Lab data activity sheet.Please help me with my Lab data activity sheet. This is based off of the TableTop Science lab website in Lab 6: Genetics. I’ve included the Background and Procedure as pdf files.
11/27/2020 Geneticshttps://labs-tabletopscience.b9ad.pro-us-east-1.openshiftapps.com/umuc/UMUCBIOSCI/act-genetics/ 1/9Learning ObjectivesDefine inheritanceDefine gene, allele and traitDescribe the role of DNA in protein synthesis (Central Dogma)Define genotype and phenotypeExplain the relationship between genes, alleles and physical traits (phenotype)Define dominant and recessive traitsConstruct a Punnett square for a dominant and recessive trait for one generationExplain how traits are inheritedEvaluate how population genetics is used to study the evolution of populationsExplain expectations about allele frequencies in a stable population (i.e., a population that is not experiencingevolutionary forces and that meets Hardy-Weinberg assumptions)Compare and contrast the mechanisms of evolutionary forces (mutations, natural selection, gene flow, and geneticdrift)Discuss the influence an evolutionary force can have on allele frequencies within a populationDNA, Genes and InheritanceHow is it that offspring resemble their parents? You are probably aware that the answer somehow relates to DNA andgenes. In this set of laboratory exercises, you will explore the transfer of traits from parents to offspring and seeexpressions of variations in genes.The theory of chromosomal inheritance1 states that DNA nucleotides are the code for genes that determine all physicaltraits such as proteins like insulin or tissues like skin. DNA make up chromosomes. Genes are sequences of DNA thatserve as the code for proteins. The code is first read and a strand of nucleotides complementary to DNA (calledmessenger RNA, or mRNA) is synthesized in the cell nucleus then the complementary code is read by ribosomes in thecell cytoplasm. The second reading sets the stage to produce proteins according to the nucleotide sequence in the gene.This sequence from genes to proteins is referred to as the Central Dogma of genetic code of life.Central Dogma of Life SciencesGenes are made up of DNA and vary in length from a few hundred DNA nucleotide bases to more than two millionnucleotide bases. Humans have about 20,000 genes in the human genome.2A gene is responsible for all parts of an organism. Products of a gene can be thought of as ‘traits.’ (Note: Traits can be aphysical appearance at a microscopic or macroscopic level, or they can be behaviors.) Physical traits or behavioral traitsare also known as the phenotype of the organism, whereas the term ‘genotype’ refers to genes that code for those traits.For a living organism to exist, a gene was necessary to code for every macromolecule, protein, or enzyme that makes upthat individual organism. A gene can be described as a segment of DNA that is “the fundamental physical and functionalunit of heredity, which carries information from one generation to the next.”3Deoxyribonucleic acid, or DNA, is the genetic blueprint for living organisms. The information in the blueprint is a codemade of four chemicals called bases. Bases are attached to a sugar molecule and a phosphate molecule and, together,form a nucleotide.4 Genetic material consists of strands of the nucleotides adenine (A), thymine (T), cytosine (C), andguanine (G). The complementary base pairs A=T and C≡G are bound by weak hydrogen bonds. Together, the strandsform a structure known as a double helix. The two strands are bound together by hydrogen bonds that pair the nucleotideguanine (G) with cytosine (C) and the nucleotide adenine (A) with thymidine (T).The helices are coiled tightly and form chromosomes. In humans, there are 23 paired chromosomes in somatic cells—cells that are not egg and sperm. The paired nature of these chromosomes are referred to as diploid. Sex chromosomesin somatic cells are paired in the form of XX or XY pairs, depending on whether the organism is female or male,respectively. In egg and sperm cells, however, the chromosomes—more correctly, DNA—undergo meiosis whichGene (DNA) −−−−−−−−−→ Messenger RNA code Amino acids, proteinstranscribed into−−−−−−−−→translated intoGene (DNA)→transcribed intoMessenger RNA code→translated intoAmino acids, proteinsGene (Genotype) −−−−−→ Trait (Phenotype)codes forGene (Genotype)→codes forTrait (Phenotype)11/27/2020 Geneticshttps://labs-tabletopscience.b9ad.pro-us-east-1.openshiftapps.com/umuc/UMUCBIOSCI/act-genetics/ 2/9= Adenine= Thymine= Cytosine= Guanine= PhosphatebackboneDNAMother’s GeneFather’s GeneB bBBBB BbBB BbMother’s GeneFather’s GeneB bBbBB BbBb bb“unpairs” the chromosomes, making