West Seneca Central School District



Video 231. In fruit flies, the phenotype for eye color is determined by a certain locus. E indicates the dominant allele and e indicates the recessive allele. The cross between a male wild-type fruit fly and a female white-eyed fruit fly produced the following offspring.The wild-type and white-eyed individuals from the F1 generation were then crossed to produce the following offspring.Determine the genotypes of the original parents (P generation) and explain your reasoning. You may use Punnett squares to enhance your description, but the results from the Punnett squares must be discussed in your answer.Maximum 4 points F1:Xe+YXeXe+ XeXe YXeXe+ XeXe Y1 pt: Genotypes of the parents (words or symbols) Xe+ Y (or X+Y) and Xe Xe1 pt: Discuss/show how these resulted in this F1 (may be annotated Punnett)1 pt: Explain that it is a sex-linked (X-linked) gene (not just the word)1 pt: How you know which type is dominant1 pt: F2 results (may be annotated Punnett square)XeYXe+Xe+ XeXe+ YXeXe XeXe Y F2:2. Use the data table above. Use a Chi-squared test on the F2 generation data to analyze your prediction of the parental genotypes. Show all your work and explain the importance of your final answer.Maximum 4 points1 pt: Correct F2 hypothesis (1:1:1:1; or 25/genotype)1 pt: Show work (components): o e o-e (o-e)2 (o-e)2/e(or correct numbers (4/25 + 36/25 + 1/ 25 + 9/25 = 50/25 = 2; or at least the last term)1 pt: Sum: correct chi-square result ~ 2.0 or 1.851 pt: degrees of freedom = 3 (critical value is 7.82)1 pt: correct interpretation of chi-square in terms of pp = probability that the difference between the observed and the expected value is due to chance alone.This p value shows we accept our hypothesis.The null hypothesis is supported in this case(alternative: 2Χ2 tests of white vs. red males and white vs. red females)center1035685003. Students in a class are studying patterns of inheritance using genes involved in determining the body color and wing shape of Drosophila flies. Each of the genes has only two alleles, one of which is completely dominant to the other. Each student in the class performed a parental cross between a fly that is true-breeding for ebony body and vestigial wings and a fly that is true-breeding for gray body and long wings. Each student then crossed several pairs of the F1 flies and determined the phenotypes of 500 of the resulting F2flies with respect to body color and wing shape. The students in the class averaged their data for the frequencies of the four possible phenotypes (Table 1). Table 1. Averaged phenotypic data of F2 flies P= ebony vestigial X gray longGray dom / long domEbony rec / vestigial recThe students performed a second cross. The parental cross was between flies that are true-breeding for gray bodies and long wings and flies that are true-breeding for ebony bodies and curly wings. They crossed pairs of F1 flies and determined the phenotypes of the resulting F2 flies. The students found an approximate 3:1ratio of flies with the dominant phenotype (gray bodies and long wings) to flies with the recessive phenotype (ebony bodies and curly wings). Only a few of the flies expressed the dominant phenotype of one trait and the recessive phenotype of the other trait.(a) In the first analysis, all of the F1flies from the students’ crosses have the identical phenotype with respect to body color and wing shape, but the F2 flies have four different phenotypes. Describe how fertilization contributes to this genetic variability. fertilization joins gametes with different allele combinations of the genes for body color and wing shape.90170508000(b) Using the template, construct an appropriately labeled graph, including error bars, to represent the data in Table 1. Based on the data in Table 1, determine whether there is a significant difference between the number of flies in each of the four phenotypes.Axis labelScale unitPlot bar graphthe number of flies with the phenotype ebony body and long wings is the same as the number of flies with the phenotype gray body and vestigial wings. Based on the error bars, the numbers of flies with the two other phenotypes are significantly different from each other and from those of the first two phenotypes.