Concepts of Genetics (11th Edition)
11th Edition
ISBN: 9780321948915
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 3, Problem 18PDQ
The following are F2 results of two of Mendel’s monohybrid crosses.
For each cross, state a null hypothesis to be tested using χ2 analysis. Calculate the χ2 value and determine the p value for both. Interpret the p values. Can the deviation in each case be attributed to chance or not? Which of the two crosses shows a greater amount of deviation?
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Shown are F₂ results of a monohybrid cross performed by Mendel.
a) Calculate the expected numbers of each type of pods.
Full pods
Constricted pods
Total
882
298
1180
0.84
b) What do these p-values mean with regards to your null hypothesis? (Please
choose either reject or fail to reject.)
shown are the F2 results of two Mendel's monohybrid crosses. Mendel crossed 2 heterozygote parents. If he were to expect segregation to happen normally, do the results match his expectations? state a null hypothesis that you will test using chi-square analysis. Calculate the chi-square value and determine the p value for both crosses
a. Full pods: 882; constricted pods: 299 (full pods are dominant over constricted)
b. violet flowers: 705; white flowers: 500 (violet dominant over white)
To assess Mendel’s law of segregation using tomatoes, a true-breeding tall variety (SS) is crossed with a true-breeding short variety (ss). The heterozygous F1 tall plants (Ss) were crossed to produce two sets of F2 data as follows.
Set 1
Set 2
30 tall
300 tall
5 short
50 short
Using the chi-square test, analyze the results for both datasets. Calculate the chi-square values and estimate the p-values in both cases.
From the above analysis, what can you conclude about the importance of generating large datasets in experimental condition?
Chapter 3 Solutions
Concepts of Genetics (11th Edition)
Ch. 3 - Pigeons may exhibit a checkered or plain color...Ch. 3 - Considering the Mendelian traits round versus...Ch. 3 - Using the forked-line, or branch diagram, method,...Ch. 3 - In one of Mendels dihybrid crosses, he observed...Ch. 3 - The following pedigree is for myopia...Ch. 3 - Prob. 1CSCh. 3 - Thomas first discovered a potentially devastating...Ch. 3 - Prob. 3CSCh. 3 - Prob. 4CSCh. 3 - HOW DO WE KNOW? In this chapter, we focused on the...
Ch. 3 - CONCEPT QUESTION Review the Chapter Concepts list...Ch. 3 - Albinism in humans is inherited as a simple...Ch. 3 - Which of Mendels postulates are illustrated by the...Ch. 3 - Discuss how Mendels monohybrid results served as...Ch. 3 - What advantages were provided by Mendels choice of...Ch. 3 - Mendel crossed peas having round seeds and yellow...Ch. 3 - Based on the preceding cross, what is the...Ch. 3 - Which of Mendels postulates can only be...Ch. 3 - In a cross between a black and a white guinea pig,...Ch. 3 - What is the basis for homology among chromosomes?Ch. 3 - In Drosophila, gray body color is dominant to...Ch. 3 - How many different types of gametes can be formed...Ch. 3 - Mendel crossed peas having green seeds with peas...Ch. 3 - In a study of black guinea pigs and white guinea...Ch. 3 - Mendel crossed peas having round green seeds with...Ch. 3 - Prob. 17PDQCh. 3 - The following are F2 results of two of Mendels...Ch. 3 - In assessing data that fell into two phenotypic...Ch. 3 - Prob. 20PDQCh. 3 - Consider the following pedigree. Predict the mode...Ch. 3 - Draw all possible conclusions concerning the mode...Ch. 3 - Prob. 23PDQCh. 3 - Prob. 24PDQCh. 3 - For decades scientists have been perplexed by...Ch. 3 - A wrongful birth case was recently brought before...Ch. 3 - TaySachs disease (TSD) is an inborn error of...Ch. 3 - Datura stramonium (the Jimsonweed) expresses...Ch. 3 - The wild-type (normal) fruit fly, Drosophila...Ch. 3 - Prob. 31ESPCh. 3 - To assess Mendels law of segregation using...Ch. 3 - Albinism, caused by a mutational disruption in...Ch. 3 - (a) Assuming that Migaloos albinism is caused by a...Ch. 3 - Prob. 35ESPCh. 3 - Prob. 36ESP
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- What can you conclude based on the value of the computed Chi-square? How can you relate the two principles of Mendel to Chi-Square Values?arrow_forwardA dihybrid cross is performed between two heterozygous individuals (heterozygous for two traits). The resulting offspring had 62 individuals that were dominant for trait 1 and 2 (D/D), 7 individuals that were R/R, 21 individuals that were R/D, and 25 individuals that were D/R. Using Mendelian inheritance as the null hypothesis, use χ2 analysis to determine if the trait follows Mendelian inheritance. A. How many D/D phenotype offspring are expected? B. How many R/D phenotype offspring are expected? C. How many degrees of freedom are there? D. What is your calculated χ2 value? E. What is the critical value if using a probability of 0.05? F. Does the trait follow Mendelian inheritance?arrow_forwardIn a series of three experiments, M. G. Addo sought to use chi-square test to determine the goodness of fit at 5% for each of his results below. Carry out the calculations and determine whether the results obtained in each case are consistent with the 3:1 or 1:1:1:1 ratio he predicted in his hypothesis. Comment on your results. CROSS Tall x short Purple x white Round yellow (F1) x Wrinkled green (F1) RESULTS HΥΡΟΤHESIS 712:352 3:1 705 : 224 3:1 31:26:27:26 1:1:1:1arrow_forward