them haploid (non-paired) strandsof chromosomes.Dominant and Recessive TraitsAll living organisms inherit traits in the form of DNA in alleles in sex chromosomes of their parents’ DNA. Alleles arevariations in the nucleotide sequence of a gene; genes are in the same location on the chromosome for the parents’alleles. Offspring inherit genes and specific variations in each gene (alleles) from their parents. For example, GregorMendel studied the gene for the color of peas. One allele of the gene resulted in yellow peas while another allele resultedin green peas.Some alleles dominate and a particular trait might be more common in a population dueto a dominant allele. Examples of dominant genes in the human population are righthandedness, dark hair, and brown eyes. Other alleles are recessive and their trait is notvisible unless both parents carry the recessive allele and the offspring inherits therecessive alleles from mother and father.Let’s consider the Punnett square shown on the left. In the male, the genotype for thegene in question is heterozygous; that is, one of his alleles is dominant and the other isrecessive. For the female, the genotype for the gene in question is homozygous; that is,both of her alleles are dominant. If a large number of offspring are born to these parents,none (0%) of the offspring in the first generation would have the physical featuresassociated with the recessive trait (phenotype). Confirm that conclusion by studying thePunnett square.In the case of two heterozygous parents (as in the Punnett square on the left), such asmight occur in later generations and in descendants of these two parents, statisticallyspeaking, 25% of the offspring would show the physical features associated with therecessive allele (see the Punnett square on the left). As with all statistical probabilities,the genotypes of the actual offspring will more closely resemble the expectedpercentages in the Punnett square as the number of births grow. In a realistic simulationas in this set of laboratory exercises, you will be able to run many trials and collect thedata for several “births” of offspring, which will permit you to collect data to testhypotheses based on your expectations about the proportion of the alleles for thegenotypes of the offspring.Genes (and alleles) in the egg and sperm cells of parents essentially re-combine andsort randomly during meiosis and fertilization so that no two offspring are genetically identical (except for identical twins).Random assorting drives the statistical nature of the Punnett square. Consider a Punnett square of allele probabilities foroffspring for as many as 20,000 different genes—which is the estimated number of genes in the human genome. Thepossible phenotypic outcomes would vary accordingly. Random sorting of genes between two parents as well as randomsorting of genes in a whole population of a species lead to genetic and phenotypic diversity.11/27/2020 Geneticshttps://labs-tabletopscience.b9ad.pro-us-east-1.openshiftapps.com/umuc/UMUCBIOSCI/act-genetics/ 3/9The reproductive success of parents or a given population of a species drives the opportunity to contribute to generationsfar into the future. Parents and populations that survive to reproduce in an environment where they live and who haveoffspring who repeat that cycle will contribute to future generations.Population Genetics and the Hardy-Weinberg EquilibriumAll females and males of the same species together make a population. The genes of populations typically do not varymuch. However, changes in the genes in a population can occur through random mutation. A random mutation in thenucleotide sequence of gene in an egg or sperm cell could be the origins of a new allele in the population if the offspringsurvive and reproduce and pass on that new allele to future generations. A mutation in the nucleotide sequence of a geneis the only way a new allele can occur. Otherwise, parents simply pass on their unmutated allele—an allele common toall members of the entire, stable population—to their offspring.Most populations of a given species have a stable set of genes and alleles. The primary origin of a new, unique allele,different from ones already in the population, is a random mutation. In fact, most random mutations in the nucleotidesequence of genes lead to death of the organism.Scientists study how and why alleles vary within populations of organisms. The following characteristics apply to allelesand genotypes in stable populations:
Individuals of all genotypes within a given population have equal rates of survival and equal reproductive success. Noselection process is at work in the stable population.