(c) Based on the data in Table 1, describe why the dominant alleles for body color and wing shape are the alleles that produce a gray body and long wings, respectively. Based on the data, describe why the two genes are most likely on different chromosomes or why they are most likely on the same chromosome. Calculate the probability of producing flies that have gray bodies and vestigial wings if a cross is performed between one of the F1 flies from the first analysis and a fly that is homozygous for a gray body and vestigial wings.G= gray L= longG= ebony l= vetigialGray dom / long dom Ebony rec / vestigial rec(1pt) flies in the largest fraction of the F2 generation have these two traits, suggesting that the alleles for these traits are dominant.(1pt) 9:3:3:1, which is characteristic of a dihybrid cross with two genes that are on separate chromosomes. 293:98:81: 28(1pt) probability is 0.5. gray vestigial ? P= GgLl X GGllThe probability of flies having gray bodies is 1.0, because the gray color is dominant and one of the flies in the cross is homozygous for a gray body. Gg x GG= GG or Gg all are grayThe probability of flies having vestigial wings is 0.5, because vestigial wings are recessive and one of the flies is homozygous for vestigial wings, while the F1 fly is heterozygous for this trait. Ll x ll= Ll/ll half normal half vestigialThe probability of flies having the two traits is calculated by multiplying the two individual probabilities together: 1×0.5=0.5.(d) Predict the most likely cause of the F2 ratio obtained by the students in the second analysis between parental flies that are true-breeding for gray bodies and long wings and flies that are true-breeding for ebony bodies and curly wings. Provide reasoning to justify your prediction.(1pt) second analysis the genes for gray or ebony body color and long or curly wings are linked on the same chromosome.(1pt) if the genes are close together/linked, the combination of parental alleles will remain unchanged except for a small percent of new combinations that result from limited crossing over in the F1 flies as they produce gametes. The F1 flies are heterozygous, with one chromosome that has both dominant alleles and one chromosome that has both recessive alleles. A cross between them is like a monohybrid cross. Approximately 1/4 of the F2 flies will be homozygous dominant for both genes,1/2 will be heterozygous for the two genes, and ? will be homozygous recessive for the two genes. This gives a 3:1phenotypic ratio of dominant to recessive for both alleles.4. Fruit flies (Drosophila melanogaster) with a wild-type phenotype have gray bodies and red eyes. Certain mutations can cause changes to these traits. Mutant flies may have a black body and/or cinnabar eyes. To study the genetics of these traits, a researcher crossed a true-breeding wild-type male fly (with gray body and red eyes) with a true-breeding female fly with a black body and cinnabar eyes. All of the F1 progeny displayed a wild-type phenotype.189357058928000Female flies from the F1 generation were crossed with true-breeding male flies with black bodies and cinnabar eyes. The table below represents the predicted outcome and the data obtained from the cross. Explain the difference between the expected data and the actual numbers observed.4 points maximum. Student explanations include the following: ? Prediction of a 1:1:1:1 ratio with correct phenotypes based on independent assortment.? Support for prediction with a diagram of the cross of BbEe x bbee. ? Correct application of chi-square analysis to show that observed results do not conform to expected Mendelian frequencies. ? Identification of body color and eye color as linked genes/loci. ? Use of ratios to show linkage and independent assortment of wing type versus linked traits. ? Identification of the bottom two phenotypes as products of crossing over (recombinant chromosome). ? Mentioning that crossover rate is approximately 9–10 percent. 