- Mendel crossed two Pea plants for plant height and flower color Tall plant (T) is dominant to Short Plant (t). Purple Flower (P) is dominant to white flower (p). Using the following information perform the dihybrid cross using punnett squares that will predict all possible genotypes of the offspring and list the number and description of the phenotypes of the offspring. A. One plant homozygous dominant for plant height and flower color crossed with another plant homozygous recessive for plant height and heterozygous for flower color.arrow_forwardIn Mendel's experiments, he obtained the following results: Which phenotypes can be said to be dominant?arrow_forwardGregor Mendel followed specific steps when breeding pea plants to determine the underlying cause and mathematical ratio of specific traits. Drag and drop the four tiles that describe the order of his process and findings. 1 Mendel crossed two identical, true-breeding plants for a particular characteristic. :: Mendel crossed two contrasting true-breeding plants with two different traits for a given characteristic. ::: Mendel labeled the offspring of the first plants bred F1, for first flower generation. Mendel labeled the offspring of the first plants bred F1, for first filial generation. ::: Mendel cross-bred F1 plants with F2 plants. ::: Mendel cross-bred F1 plants with each other. The F2 population produced flowers in a 3:1 ratio of dominant to recessive traits. The F1 population produced flowers in a 3:1 ratio of dominant to recessive traits. DELL & 6 7 1 8 9. 2 earrow_forward
- In a dihybrid cross of two bi-allelic Mendelian genes, A (two alleles – completely dominant A and recessive a), and B (two alleles – completely dominant B and recessive b), such that the parental generation comprises of pure-bred homozygotes (i.e. AABB with aabb), what are the expected genotype and phenotype ratios in the F1 and F2, if the F1 cross is (a) an intercross and if the F1 cross is (b) a testcrossarrow_forwardIn classical Mendelian genetics, how can one check the genotype of a parent (A) expressing the characters of a dominant allele? Select one: a. By performing a back cross with a recessive homozygote parent (B). If the A parent is homozygote for the dominant allele, then all the individuals from the F1 will display the dominant character. If the parent A was, instead, a heterozygote, then 50% of the F1 progeny will express the recessive character (homozygote recessive) and 50% the dominant one (heterozygotes). b. It is impossible to check such genotype without using specific molecular assays. c. By performing a back cross with a dominant homozygote parent (B). If the A parent is homozygote for the dominant allele, then all the individuals from the F1 will display the dominant character.arrow_forwardWork out the Chi-Squared Goodness of Fit for Mendel's actual data on stem length: "Expt. 7: Length of stem. –– Out of 1,064 plants, in 787 cases the stem was long, and in 277 short. Hence a mutual ratio of 2.84:1. In this experiment the dwarfed plants were carefully lifted and transferred to a special bed. This precaution was necessary, as otherwise they would have perished through being overgrown by their tall relatives. Even in their quite young state they can be easily picked out by their compact growth and thick dark–green foliage." F2: O E O-E (O-E)^2 (O-E)^2/E # short 277 266 11 121 0.455 # tall 787 798 -11 121 0.152 sums: 1,064 1064 0.607 degrees of freedom = 2-1=1 Chi-squared value = .607 p-value: 0.25 < p < 0.50 Work out the Chi-Squared Goodness of Fit for Mendel's actual data on flower position: "Expt. 6: Position of flowers. –– Among 858 cases 651…arrow_forward
- What are the expected phenotypic ratios from the following cross:Tt Rr yy Aa × Tt rr YY Aa, where T = tall, t = dwarf, R = round,r = wrinkled, Y = yellow, y = green, A = axial, a = terminal; T, R,Y, and A are dominant alleles. Note: Consider using the multiplication method in answering this problemarrow_forwardWhich of the following statements is true about the molecular basis of Mendel's second law? OThe random attachment of the sister chromatids and the split of them into different daughter cells during the second round of cell division in Meiosis ensure Mendel's second law of independent assortment of alleles. OThe random attachment of the sister chromatids and the split of them into different daughter cells during the first round of cell division in Meiosis ensure Mendel's 2nd law. O The molecular basis of independent assortment of alleles is the segregation of sister chromatids during the second round of cell division in Meiosis. O The molecular basis of independent assortment of alleles is the segregation of homologous chromosomes during the second round of cell division in Meiosis. O The molecular basis of independent assortment of alleles is the segregation of homologous chromosomes during the first round of cell division in Meiosis.arrow_forwardIn corn plants, the allele for purple kernel (P) is dominant over the allele for yellow kernel (p), and the allele for smooth kernel (S) is dominant over the allele for rough kernel (s) . Obtain an ear of corn from the supply table. You will be examining the results of the cross PpSs x PpSs. Complete the Punnett Square:arrow_forward
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