Consider the maternal effect gene that controls snail coiling where D= dextral, dominant, d=sinistral, recessive. Indicate the genotype(s) and phenotype(s) of offspring from the following crosses - females are written on the left, males on the right for each cross. Progeny Phenotype(s) (Include ratio) 1. Dd x Dd 2. dd x Dd 3. Dd x DD 4. Dd x dd
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- Pigment of chicken feathers is regulated by two genes, the gene for feather pigment C, and a gene that inhibits pigment production, I. A mother chicken, who is homozygous dominant for gene C, and heterozygous for gene I, is crossed with a father chicken, who is heterozygous for the pigment gene, and heterozygous for gene I.Inhibition is dominant to uninhibited, pigmented feathers are dominant to unpigmented feathers.In a population of 150 chickens, produced from the same parents, how many chickens would you expect to have pigmented feathers?Pigment of chicken feathers is regulated by two genes, the gene for feather pigment C, and a gene that inhibits pigment production, I. A mother chicken, who is heterozygous for gene c, and heterozygous for gene I, is crossed with a father chicken, who is homozygous recessive for the pigment gene, and heterozygous for gene I. Inhibition is dominant to uninhibited, pigmented feathers are dominant to unpigmented feathers. The inhibitor gene (I) will prevent expression of pigmented chicken feathers. Here is a DihybridCrossReview(8).pdf Download DihybridCrossReview(8).pdfreviewing the FOIL method for determining gametes and performing a dihybrid cross if you need a refresher. Use the letters C, c, I, and i for the genes. C is for color and I is for inhibitor Enter your answer as a list of haplotypes separated by spaces such as AB ab Ab aB. Do not write a sentence. Do not list duplicate haplotypes. Alphabatize your haplotypes, e.g. Ab, not bA. A) What are the potential genotypes of…In Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1
- 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…As seen in the photo, Labradors come in three colors-- black, brown and yellow. What is the genetic basis for these different coat colors? One gene produces melanin, a pigment which is deposited in the dog's fur and makes the color dark. With this gene, allele B (black) is dominant to allele b. Only in the case of a recessive homozygote (bb) will the dog's phenotype be brown. The regulatory gene is separate from the melanin gene but it acts as a switch, either turning the melanin gene on or turning it off. Allele E is dominant and allows for the melanin to be deposited in the dog's fur ("on" switch), but if the switch gene is a recessive homozygote, the melanin is blocked ("off" switch) and a yellow dog is the result! Review the information on Labrador retrievers above. What are the phenotypic ratios of the F1 generation offspring of two dihybrids? Make sure to match the numbers with coat colors (e.g, which number in the ration goes with which color). Use a Punnett square to…As seen in the photo, Labradors come in three colors-- black, brown and yellow. What is the genetic basis for these different coat colors? One gene produces melanin, a pigment which is deposited in the dog's fur and makes the color dark. With this gene, allele B (black) is dominant to allele b. Only in the case of a recessive homozygote (bb) will the dog's phenotype be brown. The regulatory gene is separate from the melanin gene but it acts as a switch, either turning the melanin gene on or turning it off. Allele E is dominant and allows for the melanin to be deposited in the dog's fur ("on" switch), but if the switch gene is a recessive homozygote, the melanin is blocked ("off" switch) and a yellow dog is the result! 1. Two other Labradors mate and produce puppies. Their genotypes are Bbee and BbEe. What color are each parent and what are the phenotypic rations of their offspring in the F1 Generation? Show your work with a Punnett square.
- Radishes may be long, round, or oval, and they may bered, white, or purple. You cross a long, white variety witha round, red one and obtain an oval, purple F1. TheF2 shows nine phenotypic classes as follows: 9 long, red;15 long, purple; 19 oval, red; 32 oval, purple; 8 long,white; 16 round, purple; 8 round, white; 16 oval, white;and 9 round, red.a. Provide a genetic explanation of these results. Besure to define the genotypes and show the constitutionof the parents, the F1, and the F2.b. Predict the genotypic and phenotypic proportionsin the progeny of a cross between a long, purple radishand an oval, purple one.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…
- Scenario A In goats, black fur is dominant to red fur. The gene for this trait can be described as Bb (if you forgot why this is so, go back and reread the background information at the beginning of this exercise). A homozygous black female goat (nanny) is crossed with a heterozygous male (billy). 1 What is the genotype of the nanny? What is the phenotype of the nanny?In goats, black fur is dominant to red fur. The gene for this trait can be described as Bb (if you forgot why this is so, go back and reread the background information at the beginning of this exercise). A homozygous black female goat (nanny) is crossed with a heterozygous male (billy). 1 What is the genotype of the nanny? Fill input: What is the phenotype of the nanny?In mice, black fur (W) is dominant over white fur (w), and black eyes are dominant (R) over red eyes (r). The two genes are located on different chromosomes.A male mouse with black fur and black eyes mated with a female mouse with white fur and red eyes. Out of 64 offspring, 30 had black fur and black eyes, and 34 had white fur and black eyes. What is the genotype of the male parental mouse? Select one: a. WwRr b. WwRR c. WWRR d. WWRr