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A planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three
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- Dr. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits.He mates these two strains with each other. Dr. Disney then takes phenotypically wild- type females from this cross and mates them with double recessive males. In the resulting testcross progeny, he observes 500 flies that are of the following makeup:41 with baby blue eyes and pink wings207 with baby blue eyes only210 with pink wings only42 with wild-type phenotype14. Assuming the wild-type alleles for these two genes are b+ and pw+, what is the correct testcross of the F1 flies?A) b+ pw+/b pw ⋅ b pw/b pwB) b+ pw+/b pw ⋅ b pw+/b+ pwC) b+ pw/b pw+ ⋅ b pw/b pwD) b+ pw/b pw+ ⋅ b+ pw+/b pwE) b+ pw+/b pw ⋅ b+ pw/b…arrow_forwardIn the world of Weeping angels, those with pointy wings are (A_), while those with round wings are (aa). In the efforts to explore this extraordinary species, you wish to determine the genotype of a pointy winged angel. Luckily, you have trapped a round winged angel that you can cross your pointy winged angel with. If the pointy winged angel is homozygous what phenotypic ratio do you expect from their offspring? (Pointy wing: round wing) Group of answer choices a. 1:0 b. 2:1 c. 1:1 d. 3:1arrow_forwardAs 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 progenyarrow_forward
- Can you please not type the answer can you write it on a paperarrow_forwardThere are two genetically-determined traits that are seen in a species of salamander. The S-gene codes for spotting: SS or Ss = white spots; ss = no spots. The W-gene codes for skin texture: WW or Ww=wrinkly; ww = smooth. If a cross is done between SSWW and ssww parents, what genotype(s) would you expect to see in the offspring of this dihybrid cross? NOTE: Assume no recombination. O Ss, ss, WW. ww O SW, sw, Sw, sw O SsWw O SSWW, sswwarrow_forwardIn the video game Animal Crossing: New Horizons, flowering breeding is based in genetics. Each flower's color is determined by the genotype at three or four unlinked genes: R, Y, W, and S. The genotype of the elusive blue rose is RR YY ww ss. In the game, one way to get a blue rose is to cross two roses with the Rr Yy Ww ss genotype. A) What types of gametes and in what proportions will a Rr Yy Ww ss rose produce? B) In a cross Rr Yy Ww ss x Rr Yy Ww ss what are the possible offspring genotypes and at what frequency will they each appear? Show your work. C) What proportion of the offspring of the cross will be blue roses?arrow_forward
- Part A) You cross a fly with straight wings with a fly with curved wings to produce the F1 generation, all of which exhibit straight wings. In the F2 generation, you observe 65 straight-winged flies and 16 curved-wing flies. Based upon the results, which phenotype is recessive? Part B) You think this trait is controlled by a single gene, but the F2 numbers don’t quite match a 3:1 expected ratio. Perform a chi-square analysis to determine if the variation you observe is due to chance or not. Based upon this determination, would you keep or reject the “null hypothesis”? Show your work and explain your reasoning.arrow_forward7) Three years later, one of Rocky's and Ruby's sons, who is heterozygous for his curly fur, meets another dog that is also heterozygous for her curly fur. Create a Punnett square to show the possibilities that would result if they had puppies. (4) a) Fill in the Punnett square, then list the possible genotypes and phenotypes for the puppies. b) What are the chances of a puppy with curly fur? out of c) What are the chances of a puppy with straight fur? out of BLOMU CAGA (C 3LS gousur co BeGAGE (p) BROM ncrinoh (e) etone 2po BLOMarrow_forwardIn humans, Cystic Fibrosis an autosomal recessive trait, currently with medicine advances people affected by this disease can live until the 40s and 50s even. A couple in which the mother is a known carrier for this disease plans to have 4 children with a man who is not affected either but had his father who had the disease. A) Please state the genotypes of the parents. Use C to denote the dominant allele and c as the recessive B) Determine the probability that the first 2 children born have Cystic Fibrosis and the 2 last children are not affected. C) In general, what proportion of the children are expected to be female carriers of Cystic Fibrosis trait?arrow_forward
- On the planet of Caracas, in the Stellar Solorais KaChunka Galaxy, there is a population of Nuggetarians. A phenotype of interest is Peanutus02. There are two alleles for the autosomal gene associated with Peanutus02. A true-breeding cross between a female with the Peanutus02 (disease) phenotype and a male with the wild-type phenotype produces the following offspring counts: Phenotype Number of Males Number of Females Wild-Type 550 562 Peanutus02 0 0 Of the following choices, the most likely mode of inheritance for the Peanutus02 phenotype is: On the planet of Caracas, in the Stellar Solorais KaChunka Galaxy, there is a population of Nuggetarians. A phenotype of interest is Peanutus02. There are two alleles for the autosomal gene associated with Peanutus02. A true-breeding cross between a female with the Peanutus02 (disease) phenotype and a male with the wild-type phenotype produces the following offspring counts: Phenotype Number of Males Number of Females Wild-Type…arrow_forwardSearch the menus (Alt+/) A, ? 100% Normal text Calibri + в I UА 11 I 1 2 | 3 5 | 6 7 8 9 10 11. In cockroaches, red exoskeletons (bodies) are recessive to black exoskeletons. A red male cockroach is crossed with a heterozygous black female cockroach. Twenty-four "baby" cockroaches are hatched from the egg. How many of these do you expect to be black? How many do you expects to be red? Allele and Phenotype All Genotype and Phenotype Possibilities Parent Genotypes Punnett Square Answer Letter Phenotype Genotype Phenotype Dad Momarrow_forwardIn your experiments with Drosophila, you found that the wildtype allele that codes for grey adult insects is sometimes mutated, with mutants having a black colour. Similarly, you find that your population has alleles that code for vestigial wings instead of the wildtype normal wings. (Both mutations are recessive.) You want to know whether the two genes for body colour and wings are linked. In your preliminary crosses you have established a large number of individuals that you require for testcrosses. 9a. Which two genotypes will you use for the testcross? (Hint: One of the two genotypes should be a heterozygous dihybrid.) Use the common wildtype notation (e.g., “ar+ tg”) rather than the allele notation (VvZZ). For simplification, disregard the gender.arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning