Genetics final

For E. coli strains with the lac genotypes shown below, use a plus sign (+) to indicate the synthesis of β-galactosidase and permease and a minus sign (−) to indicate no synthesis of the proteins.

Lactose absent Lactose present β-Galactosidase Permease β-Galactosidase Permease lacI+lacP+lacO+lacZ+lacY+ - - + + lacI-lacP+lacO+lacZ+lacY+ + + + + lacI+lacP+lacOclacZ+lacY+ + + + + lacI-lacP+lacO+lacZ+lacY- + - + - lacI-lacP-lacO+lacZ+lacY+ - - - -

What is difference between positive and negative control? What is the difference between inducible and repressible operons? Most operons are inducible or repressible?

Positive control - requires a regulatory molecule to stimulate RNA synthesis (default state of gene = off) Negative control - a regulatory molecule inhibits/represses RNA synthesis and must be removed (default state of gene = on) Inducible - operon normally not transcribed, requires an inducer negative system - inducer inactivates repressor positive system - inducer stimulates activator Repressible - operon normally transcribed negative system - repressor becomes active positive system - activator becomes inactive most operons are repressible

Describe some of the methods that have been used to study variation in DNA.

Three different types of DNA variation: (1) Restriction site variation - analyzed by restriction endonuclease digestion of DNA, and analysis of resulting DNA fragments by gel electrophoresis (2) Microsatellite length variation - analyzed by gel electrophoresis to separate different lengths (3) Variation in DNA sequences - sequencing of DNA from the locus in question

18.15 List at least three different types of DNA repair and briefly explain how each is carried out.

(1) Mismatch repair. Replication errors that are the result of base-pair mismatches are repaired. Mismatch-repair enzymes recognize distortions in the DNA structure due to mispairing and detect the newly synthesized strand by the lack of methylation on the new strand. The bulge is excised and DNA polymerase fills the gap and DNA ligase seals the repair. (2) Direct repair. DNA damage is repaired by directly changing the damaged nucleotide back to its original structure. As for example the repair done by a photolyase or methyltransferase. (3) Base-excision repair. After the damaged base is removed by glycosylases, the phosphodiester bond is excised by AP endonucleases and other enzymes remove the deoxyribose sugar; then the entire nucleotide is replaced by DNA polymerase and the nick is sealed by DNA ligase. (4) Nucleotide-excision repair. Repair enzymes recognize distortions of the DNA double-helix. Damaged regions are excised by enzymes (nucleases and helicases), which separate the strands of DNA and cut phosphodiester bonds on either side of the damaged region. The gap generated by the excision step is filled in by DNA polymerase and sealed by DNA ligase.

Briefly list some of the ways in which siRNAs and miRNAs regulate genes.

- Through cleavage of mRNA sequences through "slicer activity": binding of RISCs containing either siRNA or miRNA to complementary mRNA molecules stimulates cleavage of the mRNA followed by further degradation of the cleaved mRNA. - Through binding of complementary regions with the mRNA molecule by miRNAs to prevent translation: miRNAs, as part of RISC, bind to complementary mRNA preventing either translation initiation or elongation. - Through transcriptional silencing due to methylation of either histone proteins or DNA sequences: siRNA binds to complementary DNA sequences and stimulates methylation of histones. Methylated histones bind DNA more tightly preventing transcription factors from binding the DNA. miRNA molecules bind to complementary DNA sequences and stimulate DNA methylases to directly methylate the DNA sequences, which results in transcriptional silencing. - Through slicer-independent mRNA degradation stimulated by miRNA binding to complementary regions in the 3' UTR of the mRNA: miRNA binds to the AU rich element in the 3' UTR of the mRNA stimulating degradation using RISC and dicer.

What are some of the advantages of using molecular data in evolutionary studies?

- changes in DNA and protein sequences are inherited - they can be studied and compared in all organisms - genomes provide a vast and growing amount of data in public databases - molecular differences are quantifiable - comparisons of DNA and protein sequences provide information about evolutionary history and may be used to infer phylogenetic relationships

Why does the response to selection often level off after many generations of selection?

After many generations, the response to selection plateaus because of two factors. First, the genetic variation may be depleted—all the individuals in the population now have the alleles that maximize the quantitative trait; with no genetic variation, there can be no selection or response to selection. Second, even if genetic variation persists, artificial selection may be limited by an opposing natural selection.

The blob operon produces enzymes that convert compound A into compound B. The operon is controlled by a regulatory gene, S. Normally, the enzymes are synthesized only in the absence of compound B. If gene S is mutated, the enzymes are synthesized in the presence and in the absence of compound B. Does gene S produce a repressor or an activator? Is this operon inducible or repressible?

Because blob operon is transcriptionally inactive in the presence of B, gene S most likely codes for a repressor protein that requires compound B as a corepressor. The data suggest operon is repressible: - inactive in the presence of compound B - active when compound B is absent

List/define as many different levels and/or mechanisms that cells use to regulate gene expression.

Chromatin remodeling - Transcriptional control - Regulating transport of RNA from nucleus to cytoplasm - mRNA maturation, including alternative splicing and polyA addition - mRNA stability - Translational control - Protein stability - Protein modification/activation -

Briefly describe the differences between directional selection, overdominance, and underdominance. Describe the effect of each type of selection on the allelic frequencies of a population.

Directional selection = when individuals with a particular allele have greater fitness than other individuals without that allele. Þallele conferring greater fitness to increase in frequency and eventually reach fixation Overdominance = when the heterozygote fitness > either of the homozygotes. Þestablishes a balanced equilibrium that maintains both alleles Underdominance = when the heterozygote fitness < either of the homozygotes. => an unstable equilibrium that will degenerate once disturbed and move away from equilibrium until one allele is fixed

Seed size in a plant is a polygenic characteristic. A grower crosses two pure- breeding varieties of the 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. Explain why the backcross progeny are more variable.

F1 progeny all have the same genetic makeup: all heterozygotes for the loci that differ between the two pure-breeding strains backcross progeny will have much greater genetic diversity as a result of the genetic diversity from meiosis of the F1 heterozygotes

For the lac genotypes shown in the accompanying table, predict whether the structural gene (lacZ) is constitutive expressed, permanently repressed, or inducible in the presence of lactose

Genotype Constitutive Repressed Inducible I+O+Z+ - - + I-O+Z+ + - - I-OCZ+ + - - I-OCZ+/F'O+ + - - (I-O+Z+/F'I+) - - + I+OCZ+/F'O+ + - - ISO+Z+ - + - ISO+Z+/F'I+ - + -

Define inbreeding and briefly describe its effects on a population.

Inbreeding is preferential mating between genetically related individuals. Inbreeding increases homozygosity and reduces heterozygosity in the population.

Predict the level of genetic activity of the lac operon as well as the status of the lac repressor and CAP protein under the cellular conditions listed in the accompanying table.

Lactose Glucose operon repressor CAP (a) - - off, bound to O, bound to DNA (b) + - on (strong), inactivated, bound to DNA (c) - + off, bound to O, not bound (d) + + on (weak), ≅ off inactivated, not bound

What assumptions must be met for a population to be in Hardy-Weinberg equilibrium?

Large population, random mating, and not affected by migration, selection, or mutation

Why do polygenic characteristics have many phenotypes?

Many genotypes are possible with multiple genes. Even for the simplest two-allele loci, the number of possible genotypes is equal to 3n, where n is the number of loci or genes. 3 genes => 27 genotypes; 4 genes => 81 genotypes; etc. If each genotype corresponds to a unique phenotype, then = numbers of phenotypes. Finally, the phenotype for a given genotype may be influenced by environmental factors, leading to an even greater array of phenotypes.