Changes in a nucleotide sequence of a gene (a mutation) do not create a new allele
Individuals do not migrate into or out of the population
The population is large enough that random changes in alleles are on a small scale and do not come to dominatewithin the population
Individuals within the population are not selective about their mates; they mate randomly5These characteristics are central tenets of the Hardy-Weinberg formulation which is a mathematical representation ofalleles (and traits) within a stable population of the same species. The allele frequency for a given gene within a stablepopulation can be readily quantified using the Hardy-Weinberg equation (shown below).where represents the frequency of the dominant allele of a gene within a population, represents the frequency ofthe recessive allele within a population, and represents the frequency of heterozygotic individuals within thepopulation. The Hardy-Weinberg model relies on probability as the computational strategy to determine how prevalent anallele (or trait) is within a population. The proportional total of all frequencies of homozygotic genotypes (dominant andrecessive, such as BB and bb) and all frequencies of heterozygotic genotypes (a mixture of dominant and recessive suchas Bb) totals to 1.To gain further insight into the Hardy-Weinberg equation, consider the Punnett square below where the allele frequenciesof a gene are presented along with the genotype of the offspring.Mother’sGeneFather’s GeneFrequency B = p Frequency b = qFrequency B = pFrequency b = qFrequency BB = p 2 Frequency Bb = pqFrequency Bb = pq Frequency bb = q 2Example: Consider two alleles of a gene, B and b, in a population where the frequency of B is and thefrequency of b is . Notice that the total frequency is , which means that there are only twodifferent alleles for this gene in the population; they are all accounted for.p + 2pq + = 1,2 q2p2+2pq+q2=1,pp qq2pq2pqp = 0.7 p=0.7q = 0.3 q=0.3 p + q = 1 p+q=111/27/2020 Geneticshttps://labs-tabletopscience.b9ad.pro-us-east-1.openshiftapps.com/umuc/UMUCBIOSCI/act-genetics/ 4/9An assumption for a stable population is that mating is random. Mating probabilities can be easily computed using theprobability formula: .Probabilities:Both parents have B alleles, which means a frequency of occurrence of BB genotype offspring in the population is, or frequency of BB genotype.Both parents have b alleles, which means a frequency of occurrence of bb genotype offspring in the population is, or frequency of bb genotype.One parent contributes a B allele and the other parent contributes a b allele, which means a frequency of occurrence ofBb genotype offspring is or frequency of Bbgenotype.Mother’sGeneFather’s GeneFrequency B = 0.7 Frequency b = 0.3Frequency B = 0.7Frequency b = 0.3Frequency BB = 0.49 Frequency Bb = 0.21Frequency Bb = 0.21 Frequency bb = 0.09Notice that the Hardy-Weinberg equation is satisfied, that is,.Evolutionary Forces that Disrupt the Hardy-Weinberg EquilibriumTo use the Hardy-Weinberg equation, it is necessary to assume that populations are stable and that allele frequency doesnot change over time. These assumptions are generally accurate for large populations of a given species; however, thereare natural disruptions that occur which break these assumptions. Consider the scope of life on Earth if populations neverchanged. Would many of the species on Earth today exist if there were no changes to allele frequency within largepopulations? Biodiversity is completely due to natural selection, and natural selection is due to several specificprocesses that involve dynamic changes in the allele frequencies for a species.Processes that contribute to natural selection—that is, changes in allele frequency within a population—include (1)random mutations in the nucleotide sequence of gene, (2) the tendency of some members of a population who happen tohave a specific set of alleles to become physically separated from larger populations (gene flow or migration), and (3) thesurvival and reproductive success of a small sub-population whose gene pool does not reflect the same alleliccomposition of the larger population (genetic drift).Consider the Punnett Square allele frequency values above for the Hardy-Weinberg allele frequencies in a specialcontext in which, for example, a natural disaster causes the physical separation—such as by crevasse—of a smallerpopulation having the allele bb from the larger population with dominant allele BB. The offspring of the smaller populationwill have a new set of alleles—all alleles would be bb and the formerly recessive allele would become dominant—morefrequent—in this splinter population.The model system discussed in the next section (below) consists of (simulated) bugs where the alleles (genotype) are BBand Bb and the phenotype is the expression of the color blue for the exoskeleton. A Punnett square can be constructedfor a mating pair or for an entire population of mating pairs. The allele frequencies of the offspring across all mating pairsin the population vary according to the potential influence of processes that govern natural selection. If the population isstable, then the allele frequencies stay the same (Hardy-Weinberg assumptions hold). If natural selection (mutation, geneflow, or genetic drift) exert an influence, allele frequencies will change.Orientation to the Model Systemp + 2pq + = 1 p2+2pq+q2=12 q2p × p = 0.7 × 0.7 = 0.49 p×p=0.7×0.7=0.49 49%49%q × q = 0.3 × 0.3 = 0.09 q×q=0.3×0.3=0.09 9%9%2 × p × q = 2 × 0.7 × 0.3 = 0.42 2×p×q=2×0.7×0.3=0.42 42%42%p + 2pq + = 0.49 + 0.42 + 0.09 = 1 p2+2pq+q2=0.49+0.42+0.09=12 q211/27/2020 Geneticshttps://labs-tabletopscience.b9ad.pro-us-east-1.openshiftapps.com/umuc/UMUCBIOSCI/act-genetics/ 5/9Model Bug Key:The simulation shown in the image below is representative of the genetics experiments you will conduct in this set oflaboratory exercises. For one set of exercises, you select the alleles of the parents and will examine the alleles ofoffspring from mating pairs (“Baby Bugs”). In the other set of exercises, you will explore the impact of natural selection onmany mating pairs (the population) over several generations (“Bug Pop”).Two forms of data will be provided at theconclusion of each data run:(1) A tally of the number of offspring for thethree different allele possibilities and(2) a corresponding display of offspringphenotypes that represent a visualcounterpart of the allele distribution for theoffspring.The allele (gene) that will be of interest inthis set of experiments is the rim (or outerring) color of each bug. There are twoalleles, blue and yellow. Our model bugsare shown below. Bugs can have a blueouter rim (i.e. blue-rimmed) or a yellowouter rim (i.e. yellow-rimmed). Rim coloris determined by the genotype (allele) ofthe bug. The blue-rimmed trait is dominant (B allele).Phenotype (trait): blue-rimmed, Genotype: BB, alleles: B and BPhenotype (trait): yellow-rimmed, Genotype: bb, alleles: b and bPhenotype (trait): blue-rimmed, Genotype: Bb, alleles: B and bIn your second exercise several generations of bugs will be born and die.Dead bugs of parent generations: Gray ghost-like rims and bodiesProcedure I OverviewWhen you conduct the “Baby Bugs” activities, you will be working with a blue allele that is dominant and noted by B anda yellow allele that is recessive and noted by b. This activity will simulate breeding between bugs. You will investigatethe offspring that result from specific breeding pairs. Your work will include comparing probabilities calculated from yourdata and from Punnett Squares.Procedure II OverviewWhen you conduct the “Bugs Pop” activities, you will explore what happens when evolutionary forces act on apopulation, and the dominance and recessive qualities of the alleles are not driving the survival and reproductivesuccess of the parents and offspring. For this activity, you will observe potential composition changes to a small isolatedbreeding population of bugs over time. Your