What are the two missing genotypes in the Table above ? What is the map distances between the bd and ps loci ? Assuming the map distances are additive, what is the interference ?
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What are the two missing genotypes in the Table above ?
What is the map distances between the bd and ps loci ?
Assuming the map distances are additive, what is the interference ?
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- Using the punnet square (di-hybrid cross) solve the following: In a certain breed of dog, the dominant, B, is required for black fur; and it’s recessive, b, produces brown fur. However, the dominant, I is the epistatic gene to the color locus and can inhibit pigment formation. The epistatic recessive allele, i, on the other hand, permits pigment deposition in the fur. -Cross the following parents: bbIi x BBIi 4th Question: How many of the offspring could be albino (white)?Using the punnet square (di-hybrid cross) solve the following: In a certain breed of dog, the dominant, B, is required for black fur; and it’s recessive, b, produces brown fur. However, the dominant, I is the epistatic gene to the color locus and can inhibit pigment formation. The epistatic recessive allele, i, on the other hand, permits pigment deposition in the fur. -Cross the following parents: bbIi x BBIi 1st Question to answer ---- How many different possible genotypes are there among the offspring? 2nd Question to answer ---- How many of the offspring could be black? 3rd Question to answer ---- How many of the offspring could be brown? 4th Question to answer ---- How many of the offspring could be albino (white)? Different Possible Genotypes Black Offspring Brown Offspring Albino/White OffspringA sex-influenced trait is dominant in males and causes bushy tails.The same trait is recessive in females. Fur color is not sex influenced.Yellow fur is dominant to white fur. A true-breeding femalewith a bushy tail and yellow fur was crossed to a white male withouta bushy tail (i.e., a normal tail). The F1 females were thencrossed to white males without bushy tails. The following resultswere obtained:A. Conduct a chi square analysis to determine if these two genesare linked.B. If the genes are linked, calculate the map distance betweenthem. Explain which data you used in your calculation.
- Considering hybridization in a trait like the color of the flowers of a given plant species (red dominant/ yellow recessive) conditioned by a pair of different alleles, what are the phenotypical results of the first generation (F1) and the phenotypical results of the second generation (F2, formed by crossing among F1 genotypes)? What are the phenotypical proportions in F1 and F2?A cross was performed using Drosophila melanogaster involving a female known to be heterozygous for both ebony body and sepia eyes and a male known to be homozygous for both of these recessive traits. The following data was produced from the cross. Test these data to determine if they are significantly different from the expected phenotypic ratio. Remember to use the 5% level of significance Wild eye Wild body – 102, Wild eye Ebony body – 94, Sepia eye Wild body – 100, Sepia eye Ebony body – 93. Your answer should include the hypothesized cross in genotypes, the Chi-squared value, the critical value and whether you reject or do not reject.Tomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Part A: Record the phenotypes and the phenotype ratio in lowest terms. Part B: Convert the expected phenotypic ratio from Part A into the expected probability for each of the four phenotypes and record them in the table below. Calculate the probability for each of the four phenotypes observed in the cross from the data presented at the beginning of the question by dividing the number of progeny in each class by the total number of progeny and record these in the table below. (Probability = Number of Progeny in Phenotype…
- Tomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Part A: Record the phenotypes and the phenotype ratio in lowest terms. Part B: Convert the expected phenotypic ratio from Part A into the expected probability for each of the four phenotypes and record them in the table below. Calculate the probability for each of the four phenotypes observed in the cross from the data presented at the beginning of the question by dividing the number of progeny in each class by the total number of progeny and record these in the table below. Probability = Number of Progeny in Phenotype…. Consider the genotypes of two lines of chickens: thepure-line mottled Honduran is i/i ; D/D ; M/M ; W/W, andthe pure-line leghorn is I/I ; d/d ; m/m ; w/w, whereI = white feathers, i = colored feathersD = duplex comb, d = simplex combM = bearded, m = beardlessW = white skin, w = yellow skinThese four genes assort independently. Starting withthese two pure lines, what is the fastest and mostconvenient way of generating a pure line that has coloredfeathers, has a simplex comb, is beardless, and has yellowskin? Make sure that you showa. the breeding pedigree.b. the genotype of each animal represented.c. how many eggs to hatch in each cross, and why thisnumber.d. why your scheme is the fastest and the mostconvenientTomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Determine the recombination frequency between the genes for fruit colour and fruit shape using the following formula (in picture). Show your work. Record your answer as a whole number percentage.
- Tomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Complete a Punnett square to determine the expected testcross progeny on the basis of Mendel’s Law of Independent Assortment. Record the phenotypes and the phenotype ratio in lowest terms.Finish these crosses below:(assume all parents are homozygous for alleles) Cross 1:These are the parents: Sepia Male, Wild Type FemaleWhat are the alleles of the parents? (like Aa and AA / or like XY and XWXW+) Use a Punnett square below to determine the F1 generation phenotypic ratios. What is the phenotypic ratio of the F1? If two F1 individuals mated, what would the F2 phenotypic ratios be?The question should be whether you reject or do not reject the Fisher test null hypothesis using data for all three institutions, not for two. Test that Mode ofInheritance is Autosomal Recessive forexperiments at each of three researchinstitutions At two research instructions, the parental cross is a affected(disease) male with an unaffected (wild‐type) female. Thecross of a male and female in the F 1generation produces thefollowing counts in the F 2generation: Question to be answered: Computethe Fisher Combined P‐value test using the data fromthe two sites, and draw a conclusion (reject, do notreject) the null hypothesis for the Fisher test, namelythat the specified MOI at each site is correct. https://onlinelibrary.wiley.com/doi/10.1002/aff2.68(only abstract) https://bigislandnow.com/2022/10/23/native-predatory-fish-help-control-invasive-species-in-hawaiian-fishpond-on-o%CA%BBahu/#:~:text=Jacks%20and%20barracuda%20in%20He%CA%BBeia,on%20the%20invasive%20mullet%20species.