Compare and contrast the effects of mutation, migration, genetic drift, and natural selection on genetic variation within populations and on genetic divergence between populations.

Mutation increases genetic variation w/in populations and increases divergence between populations because different mutations arise in each population. Migration increases genetic variation w/in a population by introducing new alleles from other populations, but decreases divergence between populations. Genetic drift decreases genetic variation w/in populations because it causes alleles eventually to become lost or fixed, but it increases divergence between populations because drift takes place differently in each population. Natural selection may either decrease or maintain genetic variation w/in populations depending on whether the selection is directional or balanced (overdominance). It may increase or decrease divergence between populations depending on whether the different populations have similar or different selection pressures.

Define natural selection and fitness.

Natural selection is the differential reproductive success of genotypes. Fitness is the relative reproductive success of a genotype, defined as the mean number of offspring produced by that genotype, divided by the mean number of offspring produced by the most prolific genotype.

What is regression? How is it used?

Regression is a mathematical relationship between correlated variables. Regression is used to predict the value of a variable from the value of a correlated variable.

Distinguish between the cis-acting regulatory elements referred to as promoters and enhancers.

Promoters: - absolutely required - DNA sequences that occupy a specific location relative to the gene it regulates (immediately 5' or upstream of the transcriptional start site) - specify which strand to transcribe and where to start transcription - where polymerase binds and initiates transcription Enhancers: - not required but can upregulate the activity of polymerase at promoters - are position and orientation independent - are not necessarily gene specific

Present an overview of the manner in which chromatin can be remodeled. In what ways do each of these remodeling processes influence transcription?

Remodeling of the chromatin, i.e. changes in the organization/architecture of chromosomes and the proteins that bind them, can be effected by: - modification of the DNA (methylation), or - modification of the histone components of nucleosomes (substitution of histone proteins, acetylation, or repositioning of whole or parts of the nucleosome) Any or all of these processes can lead to increases or decreases of transcription rates Increases: acetylation or repositioning of histones Decreases: deacetylation of histones, methylation of DNA

. How are genetic drift and inbreeding similar in their effects on a population? How are they different?

Similar: - for different reasons, both will lead to populations with higher levels of homozygosity (decreased genetic diversity) - both mechanisms are likely reasons for changes in allele frequencies when populations are small Different: - inbreeding can take place even within large populations and have an effect on allele frequencies, but drift only plays a role in small populations - inbreeding can increase the frequency of both homozygous states simultaneously (eliminating the heterozygous state), while drift will tend to fix one or the other of a particular allele producing a single homozygous state

Among a population of tadpoles, the correlation coefficient for size at metamorphosis and time required for metamorphosis is -0.74. On the basis of this correlation, what conclusions can you make about the relative sizes of tadpoles that metamorphose quickly and those that metamorphose more slowly?

Size at metamorphosis and time required for metamorphosis are inversely correlated (negative correlation coefficient). The greater the time required for metamorphosis, the smaller the size, and vice versa. Therefore, tadpoles that metamorphose quickly are larger than tadpoles that metamorphose slowly.

How do broad-sense and narrow-sense heritabilities differ?

The broad-sense heritability is the portion of phenotypic variance that is due to all types of genetic variance, including additive, dominance, and genic interaction variances. The narrow-sense heritability is only that portion of the phenotypic variance that is due to additive genetic variance.

Describe a tautomeric shift and how it may lead to a mutation.

The four bases used in DNA can exist in two stable states, and it is possible for each base to shift between its two respective states by shifting a proton between two atoms within that base. Normally thymine and guanine exist in their keto forms and cytosine and adenine exist in their amine forms. When they do, they base pair properly with their respective partners (A=T and G C). If they undergo tautomeric shifts into the rarer enol form (for thymine and guanine) or imine form (for cytosine and adenine) then that base will improperly basepair with the other purine or pyrimidine (A=C or G T). If this shift occurs during replication, then the wrong base will be inserted into in the new strand of DNA.

Briefly describe the lac operon and how it controls the metabolism of lactose.

The lac operon consists of three structural genes involved in lactose metabolism. The lacZ gene codes for the enzyme β- galactosidase, which breaks the disaccharide lactose into galactose and glucose and converts lactose into allolactose. The lacY gene, located downstream of the lacZ gene, codes for lactose permease, necessary for the passage of lactose through the E. coli cell membrane. The lacA gene, located downstream of lacY, encodes the enzyme thiogalactoside transacetylase whose function has not yet been determined. All of these genes share a common overlapping promoter and operator region. Upstream from the lactose operon is the lacI gene that encodes the lac operon repressor. The repressor binds at the operator region and inhibits transcription of the lac operon by preventing RNA polymerase from successfully initiating transcription. When lactose is present in the cell, β-galactosidase converts some of it into allolactose. Allolactose binds to the lac repressor, altering its shape and reducing the repressor's affinity for the operator. Since this allolactose-bound repressor does not occupy the operator, RNA polymerase can initiate transcription of the lac structural genes from the lac promoter.

What information does the correlation coefficient provide about the association between two variables?

The magnitude or absolute value of the correlation coefficient reports how strongly the two variables are associated. A value close to +1 or -1 indicates a strong association (direct or inverse, respectively); values close to zero indicate weak association. It should be noted that correlation does not mean causation.

What information do the mean and variance provide about a distribution?

The mean is the center of the distribution. The variance is how broad the distribution is around the mean.

What factors affect the magnitude of change in allelic frequencies due to migration?

The proportion of the population due to migrants (m) and the difference in allelic frequencies between the migrant population and the original resident population.

Jean Manning, Charles Kerfoot, and Edward Berger studied genotypic frequencies at the phosphoglucose isomerase (GPI) locus in the cladoceran Bosmina longirostris (a small crustacean known as a water flea). At one location, they collected 176 of the animals from Union Bay in Seattle, Washington, and determined their GPI genotypes by using electrophoresis (J. Manning, W. C. Kerfoot, and E. M. Berger. 1978. Evolution 32:365-374). Genotype Number S1S1 4 S1S2 38 S2S2 134 Determine the genotypic and allelic frequencies for this population.

Genotypic frequencies: f(S1S1) = 4/176 = 0.023f(S1S2) = 38/176 = 0.216f(S2S2) = 134/176 = 0.761 Allelic frequencies: f(S1) = f(S1S1) + f(S1S2)/2 = 0.685 + 0.143 = (4 + 38*0.5)/176 = 0.131 f(S2) = f(S2S2) + f(S1S2)/2 = 0.029 + 0.143 = (134 + 38*0.5)/176 =0.87

Voles (Microtus ochrogaster) were trapped in fields in southern Indiana and genotyped for a transferrin locus. The following numbers of genotypes were recorded, where TE and TF represent different alleles. TETE TETF TFTF 407 170 17 Calculate the genotypic and allelic frequencies of the transferrin locus for this population.

Genotypic frequencies: f(TETE) = 407/594 = 0.685f(TETF) = 170/594 = 0.286f(TFTF) = 17/594 = 0.029 Allelic frequencies: f(TE) = f(TETE) + f(TETF)/2 = 0.685 + 0.143 = 0.828 f(TF) = f(TFTF) + f(TETF)/2 = 0.029 + 0.143 = 0.172 or f(TE) = (407 +170*0.5)/594 = 0.828 f(TF) = (17 +170*0.5)/594 = 0.172

In a large, randomly mating population, the frequency of an autosomal recessive lethal allele is 0.20. What will the frequency of this allele be in the next generation if the lethality takes place before reproduction?

