA processing lead to a greater difference in the proteome than the genome between the two cells. Explanation: -The genome is the genetic material of a cell, whereas the proteome is the complete set of proteins found within a cell.

Cotyledon orbiculata, or pig's ear, is a succulent fynbos plant from South Africa whose juice is used as a treatment for epilepsy. Suppose three South African populations of C. orbiculata were collected and characterized by DNA sequencing across a sequential 8‑kb interval of the nuclear genome. The data was summarized into haplotype data for each of the three populations and shows the number of individuals with each haplotype in each population and the SNPs at each collected loci. Select all of the haplotypes that are unique to a single population of C. orbiculata. Suppose the sequence characterized previously contains part of the unknown gene responsible for the medicinal properties of C. oribulata that allow the plant to treat epilepsy.

Answer: -East_SA-2 -Central_SA-3 -West_SA-4 Explanation: -A haplotype is a set of DNA polymorphisms that tend to be inherited together. Haplotypes are often used in genome wide association studies to locate approximately where SNPs are found on a chromosome.

Both genetic and environmental factors contribute to variation in phenotypes of individuals within a population. Match the appropriate definition with each of the terms that relate to phenotypic variance.

Answer: -Genetic interaction variance: variance due to the interplay among alleles at multiple loci -Phenotypic variance: Sum of all genetic and nongenetic factors that affect expression of a trait within a population -Additive genetic variance: variance resulting from the mean effects of single alleles -genetic-environmental interaction variance: variance due to differential effects of nongenetic factors on specific genotypes -dominance genetic variance: variance due to the interaction between two alleles at a single locus -environmental variance: variance in response to external, nongenetic factors -genetic variance: variance in the alleles among individuals within a population Explanation: -Phenotypic variance is the cumulative effect of additive genetic variance, dominance genetic variance, genic interaction variance, environmental variance, and genetic-environmental interaction variance.

Suppose a farmer is interested in developing a breeding program on his chicken farm. The farmer would like to artificially select for egg eight, egg shape, shell color, and shell thickness. In a large population of his chickens, the farmer has measured these four continuous egg traits of interest and calculated their variances, which are shown in the table. (Narrow-sense heritability) Which trait would best respond to artificial selection by the farmer? A: Shell color (59/112= .53) Select all of the statements that identify how the farmer might utilize the information gained from his data.

Answer: -Identify quantitative trait loci correlated to each trait examined in the chicken population after obtaining genetic sequence data for his chicken population. (Eventually. Identify IQTL if he has a mapping population, and genetic sequence data to map the phenotype of the different individuals) -Design an experiment to improve the shell thickness trait in the current chicken population by changing the farm environment (reducing the environmental variance) Explanation: -Cannot: -Recommend a selective breeding program for a population of chickens from another country based on his results (another country is another environment) -

Suppose a new species of thermophilic bacteria is discovered that can live in undersea volcanic vents. Researchers perform a microarray in order to determine the genetic similarities and differences between sulfur and high temperature tolerance. The results below show high gene expression in the experimental condition compared to the control as pink, low gene expression compared to the control as cyan, and no change in gene expression compared to the control as white. Each letter represents a different gene. (table here) What steps could the researchers do next in order to better understand the genes they identified?

Answer: -Identify shared 5' regulatory elements in gene groups and generate reporters using the elements -Group genes in which both experimental conditions produce the same type of response -Group genes that increase in one experimental condition and decrease in the other condition Explanation: -Consider how genes can be grouped to help draw conclusions. What further testing would give you insight into how the expression of the genes is controlled?

Suppose there are five genes in pumpkins that work additively to affect pumpkin size. A homozygous lowercase allele pumpkin, aabbccddee, weighs 20 lb. Each capitalized allele additively affects the final phenotype by either adding or subtracting weight from the base pumpkin size. The effect of a single capitalized allele is listed in the table. Allele. Effect A. +5 B. +3 C. −2 D. −4 E. +2 What would the potential size range be for pumpkin offspring produced from the cross between a parent with genotype AaBBCcddEE and a parent with genotype AABbccDDEe? Largest pumpkin? Smallest pumpkin?

Answer: -Largest: 36 lb -Smallest: 24 lb Explanation: -Use the information about additive effects to identify the genotypes for each gene that will produce the largest and the smallest offspring. Considering that the question is asking about a range, you only need to focus on the two extreme genotypes and their resulting phenotypes. -The genotype for the largest offspring will be AABBccDdEE. Sum the net change in weight for the AABBccDdEE genotype with the base pumpkin weight. -Large: 20+10+6+0+-4+4=36 -The genotype for the smallest offspring will be aABbCcDdEe. Sum the net change in weight for the aABbCcDdEe genotype with the base pumpkin weight. -Small: 20+5+3+-2+-4+2=24

Suppose a researcher studied gut tissue samples from patients with Crohn's disease, an inflammatory gut disorder, and normal controls. He found that the identity and relative abundance of gut bacterial species varied within both the Crohn's and the control populations. He also discovered general differences between the two populations. The guts of Crohn's patients had reduced levels and diversity of bacteria from the phylum Firmicutes compared to those of controls. The guts of Crohn's patients also had higher levels of bacteria from the phylum Proteobacteria than those of controls. Which type of study did the researcher perform? Which conclusions might you be able to draw from the data?