work will include comparing probabilities calculated from your data and fromthe Hardy-Weinberg equation.Summary of Formulas Needed for CalculationsCalculating Average Values from DataExample: Determine the average BB baby count from ten data run values: 3, 6, 4, 4, 3, 3, 5, 6, 4, 5.11/27/2020 Geneticshttps://labs-tabletopscience.b9ad.pro-us-east-1.openshiftapps.com/umuc/UMUCBIOSCI/act-genetics/ 6/9Calculating Probabilities (frequencies) and Percentages from DataProbabilities and percentages can be determined from a single data point or from an average of multiple data runs.Average data values tend to be more representative of a population when there is variation between individual data runs.Example: Determine the genotype probabilities (frequencies) and percentages from the following information:data averages: BB average = 4.3, Bb average = 2.3, bb average = 3.4total number of babies = 10.For the BB genotype we haveFor the Bb genotype we haveFor the bb genotype we haveExample: Determine the phenotype probabilities and percentages from the results above.For the blue phenotype we haveFor the yellow phenotype we haveDetermine Percentages from a Punnett SquareBB average = = = 4.33 + 6 + 4 + 4 + 3 + 3 + 5 + 6 + 4 + 5104310BB average=3+6+4+4+3+3+5+6+4+510=4310=4.3BB probability = = = 0.430BB averagetotal number of babies4.310BB probability=BB averagetotal number of babies=4.310=0.430BB percentage = 100% × BB probability = 100% × 0.430 = 43.0%.BB percentage=100%×BB probability=100%×0.430=43.0%.Bb probability = = = 0.230Bb averagetotal number of babies2.310Bb probability=Bb averagetotal number of babies=2.310=0.230Bb percentage = 100% × Bb probability = 100% × 0.230 = 23.0%.Bb percentage=100%×Bb probability=100%×0.230=23.0%.bb probability = = = 0.340bb averagetotal number of babies3.410bb probability=bb averagetotal number of babies=3.410=0.340bb percentage = 100% × bb probability = 100% × 0.340 = 34.0%.bb percentage=100%×bb probability=100%×0.340=34.0%.blue phenotype probability = BB probability + Bb probability = 0.430 + 0.230 = 0.660blue phenotype probability=BB probability+Bb probability=0.430+0.230=0.660blue phenotype percentage = 100% × blue phenotype probability = 100% × 0.660 = 66.0%.blue phenotype percentage=100%×blue phenotype probability=100%×0.660=66.0%.yellow phenotype probability = bb probability = 0.340yellow phenotype probability=bb probability=0.340yellow phenotype percentage = 100% × yellow phenotype probability = 100% × 0.340 = 34.0%.yellow phenotype percentage=100%×yellow phenotype probability=100%×0.340=34.0%.11/27/2020 Geneticshttps://labs-tabletopscience.b9ad.pro-us-east-1.openshiftapps.com/umuc/UMUCBIOSCI/act-genetics/ 7/9Step 1Mother’s GeneFather’s GeneB bBbStep 2Mother’s GeneFather’s GeneB bBbBB BbStep 3Mother’s GeneFather’s GeneB bBbBB BbBb bbExpected probabilities of a single birth between Bb parents.25% chance (1/4) the child is genotype BB25% chance (1/4) the child is genotype bb50% chance (2/4) the child is genotype BbWhat does this mean in terms of the expected phenotypefrom a single birth between these parents?BB and Bb genotypes share the same phenotype and sothere is a 75% chance (3/4) of a child being born into thatphenotype.Likewise, there is a 25% chance (1/4) of a child being borninto the phenotype associated with the bb genotype.Expected probabilities of births in a population composedonly of Bb parents.on average 25% of births (1/4) are genotype BBon average 25% of births (1/4) are genotype bbon average 50% of births (2/4) are genotype BbWhat does this mean in terms of the expected births in apopulation composed only of Bb parents?BB and Bb genotypes share the same phenotype and so,on average, 75% of births (3/4) are that phenotype.Likewise, on average, 25% of births (1/4) belong to thephenotype associated with the bb genotype.Example: Punnett Square for two heterozygous parents (both genotype Bb)Punnett Square Example for Two Heterozygous Parents (both genotype Bb)Both parents aregenotype Bb(heterozygous).