Given: q =0.2 => p = 0.8 After one round of random mating, genotypic frequencies = f(AA) = p2 =0.64; f(Aa) = 2pq = 0.32; f(aa) = q2 = 0.04, but die Adjusting for the death of all aa, new frequencies of the survivors (f') = f'(AA) = f(AA)/[f(AA) + f(Aa)] = 0.64/(0.64 + 0.32) = 0.67 f'(Aa) = f(Aa)/[f(AA) + f(Aa)] = 0.32/(0.64 + 0.32) = 0.33*frequencies of AA and Aa genotypes and dividing each by the total survivors new allelic frequency q' = f'(aa) + f'(Aa)/2 = 0 + 1/6 = 0.17 *note that because the genotype aa is lethal, f'(aa) = 0

What are two major mechanisms for repair of double-strand breaks? How do they differ?

Homologous recombination and nonhomologous end joining. Homologous recombination requires an identical DNA molecule (usually a sister chromatid) to repair a double-strand break. Nonhomologous end joining does not require a template.

What is catabolite repression? How does it allow a bacterial cell to use glucose in preference to other sugars?

In catabolite repression, the presence of glucose inhibits or represses the transcription of genes involved in the metabolism of other sugars. Because the gene expression necessary for utilizing other sugars is turned off, only enzymes involved in the metabolism of glucose will be synthesized. Operons that exhibit catabolite repression are under the positive control of catabolic activator protein (CAP). For CAP to be active, it must form a complex with cAMP. Glucose affects the level of cAMP. The levels of glucose and cAMP are inversely proportional—as glucose levels increase, the level of cAMP decreases. Thus, CAP is not activated.

Assume that plant weight is determined by a pair of alleles at each of two independently assorting loci (A and a, B and b) that are additive in their effects. Further assume that each allele represented by an uppercase letter contributes 4 g to weight and that each allele represented by a lowercase letter contributes 1 g to weight. (a) If a plant with genotype AA BB is crossed with a plant with genotype aa bb, what weights are expected in the F1 progeny? (b) What is the distribution of weight expected in the F2 progeny?

a) all F1 progeny = Aa Bb, so all have 4 + 1 + 4 + 1 = 10g b) dihybrid cross gives 1/16, 4/16, 6/16, 4/16, 1/16 0 additive alleles: 1/16 = 4g 1 additive allele: 4/16 = 7g 2 additive alleles: 6/16 = 10g 3 additive alleles: 4/16 = 13g 4 additive alleles:1/16= 16g

What is the biological species concept?

defines species as a group of individuals that can potentially interbreed with each other, but are reproductively isolated from members of other species

The mmm operon, which has sequences A, B, C, and D (which may be structural genes or regulatory sequences), encodes enzymes 1 and 2. Mutations in sequences A, B, C, and D have the following effects, where a plus sign (+) indicates that the enzyme is synthesized and a minus sign (−) indicates that the enzyme is not synthesized. a.Is the mmm operon inducible or repressible? Indicate which sequence (A, B, C, or D) is part of the following components of the operon: Regulator gene Promoter Structural gene for enzyme 1 Structural gene for enzyme 2

mmm absent. mmm present Enzyme 1 Enzyme 2 Enzyme 1 Enzyme 2 No mutation + + - - A - + - - B + + + + C + - - - D - - - - First line shows that enzymes made when mmm is absent but inactive when mmm is present -> repressible B D A C

In German cockroaches, curved wing (cv) is recessive to normal wing (cv+). Bill, who is raising cockroaches in his dorm room, finds that the frequency of the gene for curved wings in his cockroach population is 0.6. In his friend Joe's apartment, the frequency of the gene for curved wings is 0.2. One day Joe visits Bill in his dorm room, and several cockroaches jump out of Joe's hair and join the population in Bill's room. Bill estimates that, now, 10% of the cockroaches in his dorm room are individual roaches that jumped out of Joe's hair. What is the new frequency of curved wings among cockroaches in Bill's room?

proportion of migrants in the new population = m = 0.1. f(cv)old population = qold = 0.6; f(cv)migrants = qmigrants = 0.2 qnew = mqmigrants + (1 - m)qold = 0.1(0.2) + 0.9(0.6) = 0.56 => now calculate the new genotype frequency assuming random mating: f (cv,cv) new = qnew2 = (0.56)2 = 0.31

Two alleles of a locus, A and a, can be interconverted by mutation: A -> a at a rate equal to μ, and a -> A at a rate equal to v. μ is 6.0 x 10-7, and v is 6.0 x 10-8. What will be the frequencies of A and a at mutational equilibrium, assuming no selective difference, no migration, and no random fluctuation caused by genetic drift?

q = μ/(μ + v) = 6 x 10-7/(6 x 10-7 + 0.6 x 10-7) = 6/6.6 = 0.91 p = v/(μ + v) = 0.6 x 10-7/(6 x 10-7 + 0.6 x 10-7) = 0.66/6.6 = 0.09 or = 1 - q = 0.09

Tay-Sachs disease is an autosomal recessive disorder. Among Ashkenazi Jews, the frequency of Tay-Sachs disease is 1 in 3600. If the Ashkenazi population is mating randomly for the Tay-Sachs gene, what proportion of the population consists of heterozygous carriers of the Tay-Sachs allele?

q2 = 1/3600 = 0.000277 => q = 0.017 q = 0.017 => p = 1 - q = 0.983 2pq = 2*0.983*0.017 = 0.033

The heights of mothers and daughters are given in the following table: (a)Calculate the correlation coefficient for the heights of the mothers and daughters. (b) Using regression, predict the expected height of a daughter whose mother is 67 inches tall.

r = 3.99/[(2.2)(2.1)] = 0.88 r = covxy/[Sx*Sy] y = a + bx, where b = covxy /sx2 and a = y - bx b = 3.99/4.9 = 0.81 a = 64.3 - 0.81(62.7) = 13.5 If the mother is 67 inches tall, the regression equation y = a + bx becomes y = 13.5 + 0.81(67) = 67.8 inches

How is the response to selection related to the narrow-sense heritability and the selection differential? What information does the response to selection provide?

response to selection (R) = narrow-sense heritability (h ) × selection differential (S). value of R predicts how much the mean quantitative phenotype will change with different selection in a single generation

Ten male Harvard students were weighed in 1916. Their weights are given here in kilograms. Calculate the mean, variance, and standard deviation for these weights. 51, 69, 69, 57, 61, 57, 75, 105, 69, 63

variance = s^2 = sum(xi - x)^2/n-1

Explain why mutations in the lacI gene are trans in their effects, but mutations in the lacO gene are cis in their effects.

lacI -> repressor protein which can diffuse throughout the cell and attach to any and all operators, thereby regulating transcription on same and different DNA molecules, i.e. it can act at a distance lacO is the regulatory region, called the operator, to which the Lac Repressor binds and affects RNA polymerase activity at that site and only at that site, i.e. it acts only on the same DNA molecule on which it resides

In a population of tomato plants, mean fruit weight is 60g and (h2) is 0.3. Predict the mean weight of the progeny if tomato plants whose fruit averaged 80g were selected from the original population and interbred.

mean fruit weightpopulation = 60g; mean fruit weightparents = 80g h2 = 0.3 S = 80 - 60 = 20g R = h2 * S = 0.3*20 = 6 => mean fruit weightprogeny = 60g + 6g = 66g

Briefly explain how genes affecting a polygenic characteristic are located with the use of QTL mapping.