Answer: -Metagenomic -Two normal individuals may have different compositions of bacteria in their guts -Bacteria from the phylum Firmicutes may provide protection against Crohn's disease Explanation: -In modern biology, the suffix ‑omic refers to studies of all the components of a particular set. Consider if the research was studying the gut genes, mRNA, proteins, or microfauna in the Crohn's patients.

Suppose Jennifer is a horticulturist that specializes in the cultivation of orchids. She houses the orchids in a temperature-controlled and humidity-controlled greenhouse. A salesperson approaches Jennifer and attempts to sell her a special fertilizer that promises to increase flower production. Jennifer, who dabbles in genetics, insists that the fertilizer will have no effect on flower production. She has calculated the narrow-sense heritability of flower production in her orchids and found it to be 0.98. Jennifer claims that a narrow-sense heritability of 0.98 indicates that 98% of the variance in flower production is determined by genetic differences, so a fertilizer would have little effect on flower production in the orchids. The salesperson disagrees with Jennifer and continues to assert that the fertilizer will increase flower production. Who is correct and why?

Answer: -The salesperson is correct. High heritability of a trait within a population does not exclude new environmental factors from affecting the trait. Explanation: -Heritability describes the contribution of genetic differences on the total phenotypic variation in a population. Broad-sense heritability reflects the proportion of phenotypic variance due to any genetic variance, whereas narrow-sense heritability reflects the proportion of phenotypic variance due to additive genetic variance.

Suppose a scientist wants to clone the gene for human insulin. He uses a plasmid that contains the lacZ gene, an origin of replication, and an ampicillin resistant gene. He inserts the human insulin gene into a restriction site located within the lacZ gene. Then he transforms this plasmid into bacteria without a functional lacZ gene and spreads the appropriate dilution onto bacterial plates. When the product of the lacZ gene, beta-galactosidase, interacts with the substrate X-Gal on LB agar, colonies turn blue. Which plate indicates that the human insulin gene was successfully inserted in the plasmid?

Answer: -an LB agar plate with ampicillin and X-Gal that produces colorless colonies Explanation: -origin of replication is in all plasmids -ampicillin resistance is for selection to have bacterial cells that receive the plasmid -lacZ gene is used for a color reaction to determine which plasmids received an insert -when lacZ gene is intact and can produce product (is white) -Because it is blue, that means the plasmid is not intact and cannot carry out the rest of its functions -look for interruption of the lacZ gene

Use the Artificial Selection Interactive to increase the average flowering head size of the sunflowers and answer the questions. Why does the variance decrease as you breed towards the selected target? Which cross would you perform in the sunflowers if your goal was to increase the variance of head size?

Answer: -because the distribution of the phenotype decreases -cross sunflowers with the most phenotypic variance Explanation: -Consider that variance is a deviation from the mean, which indicates how dispersed the measurements are from the mean. -The variance does not change when the distribution remains the same. The distribution shows the range of measurements in a population. If the distribution changes, the variance also changes.

A biotechnology firm wants to produce a strain of giant Drosophila so that genetics students do not strain their eyes looking at tiny flies. A hormone called shorty substance P (SSP) normally inhibits growth in the flies. The enzyme runtase synthesizes SSP from the compound XSP in a single‑step catalytic reaction. A researcher isolated and sequenced runtase cDNA, but the researcher does not know the location of the runtase gene in the Drosophila genome. Initial attempts to delete, inactivate, or otherwise mutate this gene failed. Therefore, the researcher tries gene augmentation, or adding new genes to a cell. Which gene augmentation method could produce giant flies?

Answer: -insert a gene that expresses dsRNA for runtase, which would invoke the RNAi pathway and shut down runtase expression Explanation: -Increased runtase expression causes greater SSP production and diminished fly growth. Runtase enzymatically produces the growth-suppressing SSP hormone from the compound XSP. Thus, SSP itself cannot be genetically modified, because no gene encodes SSP.

Place the steps of generating a gene‑knockout mouse in order from the disruption of the target gene to generating homozygous knockouts. Suppose in one experiment, the stem cells used are from a black‑coated mouse, and the early mouse embryo used is a white‑coated embryo. Disrupt target gene by inserting neo+ and link tk+ to disrupted gene -> Breed mice to generate homozygous knockouts

Answer: 1) transfer disrupted gene into stem cells from a black mouse 2) allow recombination with genome to generate neo+ tk- cells 3) select for neo+ and select against tk+ to isolate cells from homologous recombination 4) inject neo+ cells into white early embryo and implant into surrogate 5) identify chimeric progeny with target-gene knockout cells Explanation: -Genetic manipulation in cell culture is required before a knockout mouse can be generated.