Insert the parentsalleles along the topand left side
Fill in the genotypesin the first row (useone allele from eachparent)
Fill in the genotypesin the second row(use one allele fromeach parent)What does the completed Punnett Square tell us? There are two equally valid interpretations.11/27/2020 Geneticshttps://labs-tabletopscience.b9ad.pro-us-east-1.openshiftapps.com/umuc/UMUCBIOSCI/act-genetics/ 8/9Using the Hardy-Weinberg EquationThe Hardy-Weinberg equation can be used to understand the proportions of alleles, genotypes, and phenotypes in astable population.Example: A small breeding population of bugs initially consists of 12 blue-rimmed bugs of genotype BB and 8 yellowrimmed bugs (8 genotype bb). After several generations of births and deaths, the population is observed to consist of 17blue-rimmed bugs (7 BB and 10 Bb) and 3 yellow-rimmed bugs (3 bb). Is this population consistent with the expectationsof the Hardy-Weinberg model, that is, is this population stable?Population DataInitial Population Composition — 12 blue bugs of genotype BB and 8 yellow bugs (8 genotype bb)Final Population Composition — 17 blue bugs (7 BB and 10 Bb) and 3 yellow bugs (3 bb)Step 1: Determine and from the initial population composition dataStep 2: Use the Hardy-Weinberg equation to predict the future population compositionThe Hardy-Weinberg equation is , where is the proportion of yellow bugs andis the proportion of blue bugs. However, once we know the proportion of yellow bugs it is easier to thedetermine the blue bug proportion using .Step 3: Determine the observed final population pheotype percentages from dataStep 4: Compare the Hardy-Weinberg predicted and observed final population phenotype percentagesBlue — predicted: ; observed:Yellow — predicted: ; observed:Conclusion: This population is consistent with the expectations of the Hardy-Weinberg model.
Klug, WS, Cummings, MR, Spencer, CA, and Palladino, MA, Concepts of Genetics, 10th edition (2012), page 5.
https://ghr.nlm.nih.gov/primer/basics/gene.
https://www.ncbi.nlm.nih.gov/books/NBK21962/.
What is DNA? Genetics Home Reference: Your Guide to Understanding Genetic Conditions from U.S. National Library of Medicine, Nationalpp qqp = = = 0.600number of B allelestotal number of alleles2440p=number of B allelestotal number of alleles=2440=0.600q = = = 0.400number of b allelestotal number of alleles1640q=number of b allelestotal number of alleles=1640=0.400p + 2pq + = 1 p2+2pq+q2=12 q2 q q22p + 2pq p2+2pq21 − q 1−q22proportion of yellow bugs = q2 = (0.400)2 = 0.160 ⇒ 16.0%proportion of yellow bugs=q2=(0.400)2=0.160⇒16.0%proportion of blue bugs = 1 − q2 = 1 − 0.160 = 0.840 ⇒ 84.0%proportion of blue bugs=1−q2=1−0.160=0.840⇒84.0%observed proportion of blue bugs = = = 0.850 ⇒ 85.0%number of blue bugstotal number of bugs1720observed proportion of blue bugs=number of blue bugstotal number of bugs=1720=0.850⇒85.0%observed proportion of yellow bugs = = = 0.150 ⇒ 15.0%number of yellow bugstotal number of bugs320observed proportion of yellow bugs=number of yellow bugstotal number of bugs=320=0.150⇒15.0%84.0%84.0% 85.0%85.0%16.0%16.0% 15.0%15.0%11/27/2020 Geneticshttps://labs-tabletopscience.b9ad.pro-us-east-1.openshiftapps.com/umuc/UMUCBIOSCI/act-genetics/ 9/9Institute of Health, https://ghr.nlm.nih.gov/primer/basics/dna. Retrieved on 7/23/2019.
Klug, WS, Cummings, MR, Spencer, CA, and Palladino, MA, Concepts of Genetics, 10th edition (2012), page 702.
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