Two homozygous, highly inbred strains that differ at many loci are crossed and the F1 are interbred. Quantitative traits are measured and correlated with the inheritance of molecular markers throughout the genome. The correlations are used to infer the presence of a linked QTL.

Contrast the various types of DNA repair mechanisms Known to counteract the effects of UV radiation. What is the role of visible light in repairing UV-induced mutations?

UV-induced damaged can be repaired in one of two ways. Photoreactivation enzyme, or photolyase, can absorb a photon of blue light and use that energy to cleave any pyrimidine dimers. The damage can also be repaired by nucleotide excision repair. NER involves cutting away a 12-13 nucleotide long stretch of the DNA strand that spans the lesion and resynthesizing a new strand of DNA in the existing gap.

What determines the allelic frequencies at mutational equilibrium?

At mutational equilibrium, the allelic frequencies are determined by the forward and reverse mutation rates.

What is the basic difference between allopatric and sympatric modes of speciation?

Allopatric speciation involves populations separated by a geographic barrier that precludes gene flow between the populations. Sympatric speciation takes place between populations occupying the same geographical area.

How is anagenesis different from cladogenesis?

Anagenesis is change in a single group of organisms over time. Cladogenesis involves splitting of a group into two or more groups that become different from each other.

What is attenuation? What is the mechanism by which the attenuator forms when tryptophan levels are high and the antiterminator forms when tryptophan levels are low?

Attenuation - termination of transcription prior to structural genes of an operon. It is a result of the formation of a termination hairpin structure (attenuator) in the mRNA. Two types of secondary structures can be formed by the mRNA 5′ UTR of the trp operon: 1) two hairpin structures from the base pairing of region 1 with region 2 and the pairing of region 3 with region 4 -> terminator forms stopping transcription of the structural genes, 2) hairpin structure formed by the pairing of region 3 with region 4 -> antiterminator that allows transcription of structural genes. Region 1 of the 5' UTR also encodes a small protein and has two adjacent tryptophan codons (UGG). Tryptophan levels affect transcription due to the coupling of translation with transcription in bacterial cells. When tryptophan levels are high, the ribosome quickly moves through region 1 and into region 2, thus preventing region 2 from pairing with region 3. Therefore, region 3 is available to form the attenuator hairpin structure with region 4, stopping transcription. When tryptophan levels are low, the ribosome stalls or stutters at the adjacent tryptophan codons in region 1. Region 2 now becomes available to base pair with region 3, forming the antiterminator hairpin. Transcription can now proceed through the structural genes.

How do base analogs lead to mutations?

Base analogs have structures similar to the nucleotides and if present, may be incorporated into the DNA during replication. Many analogs have an increased tendency for mispairing, which can lead to mutations. DNA replication is required for the base analog-induced mutations to be incorporated into the DNA.

Is the binding of a transcription factor to its DNA recognition sequence necessary and sufficient for an initiation of transcription at a regulated gene? What else plays a role in this process?

Binding is necessary, but not sufficient Transcription factor activity is also often regulated beyond just binding to DNA recognition sequences RNA polymerase as well as a host of other transcription factors all play a role in initiation of transcription--the roles may involve chromatin remodeling, formation of the initiator complex, etc.

On the basis of the phylogeny of Darwin's finches shown in Figure 26.8, predict which two species in each of the following groups will be the most similar genetically. a.Camarhynchusparvulus, Camarhynchuspsittacula b. Camarhynchus parvulus, Camarhynchus pallida, Platyspiza crassirostris c. Geospiza difficilis, Geospiza conirostris, Geospiza scandens d. Camarhynchus parvulus, Certhidea fusca, Pinaroloxias inornata

Camarhynchus parvulus and Camarhynchus psittacula Camarhynchus parvulus and Camarhynchus pallid Geospiza conirostris and Geospiza scandens Camarhynchus parvulus and Certhidea fusca

A mutation prevents the catabolite activator protein (CAP) from binding to the promoter in the lac operon. What will the effect of this mutation be on transcription of the operon?

Catabolite activator protein binds CAP site of the lac operon and stimulates RNA polymerase to bind the lac promoter, thus resulting in increased levels of transcription from the lac operon. If a mutation prevents CAP from binding to the site, then RNA polymerase will bind the lac promoter poorly -> significantly lower levels of transcription of the lac structural genes

What changes take place in chromatin structure and what role do these changes play in eukaryotic gene regulation?

Changes in chromatin structure can result in repression or stimulation of gene expression. As genes become more transcriptionally active, the chromatin structure is more open. Acetylation of histone proteins by acteyltransferase proteins results in the destabilization of the nucleosome structure and increases transcription. The reverse reaction by deacetylases stabilizes nucleosome structure. Other transcription factors and regulatory proteins, called chromatin remodeling complexes, bind directly to the DNA-altering chromatin structure without acetylating histone proteins. The chromatin remodeling complexes allow for transcription to be initiated by increasing accessibility to the promoters by transcription factors. DNA methylation is also associated with decreased transcription.

What is an enhancer? How does it affect transcription of distant genes?

Enhancers are DNA sequences that are the binding sites of transcriptional activator proteins. Transcription at a distant gene is affected when the DNA sequence located between the gene's promoter and the enhancer is looped out, allowing for the interaction of the enhancer-bound proteins with proteins needed at the promoter, which in turn stimulates transcription.

A total of 6129 North American Caucasians were blood typed for the MN locus, which is determined by two codominant alleles, LM and LN. The following data were obtained: Blood type Number M 1787 MN 3039 N 1303 Carry out a chi-square test to determine whether this population is in Hardy- Weinberg equilibrium at the MN locus.

First, HO = population is not statistically different from a population in HW equilibrium Second, calculate allele frequencies: f(LM) = (1787 + 3039/2)/6129 = 0.54 f(LN) = (1303 + 3039/2)/6129 = 0.46 Third, given these allele frequencies, calculate expected genotypic frequencies: f(LMLM) = p2 = 0.29;f(LMLN) = 2pq = 0.5;f(LNLN) = q2 = 0.21 Fourth, calculate expected # given expected frequencies f(LMLM) = p2 = 0.29 = 0.29*6129 = 1777 f(LMLN) = 2pq = 0.5 = 0.5*6129 = 3065 f(LNLN) = q2 = 0.21 = 0.21*6129 = 1287 Σ = 0.48 df = 3 - 2 = 1 (#genotypes - #alleles) According to chi-squared table, p > 0.05 so fail to reject HO => population is in HW equilibrium for this locus Extra practice, try problem 22

Briefly outline the process of allopatric speciation.

First, a population is split by a geographical barrier that prevents gene flow between the two groups on either side of the barrier. The two groups then evolve independently; they accumulate genetic differences through various evolutionary processes such as mutation, selection, and random drift. If these genetic differences lead to reproductive isolation, meaning these individuals cannot or will not interbreed, then speciation has occurred.

What are the two steps in the process of evolution?

First, genetic variation is created via mutations and recombination. Mutation causes genetic variation in a population and recombination creates new combinations of genetic variants. Second, the frequencies of the variants change over time (from generation to generation), as a result of random processes (e.g., drift) or selection.

For the genotypes and conditions (lactose present or absent) shown in the accompanying table, predict whether functional enzymes, nonfunctional enzymes, or no enzymes are made.

Functional Nonfunctional No enzyme made I+O+Z+ No lactose - - + I+OCZ+ Lactose + - - I-O+Z- No lactose - + - I-O+Z- Lactose - + - I-O+Z+/F'I+ No lactose - - + I+OCZ+/F'O+ Lactose + - - I+O+Z-/F'I+O+Z+ Lactose + + - I-O+Z-/F'I+O+Z+ No lactose - - + ISO+Z+/F'O+ No lactose - - + I+OCZ+/F'O+Z+ Lactose + - -

Why is gene regulation important for bacterial cells?

Gene regulation allows for biochemical and internal flexibility while maintaining energy efficiency by the bacterial cells.

Define genetic drift and give three ways in which it can arise. What effect does genetic drift have on a population?

Genetic drift is change in allelic frequencies resulting from sampling error. It may arise through 1) a long-term limitation on population size, 2) founder effect that occurs when the population is founded by a small number of individuals, or 3) a bottleneck effect when the population undergoes a drastic reduction in population size. Genetic drift causes changes in allelic frequencies and loss of genetic variation because some alleles are lost as other alleles become fixed. It also causes genetic divergence between populations because the different populations undergo different changes in allelic frequencies and become fixed for different alleles.

A genetics researcher determines that the broad-sense heritability of height among Southwestern University undergraduate students is 0.90. Which of the following conclusions would be reasonable? Explain your answer. (a) Since Sally is a Southwestern University undergraduate student, 10% of her height is determined by nongenetic factors. (b) Ninety percent of variation in height among all undergraduate students in the United States is due to genetic differences. (c) Ninety percent of the height of Southwestern University undergraduate students is determined by genes. (d) Ten percent of the variation in height of Southwestern University undergraduate students is determined by variation in nongenetic factors. (e) Because the heritability of height among Southwestern University students is so high, any change in the students' environment will have minimal impact on their height.

H2 = VG/VP - applies only to a particular population Not (a) - H2 does not apply to an individual Not (b) - only applies to Southwestern University undergraduates Not (c) - only applies to variance in height not absolute height, i.e. 90% of the variance in height rather than 90% of height Not (e) - only applies to the set of environmental conditions that existed while under the study; if environment changes, then H2 could change only (d) isreasonable and accurate

What are the predictions given by the Hardy-Weinberg law?

HW law states that a large population mating randomly with no effects from selection, migration, or mutation will have the following relationship between the genotype frequencies and allele frequencies: f(AA) = p2; f(Aa) = 2pq; f(aa) = q2, where p and q equal the allelic frequencies Moreover, the allele frequencies do not change from generation to generation, as long as the above conditions hold.

Briefly outline some of the ways that heritability can be calculated.

Heritability can be calculated via elimination of variance components from the equation: VP = VG + VE + VGE. Value of VGE can be assumed to be 0 or to be a part of VG. By either eliminating genetic variance (VG = 0) with genetically identical individuals, or by eliminating VE with individuals raised in identical environments, we can determine the values of VE or VG, respectively. If VP can be determined under conditions of genotypic variance, then the missing term VG or VE can be calculated by simple subtraction. Parent-offspring regression: mean phenotypic values of the parents are plotted against the mean phenotypic values of the offspring for a series of families h2 = regression coefficient Comparison of phenotypes for different degrees of relatedness: compare phenotypes of monozygotic and dizygotic twins. Twice the difference in correlation coefficients of monozygotic and dizygotic twins ≅ H2 Response to selection: response to selection = product of the narrow-sense heritability and the selection differential

What is a Mendelian population? How is the gene pool of a Mendelian population usually described?

Mendelian population = group of sexually reproducing individuals mating with each other and sharing a common gene pool. gene pool = usually described by genotype frequencies and allele frequencies.

What is the difference between prezygotic and postzygotic reproductive isolating mechanisms? List the different types of each.

Prezygotic mechanisms operate before fertilization of the egg by sperm (or fusion of gametes) and include ecological isolation, behavioral isolation, temporal isolation, mechanical isolation, and gametic isolation. Postzygotic mechanisms operate after fertilization and include hybrid inviability, hybrid sterility, and hybrid breakdown.

Briefly explain why the relation between genotype and phenotype is frequently complex for quantitative characteristics.

Quantitative characteristics are polygenic, so many genotypes are possible. Moreover, most quantitative characteristics are also influenced by environmental factors. Therefore, the phenotype is determined by complex interactions of many possible genotypes and environmental factors.

A rancher determines that the average amount of wool produced by a sheep in her flock is 22 kg per year. In an attempt to increase the wool production of her flock, the rancher picks five male and five female sheep with the greatest wool production; the average amount of wool produced per sheep by those selected is 30 kg. She interbreeds these selected sheep and finds that the average wool production among the progeny of the selected sheep is 28 kg. What is the narrow-sense heritability for wool production among the sheep in the rancher's flock?

R = h2 x S R = 28kg - 22kg = 6kg and S = 30kg - 22kg = 8kg so h2 = R/S=6kg/8kg=0.75

The following nucleotide sequence is found on the template strand of DNA. First, determine the amino acids of the protein encoded by this sequence by using the genetic code provided in Figure 15.10. Then, give the altered amino acid sequence of the protein that will be found in each of the following mutations:

Sequence of DNA template: 3'-TAC TGG CCG TTA GTT GAT ATA ACT-5' Nucleotide number -> 1 24 Original Sequence: 3'-TAC TGG CCG TTA GTT GAT ATA ACT-5' mRNA sequence: 5'-AUG ACC GGC AAU CAA CUA UAU UGA-3' Amino acid sequence:Amino-Met ThrGlyAsnGlnLeu Tyr Stop-Carboxyl Mutant 1: A transition at nucleotide 11 The transition results in the substitution of Ser for Asn.Original sequence: 3'-TAC TGG CCG TTA GTT GAT ATA ACT-5' Mutated sequence: 3'-TAC TGG CCG TCA GTT GAT ATA ACT-5' mRNA sequence: 5'-AUG ACC GGC AGU CAA CUA UAU UGA-3' Amino acids: Amino-Met Thr Gly Ser Gln Leu Tyr-Carboxyl Mutant 2: A transition at nucleotide 13 The transition results in the formation of a UAA nonsense codon. Original sequence: 3'-TAC TGG CCG TTA GTT GAT ATA ACT-5' Mutated sequence: 3'-TAC TGG CCG TTA ATT GAT ATA ACT-5' mRNA sequence: 5'-AUG ACC GGC AAU UAA CUA UAU UGA-3' Amino acid sequence: Amino-Met Thr Gly Asn-Carboxyl Mutant 3: A one-nucleotide deletion at nucleotide 7 The one-nucleotide deletion results in a frameshift mutation.Original sequence: 3'-TAC TGG CCG TTA GTT GAT ATA ACT-5' Mutated sequence: 3'-TAC TGG CGT TAG TTG ATA TAA CT-5' mRNA sequence: 5'-AUG ACC GCA AUC AAC UAU AUU GA-3' Amino acids: Amino-Met Thr Ala Ile Asn Tyr Ile -Carboxyl Mutant 4: A T -> A transversion at nucleotide 15 The transversion results in the substitution of His for Gln in the protein. Original sequence: 3'-TAC TGG CCG TTA GTT GAT ATA ACT-5' Mutated sequence: 3'-TAC TGG CCG TTA GTA GAT ATA ACT-5' mRNA sequence: 5'-AUG ACC GGC AAU CAU CUA UAU UGA-3' Amino acids: Amino-Met Thr Gly Asn His Leu Tyr-CarboxylorMutated sequence: 3'-TAC TGG CCG TTA GTG GAT ATA ACT-5' mRNA sequence: 5'-AUG ACC GGC AAU CAC CUA UAU UGA-3' Amino acids: Amino-Met Thr Gly Asn His Leu Tyr-Carboxyl Mutant 5: An addition of TGG after nucleotide 6 The addition of the three nucleotides results in the addition of Thr to the amino acid sequence of the protein, and is an in-frame insertion. Original sequence: 3'-TAC TGG CCG TTA GTT GAT ATA ACT-5' Mutated sequence: 3'-TAC TGG TGG CCG TTA GTT GAT ATA ACT-5' mRNA sequence: 5'-AUG ACC ACC GGC AAU CAA CUA UAU UGA-3' Amino acids: Amino-Met Thr Thr Gly Asn Gln Leu Tyr-Carboxyl Mutant 6: A transition at nucleotide 9 The protein retains the original amino acid sequence.Original sequence: 3'-TAC TGG CCG TTA GTT GAT ATA ACT-5' Mutated sequence: 3'-TAC TGG CCA TTA GTT GAT ATA ACT-5' mRNA Sequence: 5'-AUG ACC GGU AAU CAA CUA UAU UGA -3' Amino acids: Amino-Met Thr Gly Asn Gln Leu Tyr-Carboxyl