2336800005. In a certain species of plant, the diploid number of chromosomes is 4 (2n = 4). Flower color is controlled by a single gene in which the green allele (G) is dominant to the purple allele (g). Plant height is controlled by a different gene in which the dwarf allele (D) is dominant to the tall allele (d). Individuals of the parental (P) generation with the genotypes GGDD and ggdd were crossed to produce F1 progeny.If the two genes were genetically linked, describe how the proportions of phenotypes of the resulting offspring would most likely differ from those of the testcross between an F1 individual and a ggdd individual.GgDd X ggdd1:1:1:1 is not linkedMore with parental combos than recombinants means it is linked Student response earns 1 of the following 1 points1 point(s) maximumIdentify difference (1 point)? The majority/greater than 50 percent would have the parental plant phenotypes? Greater than 25 percent would be green dwarf plants and greater than 25 percent would be purple tall plants? Less than 25 percent would be green tall plants and less than 25 percent would be purple dwarf plants-207645930275006. Drosophila melanogaster (D. melanogaster) is a species of fruit fly frequently used by researchers in genetic studies. Members of this species have two of each of four different chromosomes: the sex chromosome (flies have X and Y) and three autosomes (chromosomes 2, 3, and 4). Researchers studying D. melanogaster conducted genetic crosses to investigate a particular X-linked recessive trait encoded by a single gene (Table 1). Affected flies have the trait.Cross 1: Xr+Xr+ and XrYr+= domCross 2: XrXr and Xr+Yr = recCross 3: Xr+Xr and Xr+YCross 4: XrXr and XrYIdentify the genotypes of the male and female flies used in cross 2.response identifies XrXr and Xr+Y as the genotypes of the male and female flies used in cross 2.Identify the cross in which the female parent was most likely heterozygous.cross 3 as the cross in which the female parent was most likely heterozygous.The researchers hypothesize that crossing any unaffected female and an affected male will result in a 0% chance of producing an affected male offspring. Evaluate the validity of the hypothesis.hypothesis is not supported because an unaffected female could be heterozygous and pass on the recessive allele.Explain how the results exclude the possibility that the trait is encoded by a mitochondrial gene.traits encoded by mitochondrial genes are inherited from the female parent only, resulting in offspring that all have the same genotype as their mother.5251450603885007. Electrons from the oxidation of glucose and other nutrients by glycolysis and the Krebs cycle are added to the electron-carrying compound NAD+ to form NADH. To meet the continuous requirement for NAD+ by cells, NADH is reduced back to NAD+ through the electron transport chain, in the presence (+) of oxygen, or through fermentation, in the absence (?) of oxygen (Figure 1).Figure 1. A simplified model of metabolism and the recycling of NAD+MT-ND5 is a mitochondrial gene that encodes a subunit of NADH dehydrogenase, the enzyme that catalyzes the initial oxidation of its substrate NADH to NAD+ and H+ in the electron transport chain of mitochondria. A mutation in MT-ND5 is associated with a rare genetic disorder that results in a buildup of lactic acid in the body. A researcher hypothesizes that the mutated NADH dehydrogenase has decreased activity but is not completely nonfunctioning and that, by increasing the pool of NADH in cells, the activity of NADH dehydrogenase will increase.456057063500To test this idea, the researcher treated a group of individuals with this disorder with a vitamin that is similar to NADHand measured the concentration of NAD+ and lactic acid in the bloodover the course of 20 weeks. The results from a representative individual are shown in Figure 2.Figure 2. The concentration of NAD+(top) and lactic acid (bottom) in the blood of a representative treated individual(a)Describe the pattern of inheritance that is most likely associated with amutation in the MT-ND5 gene. Explain why individuals are not typically heterozygous with respect to mitochondrial genes.A description that the mutant allele is only inherited from the maternal parent.An explanation that MT-ND5 is a mitochondrial gene and mitochondria only carry a single, unpaired chromosome. (b) Identify a dependent variable measured in the researcher’s experiment. Identify one control that the researcher could use to improve the validity of the experiment. Justify the researcher analyzing blood samples at many intermediate time points instead of at only the beginning and the end of the 20-week period.An identification of an acceptable dependent variable. Acceptable dependent variables include the concentration of lactic acid in the blood and the concentration of NAD+in the blood.An identification of the control as measuring NAD+and lactic acid levels in a group of other affected individuals treated with placebo.The justification that by collecting more data, the researchers will see a more accurate trend.