J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore. 1943. Journal of Heredity 34:3–7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following crosses, producing the progeny indicated. Parent phenotypes Progeny phenotypes burnsi × burnsi 39 burnsi, 6 pipiens burnsi × pipiens 23 burnsi, 33 pipiens burnsi × pipiens 196 burnsi, 210 pipiens a. On the basis of these results, is the burnsi phenotype most likely inherited as a dominant trait or as a recessive trait? b. Give the most likely genotypes of the parent in each cross (use B for the burnsi allele and B+ for pipiens allele). c. Use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected on the basis of your proposed genotypes.

J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore. 1943. Journal of Heredity 34:3–7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following crosses, producing the progeny indicated. Parent phenotypes Progeny phenotypes burnsi × burnsi 39 burnsi, 6 pipiens burnsi × pipiens 23 burnsi, 33 pipiens burnsi × pipiens 196 burnsi, 210 pipiens a. On the basis of these results, is the burnsi phenotype most likely inherited as a dominant trait or as a recessive trait? b. Give the most likely genotypes of the parent in each cross (use B for the burnsi allele and B+ for pipiens allele). c. Use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected on the basis of your proposed genotypes.

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Description: J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore. 1943. Journal of Heredity 34:3–7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked sp

Concepts of Biology

1st Edition

ISBN:9781938168116

Author:Samantha Fowler, Rebecca Roush, James Wise

Publisher:Samantha Fowler, Rebecca Roush, James Wise

Chapter8: Patterns Of Inheritance

Section: Chapter Questions

Problem 2ACQ: Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are...