List some important differences between bacterial and eukaryotic cells that affect the way in which the genes are regulated.

Some examples (many others possible): - Bacterial genes are frequently organized into operons with coordinate regulation, and genes with operons can be transcribed as on a single long mRNA. Eukaryotic genes are not organized into operons and are singly transcribed from their own promoters. - In eukaryotic cells, nucleosome structure of the DNA is remodeled prior to transcription occurring. Essentially, the chromatin must assume a more open configuration state, allowing for access by transcription associated factors. - Activator and repressor molecules function in both eukaryotic and bacterial cells; however, in eukaryotic cells activators appear to be more common than in bacterial cells. - In bacteria, transcription and translation can occur concurrently. In eukaryotes, the nuclear membrane separates transcription from translation both physically and temporally. This separation results in a greater diversity of regulatory mechanisms that can occur at different points during gene expression.

In many eukaryotic organisms, a significant proportion of cytosine bases are naturally methylated to 5-methylcytosine. Through evolutionary time, the proportion of AT base pairs in the DNA of these organisms increases. Can you suggest a possible mechanism for this increase?

Spontaneous deamination of 5-methylcytosine produces thymine. If the subsequent repair of the GT mispairing is repaired incorrectly or, more likely, not repaired at all because the thymine is a normal base, then a GC to AT transition will result. Over time, the incorrect repairs will lead to an increase in the number of AT base pairs.

DNA damage brought on by a variety of natural and Artificial agents elicits a wide variety of cellular responses involving numerous signaling pathways. In addition to the activation of DNA repair mechanisms, there can be activation of pathways leading to apoptosis (programmed cell death) and cell-cycle arrest. Why would apoptosis and cell-cycle arrest often be part of a cellular response to DNA damage?

Stopping the cell cycle provides time for the cell to repair damage before 1) replication fixes the mutations into the DNA, or 2) mutated gene sequences are expressed that may affect growth or development of a cell Apoptosis is a cell suicide mechanism that would allow a heavily damaged cell with lots of DNA mutations to sacrifice itself for the good of the organism so that potentially lethal damage is not propagated to daughter cells

How do insertions and deletions arise?

Strand slippage that occurs during DNA replication and unequal crossover events due to misalignment at repetitive sequences have been shown to cause deletions and additions of nucleotides to DNA molecules. Strand slippage results from the formation of small loops on either the template or the newly synthesized strand. If the loop forms on the template strand, then a deletion occurs. Loops formed on the newly synthesized strand result in insertions. If, during crossing over, a misalignment of the two strands at repetitive sequence occurs, then the resolution of the crossover will result in one DNA molecule containing an insertion and the other molecule containing a deletion.

Briefly describe the process of mismatch repair in Escherichia coli. How does this repair system discriminate between the nascent DNA strand and the parental DNA strand?

This DNA is hemimethylated. Using this information, the mismatch repair enzymes cut away the strand of DNA that contains the improperly added base. The DNA that is cut away is all of the DNA between the GATC site and just to the other side of the mismatch. DNA polymerase then fills in the gap with the proper sequence.

What role does RNA stability play in gene regulation? What controls RNA stability in eukaryotic cells?

The total amount of protein synthesized is dependent on how much mRNA is available for translation. The amount of mRNA present is dependent on the rates of mRNA synthesis and degradation. Less-stable mRNAs will be degraded faster so there will be fewer copies available to serve as templates for translation. The presence of the 5′ cap, 3' poly(A) tail, the 5' UTR, 3' UTR, and the coding region in the mRNA molecule are features that can affect the stability of the mRNA molecule. Poly(A) binding proteins (PABP) bind at the 3' poly(A) tail. These proteins contribute to the stability of the tail and protect the 5' cap through direct interaction. Once a critical number of adenine nucleotides have been removed from the tail, the protection is lost and the 5' cap is removed. The removal of the 5' cap allows for 5' to 3' nucleases to degrade the mRNA. AU-rich sequence elements in the 3' UTR can also increase degradation of the mRNA.

Briefly explain how transcriptional activator proteins and repressors affect the level of transcription of eukaryotic genes.

Transcriptional activator proteins stimulate transcription by binding DNA at specific base sequences such as an enhancer or regulatory promoter and attracting or stabilizing the basal transcription factor apparatus. Repressor proteins bind to silencer sequences or promoter regulator sequences. These proteins may inhibit transcription by either blocking access to the enhancer sequence by the activator protein, preventing the activator from interacting with the basal transcription apparatus, or preventing the basal transcription factor from being assembled.

List all the components that contribute to the phenotypic variance and define each component.

VG - Component of variance due to variation in genotype VA - Component of variance due to additive genetic variance VD - Component of variance due to dominance genetic variance VI - Component of variance due to genic interaction variance VE - Component of variance due to environmental differences VGE - Component of variance due to interaction between genes and environment

Here is the double-stranded DNA sequence of a portion of a bacterial gene: 5' CACTATGCTTGCGTGGACGCATTAAC 3' 3' GTGATACGAACGCACCTGCGTAATTG 5' The bottom strand is the template strand. a.Assuming transcription starts with the first A (in the RNA) and continues to the end, what would be the sequence of the mRNA transcribed from this gene fragment? b. Assuming that translation begins with an initiator codon, what is the amino acid sequence encoded by this mRNA? c. A tautomerization to the enol form of T in the first T of the nontemplate strand during DNA replication. d. A tautomerization to the imino form of A in the second A of the template strand during DNA replication. e. Deamination of the third C of the nontemplate strand prior to DNA replication. f. Deletion of the third C in the nontemplate strand.

a. 5' ACUAUGCUUGCGUGGACGCAUUAAC 3' 3' GTGATACGAACGCACCTGCGTAATTG 5' b. Met Leu Ala Trp Thr His Stop c.Change in RNA sequence does not affect amino acid sequence d. Change in RNA sequence eliminates start codon, so no protein e. Replication must occur to change the nucleotide on the template strand, Change in RNA sequence leads to a single amino acid change Leu->Phe f. Replication must occur to change the nucleotide on the template strand, Deletion causes a frameshift so reading frame shifts and all amino acids are different

