Another trait you are studying in monster genetics is fur texture. A mommy monster (P1) who is homozygous for spiky fur and a daddy monster (P1) who is homozygous for feathery fur have a litter of monsters (F1). All of the F1 monsters have feathery fur. Using the letter, F, for your alleles, determine the following: 1. What is the genotype of the mommy monster? 2. What is the genotype of the daddy monster? 3. What are the genotypes of all of the F1 generation? 4. If two of the F1 offspring are crossed, what is the phenotypic ratio of the result? (A ratio is written as: X:X) 5. What are the genotypes (and how many of each) that result from the cross described in question 4 above?
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- As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progenyWhy Geneticists Use a Variety of Symbols for Alleles ? What are those ?In a hypothetical situation, brachydactyly is caused by a recessive allele (b). If a woman who is heterozygote with normal fingers (Nb) and a man with brachydactyly (bb) have children, what proportion of these children would you expect to have normal fingers? (You may want to draw the punnett square on a scrap piece of paper). Group of answer choices A-50% B-0% C-75% D-100% E-25%
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- .) A female poodle with green eyes (recessive) and yellow spots on her tongue (recessive) was bredwith a male poodle who was the wild-type phenotype but hybrid genotype for those same two traits.In a series of litters, this breeding pair gave birth to the following 85 puppies:39 Normal eyes, normal tongue36 Green eyes, yellow-spot tongue6 Green eyes, normal tongue4 Normal eyes, yellow-spot tonguea.) Use the symbols grn and grn+ for the green eyes gene, and yel and yel for the yellow tonguegene and give the genotypes of the parents of this cross. (Remember, each parent has twoalleles for each gene.)b.) Now give the expected genotypes of the offspring.c.) If these two genes were unlinked, given the way the cross was constructed, what ratio wouldyou expect to see among the four categories of offspring shown here?d.) Now match up the expected phenotypes (answer to 2b) with the actual offspring.e.) Which categories of dogs are “parental types” and which are recombinant types?f.) Do these…A genetic engineer is going to cross two watermelon plants to produce seeds for a spring planting. He is breeding for size, and wants to have as many watermelons with the phenotype for long shape as possible. In watermelons, the allele for short shape (R) is dominant to the allele for long shape (r). Would crossing a watermelon homozygous recessive for the trait with a watermelon heterozygous for the trait give the most long watermelons possible? Explain your answer using Punnett Squares.Draw a punnett square for each problem. Please be careful to notice whether a trait is homozygous or heterozygous. In rabbits it has been shown that albino (white) is recessive to the brown color, such as in the wild rabbit. If a homozygous albino rabbit and a homozygous brown rabbit are bred, what will be the color ratio of the first generation? Answer:___________brown _____________tan______________white 2. If a pure (homozygous) white guinea pig is bred with a hybrid (heterozygous) black guinea pig, what would be the probable color ratio of the next generation? Black is dominant. White is recessive. Answer:___________black _____________gray _____________white Thank you.
- When Gregor Mendel performed his breeding experiments on pea plants, he discovered that tallness in the plants is inherited through a simple dominant trait (coded "T") with shortness as the recessive trait (coded "t"). Imagine that Mendel bred a homozygous dominant pea plant with a pea plant heterozygous for tallness. Draw a Punnett Square to help you answer this question and the next one: In the 1st generation of offspring, we expect the genotypes (for tallness) to include… a) 50% homozygous dominant, 50% heterozygous b) 100% heterozygous c) 50% tall, 50% short d) 100% tall e) None of the above: In poultry, the genotype-phenotype relationships for comb shape are R/– P/–, walnut; R/–p/p, rose, r/r P/–, pea; and r/r p/p, single. What will be the comb characters of the offspring ofthe following crosses?a) A walnut crossed with a single produces offspring that are walnut, rose, pea, and single.b) A rose crossed with a walnut produces offspring that are walnut, rose, pea, and single.c) A rose crossed with a pea produces five walnut and six rose offspring.d) A walnut crossed with a walnut produces one rose, two walnut, and one singleoffspringA 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.