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J. A. Moore investigated the inheritance of spotting patterns in leopard frogs. The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the crosses in the table, producing the progeny indicated. (Moore, 1943). Cross Parent phenotypes Progeny phenotypes, observed 1 burnsi x burnsi 39 burnsi, 6 pipiens 2 burnsi × pipiens 23 burnsi, 33 pipiens 3 burnsi x pipiens 196 burnsi, 210 pipiens
J.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.
Coat color in mice is influenced by two genes, one for color (A) and one for the amount of pigment production (C). Mice with the wild type agouti coat color have a yellowish to brownish color. Mice also have a gene that determines the amount of pigment the hair produces. Mouse Coat Phenotype and Genotype Correlation Agouti coat AA, Aa Solid gray/black coat aa Pigment production CC, Cc Albinism cc Multiple crosses were made between male and female mice that were each heterozygous for both traits (AaCc). The data table shows the number of mice of each coat type. Calculate the average F1 generation coat color to answer the question. Coat Color Number of Mice Born in Each Trial Cross Mean Number of Mice 1 2 3 4 5 6 7 8 9 10 Agouti coat 11 8 9 9 10 8 10 7 10 9 Solid gray/black coat 5 2 1 3 4 3 4 3 3 2 Albinism 3 3 6 4 4 3 3 7 4 3 Which…
If a homozygous tall pea plant is crossed with a homozygous shortpea plant, what are the possible genotype and phenotype of theoffspring? Legend: T- tall pea plant t- short pea plant Genotypes of the parents: TT x tt Question to answer ff: 1. What is the genotypic ratio? 2. What are the possible phenotypes of the offspring? 3. What is the phenotypic ratio?
Two pure-breeding lines of petunia plants are crossed. Line 1 plants grow to a height of 54 cm, and Line 2 plants grow to a height of 18 cm. Petunia plant height is controlled by three genes, A, B and C. Line 1 has the genotype A₁A₁B₁B₁C₁C₁, and line 2 has the genotype A2A2B₂B₂C₂C₂. Assume that genotype alone determines plant height under ideal growth conditions and that the alleles of the three genes are additive. If the F1 plants are self crossed, what is the expected proportion of F2 plants with the genotype A₁A₁B₁B₁C₁C₁ 1/8 1/32 1/16 1/4 1/64
Punnett Square for a Dihybrid Cross to Investigate Coat Colour in Mice be Вс BbCc Answer: D Ova BBCC BbCc bc) bb Cc (BC) BbCC D BbCc Bbcc X BBCC bbcc BBcc bbcc Bb Cc BC O Sperm BbCC Bb Cc Bbcc bc BBCC bbcc Вс BbCc Legend Black Brown White Coat colour in mice is controlled by the interaction of two genes. Three phenotypes result: black coat, brown coat, and white coat. The expected phenotypic ratio resulting from a cross between a bbCc female mouse and BbCc male mouse, is black: brown : white Enter only the number values in the order black brown white (Do not include spaces, commas, etc.)
Using the following information perform following crosses using the forked line and predict all possible phenotypes types of the offspring. List the number and description of the phenotypes of the offspring. 1. aa BB Cc Dd Ee X aa bb Cc Dd EE 2. Ff Rr Gg dd Bb X Ff RR Gg Dd Bb 3. Dd gg JJ bb Hh Ee X DD GG Jj Bb Hh Ee
A variety of opium poppy (Papaver somniferum L.) with lacerate leaves was crossed with a variety that has normal leaves. All the F1 had lacerate leaves. Two F1 plants were interbred to produce the F2. Of the F2, 249 had lacerate leaves and 16 had normal leaves. Give genotypes for all the plants in the P, F1, and F2 generations. Explain how lacerate leaves are determined in the opium poppy.
A pea plant has a mendilian completely dominant gene for height flower colour phenotype. A pure breeding tall purple flower plant is crossed with a white dwarf plant. All offspring are purple and tall. If one of these offspring are crossed again with the white dwarf plant, what is the chances of producing a tall white flower? Explain your answer and any assumptions that you had to make. Note: You may not need the entire punnett square below or you may wish to use a different method.
A homozygous variety of opium poppy (Papaver somniferum Laciniatum) with lacerate leaves was crossed with another homozygous variety with normal leaves. All the F₁ had lacerate leaves (jagged-edged leaves). Two F₁ plants were crossed to produce the F2. Of the F2, 249 had lacerate leaves and 16 had normal leaves. How are lacerate leaves determined in the opium poppy? two genes, with a dominant allele at both loci two genes, where one allele at a single locus is recessive lethal one gene, with codominant alleles at a single locus two genes, with a dominant allele at either or both loci Give genotypes for all the plants in the P, F₁, and F2 generations. Some generations will have more than one genotype. P F₁ F₂ (lacerate) F₂ (normal) aabb A_B_ Answer Bank aaB AaBb A_bb AABB
An ebony strain of flies was discovered to be sensitive to carbon dioxide. Crossing a female sensitive strain with male resistant strain gave all sensitive offspring. The offspring of an F1 female crossed with a resistant male were all sensitive. Using the following key to your illustrations using shapes, make a reciprocal cross up to the F2 generation. Put your illustrations in the space provided below. Label the phenotypes of all individuals in the reciprocal cross. Adjust spacing, if necessary. Make sure that the complete cross(es) can fit the same page. Big blue circle - male cytoplasm Big pink circle - female cytoplasm Small half-blue-half-pink circle - F1 nucleus Small blue circle - male nucleus Small pink circle - female nucleus give a diagram please
An ebony strain of flies was discovered to be sensitive to carbon dioxide. Crossing a female sensitive strain with male resistant strain gave all sensitive offspring. The offspring of an F1 female crossed with a resistant male were all sensitive. Using the following key to your illustrations using shapes, make a reciprocal cross up to the F2 generation. Put your illustrations in the space provided below. Label the phenotypes of all individuals in the reciprocal cross. Adjust spacing, if necessary. Make sure that the complete cross(es) can fit the same page. Big blue circle - male cytoplasm Big pink circle - female cytoplasm Small half-blue-half-pink circle - F1 nucleus Small blue circle - male nucleus Small pink circle - female nucleus