Assume that the difference between a 10 decimeter (dm) tall corn plant and 26 dm tall corn plant results from four pairs of equal and cumulative multiple alleles where each capital letter contributes 2 dm to the height of a plant. (a) In a cross between a 26-dm tall plant (AA BB CC DD) and a 10-dm tall plant (aa bb cc dd) what will be the genotypes and phenotypes of F1 individuals and what will be the genotypes (with regard to only the number of additive alleles present, not all possible genotypes) and phenotypes of F2 individuals? (Hint: Use the following figure to simplify your analysis of the F2 generation.) (b) Determine the upper and lower limits of height variation in the offspring of the following crosses: I. Aa BB cc dd x Aa bb Cc dd II. Aa Bb Cc Dd x Aa bb Cc Dd

a. AA BB CC DD X aa bb cc dd 26 dm 10 dm F1: Aa Bb Cc Dd 18 dm 1 (all 8 capital letter alleles): 1/256 26 dm 8 (some combination with 7 capital letter alleles): 8/256 24 dm 28 (some combination with 6 capital letter alleles): 28/256 22 dm 56 (some combination with 5 capital letter alleles): 56/256 20 dm 70 (some combination with 4 capital letter alleles): 70/256 18 dm 56 (some combination with 3 capital letter alleles): 56/256 16 dm 28 (some combination with 2 capital letter alleles): 28/256 14 dm 8 (some combination with 1 capital letter allele): 8/256 12 dm 1(no capital letter alleles):1/25610 dm b. I. Aa BB cc dd x Aa bb Cc dd Smallest number of additive alleles (capitals) possible: aa Bb cc dd (12 dm) Largest number of additive alleles (capitals) possible: AA Bb Cc dd (18 dm) So range of plants from cross will be 12 to 18 dm in height II. Aa Bb Cc Dd x Aa bb Cc Dd Smallest number of additive alleles (capitals) possible: aa bb cc dd (10 dm) Largest number of additive alleles (capitals) possible: AA Bb CC DD (24 dm) So range of plants from cross will be 10 to 24 dm in height

In humans the presence of chin and cheek dimples is dominant to the absence of dimples, and the ability to taste the compound PTC is dominant to the inability to taste the compound. Both traits are autosomal, and they are unlinked. The frequencies of alleles for dimples are D = 0.62 and d = 0.38. For tasting, the allele frequencies are T = 0.76 and t = 0.24. a. Determine the frequency of genotypes for each gene and the frequency of each phenotype. b. What are the expected frequencies of the four possible phenotype combinations: dimpled tasters, undimpled tasters, dimpled nontasters, and undimpled nontasters?

a. D = 0.62 = p and d = 0.38 = q So DD = p2 = 0.3844; Dd = 2pq = 0.4712; dd = q2 = 0.1444 T = 0.76 = p and t = 0.24 = q SoTT = p2 = 0.5776;Tt = 2pq = 0.3648;tt = q2 = 0.0576 b. dimpled tasters D_T_ = (0.3844+0.4712)(0.5776+0.3648) = 0.8063 undimpled tasters ddT_ = (0.1444)(0.5776+0.3648) = 0.1361 dimpled nontasters D_tt = (0.3844+0.4712)(0.0576) = 0.0493 undimpled nontasters ddtt = (0.1444)(0.0576) = 0.0083

In a large, randomly mating population, the frequency of the allele (s) for sickle-cell hemoglobin is 0.028. The results of studies have shown that people with the following genotypes at the beta-chain locus produce the average numbers of offspring given: Genotype Average number of offspring produced SS 5 Ss 6 ss 0 a. What will the frequency of the sickle-cell allele (s) be in the next generation? b. What will the frequency of the sickle-cell allele be at equilibrium?

a. Given: f(s) = q = 0.028 => f(S) = p = 1 - 0.028 = 0.972 f(SS) = p2 = 0.945 f(Ss) = 2pq = 0.054 f(ss) = q2 =0.001 WSS = 5/6 = 0.83 WSs = 6/6 = 1.0 Wss= 0/6 =0.0 sSS = 1 - 0.83 = 0.17 sSs = 1.0 - 1.0 = 0.0 sss= 1.0 - 0.0 = 1.0 W = p2WSS + 2pqWSs + q2Wss = 0.838 f(SS) = (p2WSS)/W = 0.936 f(Ss) = (2pqWSs)/W = 0.064 f(ss) = (q2Wss)/W= 0.0 qnew comes solely from heterozygotes = 0.064/2 = 0.032 b. given that this is a case of overdominance q = f(s) = sSS/(sSS + sss) = 0.17/(0.17 + 1.0) = 0.17/1.17 = 0.145

In a large herd of cattle, three different characters showing continuous distribution are measured, and the variances in the following table are calculated: characters shank neck fat variance length length content phenotypic 310.2 730.4 106.0 environmental 248.1 292.2 53.0 additive genetic 46.5 73.0 42.4 dominance genetic 15.6 365.2 10.6 (a) Calculate the broad- and narrow-sense heritabilities for each character. (b) In the population of animals studied, which character would respond best to selection? Why? (c) A project is undertaken to decrease mean fat content in the herd. The mean fat content is currently 10.5%. Animals with a mean of 6.5% fat content are interbred as parents of the next generation. What mean fat content can be expected in the descendants of these animals?

a. H2 = (46.5+15.6)/310.2 (73+365.2)/730.4 (42.4+10.6)/106 = 0.2 0.6 0.5 h2 = 46.5/310.2 73.0/730.4 42.4/106.0 =0.150.10.4 b. fat content has the highest narrow-sense heritability (highest percent of total variance is due to additive genetic effects), so it has the highest potential to be affected via selection c. mean fat contentpopulation = 10.5%; mean fat contentparents = 6.5% h2 = 0.4; S = 6.5 - 10.5 = -4 R = 0.4 x -4 = -1.6=> mean fat contentprogeny = 10.5 + -1.6 = 8.9%

The human MN blood type is determined by two codominant alleles, LM and LN. The frequency of LM in Eskimos on a small Arctic island is 0.80. a. If random mating takes place in this population, what are the expected frequencies of the M, MN, and N blood types on the island? b. If the inbreeding coefficient for this population is 0.05, what are the expected frequencies of the M, MN, and N blood types on the island?

a. If f(LM) = 0.8, then f(LN) = 0.2 f(LMLM) = 0.8*0.8 = 0.64 f(LMLN) = 2*0.8*0.2 = 0.32 f(LNLN) = 0.2*0.2 = 0.04 b. f(LMLM) = p2 + Fpq = (0.8)2 + (0.05)(0.8)(0.2) = 0.648 f(LMLN) = 2pq - 2Fpq = 2(0.8)(0.2) - 2(0.05)(0.8)(0.2) = 0.304 f(LNLN) = q2 + Fqp = (0.2)2 + (0.05)(0.8)(0.2) = 0.048

New allopolyploid plant species can arise by hybridization between two species. a. If hybridization occurs between a diploid plant species with 2n = 14 and a second diploid species with 2n = 22, the new allopolyploid would have how many chromosomes? b. Is it likely that sexual reproduction between the allopolyploid species and either of its diploid ancestors would yield fertile progeny? Why or why not? c. What type of isolation mechanism is most likely to prevent hybridization between the allopolyploid and the diploid species? d. What pattern of speciation is illustrated by the development of the allopolyploid species?

