Concept explainers
Dr. Ara B. Dopsis and Dr. C. Ellie Gans are performing genetic crosses on daisy plants. They self-fertilize a blue- flowered daisy and grow
a. Use the form below to calculate chi square for the
b. Use the form below to calculate chi square for the
c. What is your conclusion regarding these two genetic hypotheses?
d. Using any of the
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- 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 wild type male. The resulting progeny were allowed to mate with one another to produce the data set. Three repetitions of the experiment were conducted. The following data were produced from the crosses. Test these data to determine if they are significantly different from the expected phenotypic ratio. Use the 5% level of significance. Your answer should include the hypothesized cross in genotypes, the Chi-squared value, the critical value and whether you reject or do not reject for each experiment. Wild eye Wild body – 112, Wild eye Ebony body – 40, Sepia eye Wild body – 35, Sepia eye Ebony body – 11arrow_forwardAll crosses involve many more than one character. So far, only one character has been examined to help simplify the principles involved. Now, two pairs of alleles or two characters will be examined at the same time. In working the following problem, each pair of characteristics is on a different chromosome and will be inherited independently of the other (Principle of Independent Assortment). Problem 1: In garden peas, a homozygous recessive male plant with wrinkled, green seeds is crossed with a plant that is homozygous dominant for round, yellow seeds. Use this information to complete the following: Male phenotype= genotype= gametes= F2 phenotype= genotype = gametes= F, phenotype= F₁ genotype = Female Figure 12.10 Punnett Square for F₁ The F, genotype should be all RrYy. The gametes are formed from various combinations of these alleles. Each gamete must have one R and one Y. For example, the F, male gametes will be RY, Ry, rY, ry. Would the female gametes be the same as the male or…arrow_forwardA 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.arrow_forward
- a) how many degree of freedom b) what is the chi square value c) what is the percentage assurance d) is your data a good fit for the ration you expected. Previous question asks in pea plants stem length is either long dominant or short recessive. 2 heterozygous long pea plants were crossed. The progeny included 402 pea plants, 308 of which were long and 94 of which were short. The ratio is 3:1arrow_forwardA tall pea plant (homozygous dominant) is crossed to a pea plant that is heterozygous for the gene for height. Create a Punnett Square and use it to answer the following questions: What is the genotypic ratio of the offspring? What is the phenotypic ratio of the offspring? Answers are written as genotype; phenotype a) 50% TT : 50% Tt; 100% tall plants b) 1 TT : 2 Tt : 1tt; 75% tall plants : 25% dwarf plants c) 1 TT : 2 Tt : 1tt; 25% tall plants : 50% medium height plants : 25% dwarf plantsarrow_forwardA homozygous tomato plant with orange fruit and white flowers was crossed with a homozygous tomato plant with red fruit and red flowers. The F1 all had orange fruit and white flowers. The F1 were testcrossed by crossing them to homozygous recessive individuals and the following offspring were obtained: Orange fruit and white flowers- 64 Red fruit and red flowers- 69 Orange fruit and red flowers- 14 Red fruit and white flowers- 13 What is the recombination frequency of these two genes?arrow_forward
- As a graduate student, you join a lab and you are looking through some old notebooks. You find a linkage map someone has drawn. G--15 mu-----T----10 mu-------B If you plan to complete a test cross with a GtB/gTb male, A) What are all of the expected progeny genotypes? B.) How many of each genotype would you expect if you collected 1000 progeny?arrow_forwardThe parental genotypes for a series of crosses are wild-type male fruit flies mated to females with white eyes (wh) and miniature (min) wings. The phenotypes of the F1 generation were wild-type females, and males with white eyes, and miniature wings. These flies were allowed to mate with each other and produced the following offspring: Red eyes, long wings White eyes, miniature wings Red eyes, miniature wings White eyes, long wings 770 716 401 318 Total 2205 A. Are these genes linked? Why or why not?arrow_forwardNicotiana glutinosa (2 n = 24) and N. tabacum (2 n = 48) are two closely related plants that can be intercrossed, but the F1 hybrid plants that result are usually sterile. In 1925, Roy Clausen and Thomas Goodspeed crossed N. glutinosa and N. tabacum and obtained one fertile F1 plant (R. E. Clausen and T. H. Goodspeed. 1925 Genetics 10:278– 284). They were able to self-pollinate the flowers of this plant to produce an F2 generation. Surprisingly, the F2 plants were fully fertile and produced viable seeds. When Clausen and Goodspeed examined the chromosomes of the F2 plants, they observed 36 pairs of chromosomes in metaphase I and 36 individual chromosomes in metaphase II. Explain the origin of the F2 plants obtained by Clausen and Goodspeed and the numbers of chromosomes observed.arrow_forward
- You carry out a trihybrid cross (a cross in which the parental plants differ for three characters) between a tall pea plant with round, yellow seeds (TT RR YY) and a short pea plant with wrinkled, green seeds (tt rr yy). The parental plants are homozygous for all of the three characters. They are crossed to produce the F1 generation. Tall, round, and yellow are the dominant traits for each character. What will be the phenotypes of the F1 generation?arrow_forwardTwo linked loci have a recombination frequency of 5%. A series of crosses is performed. The P generation includes an individual that is homozygous dominant for trait 1 and homozygous recessive for trait 2. The second individual is homozygous recessive for trait 1 and homozygous dominant for trait 2. The F1 generation is crossed with individuals that are homozygous recessive for both traits. If 400 F2 offspring are produced, how many offspring with each phenotype are expected? Fill in the table below with your answers. Phenotype Number of Offspring Predicted Recessive 1, Recessive 2 Dominant 1, Dominant 2 Recessive 1, Dominant 2 Dominant 1, Recessive 2 Total offspring 400 For the results above, determine which phenotypes are parental and which are recombinant. Phenotype Parental or Recombinant? Dominant 1, Dominant 2 Recessive 1, Dominant 2 Recessive 1, Recessive 2 Dominant 1, Recessive 2arrow_forwardJ.W. McKay crossed a stock (true-breeding) melon plant that produced tan seeds with a plant that only produced red seeds and obtained the following results (J.W. McKay. 1936. Journal of Heredity 27:110-112). Cross F1 F2 Tan x red 13 tan 93 tan, 24 red a) Explain the inheritance of tan seeds and red seeds in this plant. b) Assign symbols for the alleles in this cross and draw out the Punnett Squares for the initial cross and the F1 cross.arrow_forward
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