(c) Describe the relationship between the concentration of NAD+ in the blood and the concentration of lactic acid in the blood during the first 5 weeks of treatment with the vitamin. Based on Figure 2, calculate the average rate of change in blood NAD+ concentrations from week 5 to week 17.A description that the concentration of NAD+ increased and the concentration of lactic acid decreased.A calculation of the average rate of change in blood NAD+ as ±20?μmol/l/week.The researcher performed a follow-up experiment to measure the rate of oxygen consumption by muscle and brain cells. Predict the effect of the MT-ND5 mutation on the rate of oxygen consumption in muscle and brain cells. Justify your prediction. The researcher had hypothesized that the addition of the vitamin that is similar in structure to NADH would increase the activity of the mutated NADH dehydrogenase enzyme in individuals with the disorder. Explain how the vitamin most likely increased the activity of the enzyme.(3pts)The prediction that the mutation will decrease the rate of oxygen consumption by the electron transport chain.The justification that the mutation likely inhibits the rate of oxygen consumption because the electron transport chain will be less efficient, so less oxygen will be required as a terminal electron acceptor.The explanation that the enzyme had decreased activity but still functioned. Increasing the amount of substrate should increase the amount of product produced. Because the vitamin was similar to NADH, it could bind to the active site of the enzyme and effectively increase the substrate concentration.8.In the tongue sole fish (Cynoglossus semilaevis), sex is determined by a combination of genetics and environmental temperature. Genetically male fish have two Z chromosomes (ZZ), and genetically female fish have one Z chromosome and one W chromosome (ZW). When fish are raised at 22℃, ZZ fish develop into phenotypic males and ZW fish develop into phenotypic females. However, when fish are raised at 28℃, the Z chromosome is modified (denoted as Z*). Z*W individuals develop as phenotypic males that are fertile and can pass on the Z* chromosome to their offspring even when the offspring are raised at 22℃. A cross between a ZW female and a Z*Z male is shown in the Punnett square below.left-190500Predict the percent of phenotypic males among the F1 offspring of the cross shown in the Punnett square if the offspring are raised at 22℃.75%9. A new species of fly was discovered on an island in the South Pacific. Several different crosses were performed, each using 100 females and 100 males. The phenotypes of the parents and the resulting offspring were recorded. Cross I: True-breeding bronze-eyed males were crossed with true-breeding red-eyed females. All the F1 offspring had bronze eyes. F1 flies were crossed, and the data for the resulting F2 flies are given in the table below.Bronze = dom red= rec r+ r762008509000P= r+ r+ X r r F1= r+ rF2= bronze : red 3:1left62357000Cross II: True-breeding normal-winged males were crossed with true-breeding stunted-winged females. All the F1 offspring had stunted wings. F1 flies were crossed, and the data for the resulting F2 flies are given in the table below.s normal S stuntedP= ss X SS F1= Ss (stunted males ) and Ss(stunted females)F2= stunted: normal 3:1 SS, Ss / ssleft70104000Cross III: True-breeding bronze-eyed, stunted-winged males were crossed with true-breeding red-eyed, normal-winged females. All the F1 offspring had bronze eyes and stunted wings. The F1 flies were crossed with true- breeding red-eyed, normal-winged flies, and the results are shown in the table below.Bronze= dom red = recStunted = dom normal = recP= Bronze stunted X red normalF1= bronze stuntedWhat conclusions can be drawn from the data from cross III? Explain how the data support your conclusions.4 points maximumConclusion for cross III (1 point per bullet; 2 points maximum)Genes linkedCrossing overGenes 10 map units apartExplanation for cross III (1 point per bullet; 2 points maximum)Not a 1:1:1:1 ratio (as predicted by independent assortment)Not a 1:1 ratio/two recombinant phenotypes (unexpected).Frequency of recombinant phenotypes was 10 percent (setup equation OK)/parental phenotypes (bronze/stunted and red/normal) are represented in 90 percent of offspring. ................
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