a. Imagine a diploid gamete from each parent (species) fusing to create 4n = 14 + 22 = 36 b. The allopolyploid species would produce gametes with 18 total chromosomes (7 from first species and 11 from second species) First species would produce haploid gametes (7 total chromosomes) or diploid gametes (14 total chromosomes) Second species would produce haploid gametes (11 total chromosomes) or diploid gametes (22 total chromosomes) -> any combination leads to an aneuploid offspring (likely infertile) c. postzygotic mechanism: hybrid sterility d. sympatric speciation

Joe is breeding cockroaches in his dorm room. He finds that the average wing length in his population of cockroaches is 4 cm. He chooses six cockroaches that have the largest wings; the average wing length among these selected cockroaches is 10 cm. Joe interbreeds these selected cockroaches. From earlier studies, he knows that the narrow-sense heritability for wing length in his population of cockroaches is 0.6. (a) Calculate the selection differential and expected response to selection for wing length in these cockroaches. (b) What should be the average wing length of the progeny of the selected cockroaches?

a. R = h2 x S S = 10cm - 4cm = 6cm and h2 = 0.6 soR = (0.6)(6cm) = 3.6cm b. average wing length of progeny = average wing length of whole population + response lprogeny = lpopulation + R = 4 cm + 3.6 cm = 7.6 cm

Color blindness in humans is an X-linked recessive trait. Approximately 10% of the men in a particular population are colorblind. a.If mating is random for the color-blind locus, what is the frequency of the color-blind allele in this population? b.What proportion of the women in this population is expected to be colorblind? c.What proportion of the women in the population are expected to be heterozygous carriers of the color-blind allele?

a. for males, f(color blind) = q = 0.1 b. for females, f(color blind-homozygotes) = q2 = 0.01 c. for females, f(carriers-heterozygotes) = 2pq = 2*0.1*0.9 = 0.18 => 18% of women

A characteristic has a narrow-sense heritability of 0.6. (a) If the dominance variance (VD) increases and all other variance components remain the same, what will happen to the narrow-sense heritability? Will it increase, decrease, or remain the same? Explain (b) What will happen to the broad-sense heritability? Explain. (c) If the environmental variance (VE) increases and all other variance components remain the same, what will happen to the narrow-sense heritability? Explain. (d) What will happen to the broad-sense heritability? Explain.

a. h2 (down arrow) h2 = VA/VP so, increasing VD will increase the total phenotypic variance, VP if VA remains unchanged, then VA/VP will become smaller b. H2 up arrow VG = VA + VD + VI and VP = VA + VD + VI + VE since numerator and denominator are increasing by same increment and H2 < 1 => value of fraction will up arrow c. h2 down arrow h2 = VA/VP so, increasing VE will increase the total phenotypic variance, VP if VA remains unchanged, then VA/VP will become smaller d. H2 down arrow VP = VG + VE so, as VE increases so does VP (denominator gets larger -> decreasing value of H2)

Phenotypic variation in tail length of mice has the following components: Additive genetic variance (VA) = 0.5Dominance genetic variance (VD) = 0.3 Genic interaction variance (VI) = 0.1 Environmental variance (VE) = 0.4 Genetic-environmental interaction variance (VGE) = 0.0 (a)What is the narrow-sense heritability of tail length? (b)What is the broad-sense heritability of tail length?

a. h2 = VA/VP = 0.5/1.3 = 0.38 b. H2 = VG/VP = (VA + VD + VI) /VP = 0.9/1.3 = 0.69

Height in humans depends on the additive action of genes. Assume that this trait is controlled by the four loci R, S, T, and U and that environmental effects are negligible. Instead of additive versus nonadditive alleles, assume additive and partially additive alleles exist. Additive alleles contribute two units, and partially additive alleles contribute one unit to height. (a) Can two individuals of moderate height produce offspring that are much taller or shorter than either parent? If so, how? (b) If an individual with the minimum height specified by these genes marries an individual of intermediate or moderate height, will any of their children be taller than the tall parent? Why or why not?

a. yes, if they each pass their additive alleles to the same offspring, then the offspring will have more total additive alleles than either parent (and be taller) or pass their nonadditive alleles so that the offspring has fewer total additive alleles (and be shorter) b. no, the parent of minimum height has no extra additive alleles to pass on, so the most an offspring could get is the same number as the parent of intermediate height possesses

Full color (D) in domestic cats is dominant over dilute color (d). Of 325 cats observed, 194 have full color and 131 have dilute color. a.If these cats are in Hardy-Weinberg equilibrium for the dilution locus, what is the frequency of the dilute allele? b.How many of the 194 cats with full color are likely to be heterozygous?

a.f(dilute color) = f(dd) = q2 = 131/325 = 0.403 q = 0.635 b. q = 0.635; 1-q = p = 0.365 f(Dd) = 2pq = 2*0.365*0.635 = 0.464 0.464(325) = 151 heterozygotes

What conclusion can you draw from Figure 24.18 about the proportion of phenotypic variation in shell breadth that is due to genetic differences? Explain your reasoning. *positive correlation scatter plot"

h2 = VA/VP h2 = regression coefficient = 0.7 so, proportion of the phenotypic variance in shell breadth that is due to additive genetic variance is 0.7.

The forward mutation rate for piebald spotting in guinea pigs is 8 × 10-5; the reverse mutation rate is 2 × 10-6. If no other evolutionary forces are assumed to be present, what is the expected frequency of the allele for piebald spotting in a population that is in mutational equilibrium? q = μ/(μ+v)

where μ = 8 × 10-5 and v = 2 × 10-6 the frequency at equilibrium is then 8 × 10-5/(8 × 10-5 + 2 × 10-6) = 8/8.2 = 0.98.

In the autoregulation of tubulin synthesis, two models for the mechanism were proposed: (1) the tubulin subunits bind to the mRNA, and (2) the subunits interact with the nascent tubulin polypeptides. To distinguish between these two models, Cleveland and colleagues induced mutations into the 13-base regulatory element of the b-tubulin gene. Some of the mutations resulted in amino acid substitution, while other did not. Which of the two models is supported by the following results? What experiments would you do to confirm your answer?

wild-type met arg glu lys autoregulation AUG AGG GAA ATC + second UGG - codon GGG - mutations CGG + AGA + AGC - third GAC + codon AAC - mutations UAU - UAC- Autoregulation is preserved with mutations encoding for similar or identical amino acids suggesting that the exact mRNA sequence is not critical, but rather the particular amino acids Additional experiments include: determine if purified tubulin can bind to mRNA, and if so, can it do so in a sequence-specific manner; or induce frameshift mutations into the mRNA sequence and determine if autoregulation is maintained

For three years, Gunther Schlager and Margaret Dickie estimated the forward and reverse mutation rates for five loci in mice that encode various aspects of coat color by examining more than 5 million mice for spontaneous mutations (G. Schlager and M. M. Dickie. 1966. Science 151:205-206). The numbers of mutations detected at the dilute locus are as follows: Number of gametes Number of examined mutations detected Forward mutations 260,675 5 Reverse mutations 583,360 2 Calculate the forward and reverse mutation rates at this locus. If these mutations rates are representative of rates in natural populations of mice, what would the expected equilibrium frequency of dilute mutations be?

μ = 5/260,675 = 1.9 x 10-5 v = 2/583,360 = 3.4 x 10-6 q = μ/(μ+v) = 1.9 x 10-5/(1.9 x 10-5 + 3.4 x 10-6) = 0.85