1. It's a very well known fact that most dragons do not breathe fire, and it's also well known that alleles that enable fire breathing are recessive to the wild-type Sweet Breath phenotype. However, it's unclear how many different genes when mutated cause the fire breathing phenotype. To answer this guestion you use four different strains of dragon and perform a series of crosses. Your strains are the Hungarian Horntail (Strain A), the Norwegian Ridgeback (Strain B), the Romanian Longhorn (Strain C), and the Swedish Shortsnout (Strain D). All strains are true breeding for a single homozygous recessive mutation that allows them to breathe fire. Cross #1 Hungarian Horntail (A) X Norwegian Ridgeback (B) = F1: All Sweet Breath Cross #2 Romanian Longhorn (C) X Swedish Shortsnout (D) = F1: All Sweet Breath Cross #3 Swedish Shortsnout (D) X Norwegian Ridgeback (B) F1: All Firebreathing Cross #4 Romanian Longhorn (C) X Hungarian Horntail (A) = F1: All Firebreathing How many genes are involved in creating a firebreathing phenotype? Group the dragons (A, B, C, and D) as to which contain mutations in the different genes.

1. It's a very well known fact that most dragons do not breathe fire, and it's also well known that alleles that enable fire breathing are recessive to the wild-type Sweet Breath phenotype. However, it's unclear how many different genes when mutated cause the fire breathing phenotype. To answer this guestion you use four different strains of dragon and perform a series of crosses. Your strains are the Hungarian Horntail (Strain A), the Norwegian Ridgeback (Strain B), the Romanian Longhorn (Strain C), and the Swedish Shortsnout (Strain D). All strains are true breeding for a single homozygous recessive mutation that allows them to breathe fire. Cross #1 Hungarian Horntail (A) X Norwegian Ridgeback (B) = F1: All Sweet Breath Cross #2 Romanian Longhorn (C) X Swedish Shortsnout (D) = F1: All Sweet Breath Cross #3 Swedish Shortsnout (D) X Norwegian Ridgeback (B) F1: All Firebreathing Cross #4 Romanian Longhorn (C) X Hungarian Horntail (A) = F1: All Firebreathing How many genes are involved in creating a firebreathing phenotype? Group the dragons (A, B, C, and D) as to which contain mutations in the different genes.

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter13: Modern Understandings Of Inheritance

Section: Chapter Questions

Problem 2VCQ: Figure 13.4 Which of the following statements is true? Recombination of the body color and red/...

See similar textbooks

Similar questions

. In 1919, Calvin Bridges began studying an X-linkedrecessive mutation causing eosin-colored eyes inDrosophila. Within an otherwise true-breedingculture of eosin-eyed flies, he noticed rare variantsthat had much lighter cream-colored eyes. By intercrossing these variants, he was able to make a truebreeding cream-eyed stock. Bridges now crossedmales from this cream-eyed stock with true-breedingwild-type females. All the F1 progeny had red (wildtype) eyes. When F1 flies were intercrossed, the F2progeny were 104 females with red eyes, 52 maleswith red eyes, 44 males with eosin eyes, and14 males with cream eyes. Assume that thesenumbers represent an 8:4:3:1 ratio.a. Formulate a hypothesis to explain the F1 and F2results, assigning phenotypes to all possiblegenotypes.b. What do you predict in the F1 and F2 generations if the parental cross is between truebreeding eosin-eyed males and true-breedingcream-eyed females?c. What do you predict in the F1 and F2 generationsif the parental cross is…
. As we learned in this chapter, the white mutation ofDrosophila studied by Thomas Hunt Morgan isX-linked and recessive to wild type. When truebreeding white-eyed males carrying this mutationwere crossed with true-breeding purple-eyed females,all the F1 progeny had wild-type (red) eyes. When theF1 progeny were intercrossed, the F2 progeny emergedin the ratio 3/8 wild-type females: 1/4 white-eyedmales: 3/16 wild-type males: 1/8 purple-eyed females:1/16 purple-eyed males.a. Formulate a hypothesis to explain the inheritanceof these eye colors.b. Predict the F1 and F2 progeny if the parental crosswas reversed (that is, if the parental cross wasbetween true-breeding white-eyed females andtrue-breeding purple-eyed males).
1. Strong Avengers (Earth's Mightiest Heroes) who normally have tall stature (+), long hair (+), and love hammers (+). But three recessive mutations showed up (likely due to gamma radiation) that result in short statured (ss) Avengers that have short hair (sh) and are allergic to hammers (ah). Data from a three-point testcross are shown below: Phenotype Genotype # Scored Normal + + + 351 Short statured Ss + + 81 Short hair 39 Allergic to hammers + + ah 4 Short statured, rounded ears 1 Short statured, allergic to hammers ss + ah 45 Short hair, allergic to hammers + sh ah 78 Short statured, short hair, allergic to hammers ss sh ah 401 ТОTAL 1000 (a) List the genotypes that are blank in the Table. (b) List the parental and double-recombinant genotypes, make sure you clearly identify which is which. (c) Calculate the map distance between the following two adjacent genes: short statured (ss) and allergic to hammers (ah). (d) Are allergic to hammers (ah) and short hair (sh) genes located on…
. The normal eye color of Drosophila is red, but strains inwhich all flies have brown eyes are available. Similarly,wings are normally long, but there are strains with shortwings. A female from a pure line with brown eyes andshort wings is crossed with a male from a normal pureline. The F1 consists of normal females and short-wingedmales. An F2 is then produced by intercrossing the F1.Both sexes of F2 flies show phenotypes as follows:38 red eyes, long wings38 red eyes, short wings18 brown eyes, long wings18 brown eyes, short wingsDeduce the inheritance of these phenotypes; useclearly defined genetic symbols of your own invention.State the genotypes of all three generations and thegenotypic proportions of the F1 and F2.Unpacking Problem 31Before attempting a solution to this problem, try answering thefollowing questions:1. What does the word normal mean in this problem?2. The words line and strain are used in this problem.What do they mean, and are they interchangeable?3. Draw a simple…
3. Short-tailed pup distribution. In mice, the dominant T allele results in a short tail. Homozygous T/T genotype is lethal, which means mouse embryos with this genotype die before they are born. Homozygous t/t is normal, with normal tail. A cross between two short-tailed mice produces a litter of 5 pups. a). What are the genotype(s) of the two short-tailed mice used in the cross? b). What are the possible genotypes and phenotypes of the 5 pups? What are their perspective ratio?
1. The allele G for yellow stigma is completely dominant to green (g). Supposingtwo strains of autotetraploid plants are available and their genotypes are as follows:GGgg – in this plant the gene is close to the centromereGggg – in this plant the gene is far from the centromere If these two plants are crossed:a) provide the gametes that can be obtained from the two plants;b) provide the genotypic and phenotypic ratios of the offspring. 2. Consider the illustration below. Diagram the configuration you would observe at Anaphase I if crossing-over happens within the inversion. (IMAGE ATTACHED)
Genetic Crosses that Involve 2 Traits In rabbits, black hair is dominant to brown hair. Also in rabbits, long straight ears are dominant to floppy ears. These letters represent the genotypes and phenotypes of the rabbits: BB = black hair EE = long ears Ee = long ears ee = floppy ears %3D %3D Bb = black hair bb = brown 1. A male rabbit with the genotype BBee is crossed with a female rabbit with the genotype bbEe The square Is set up below. Fill it out and determine the phenotypes and proportions in the offspring. How many out of 16 have black hair Ве Ве Ве Ве and long ears? bE How many out of 16 have black hair and floppy ears? be How many out of 16 have brown hair and long ears? bE How many out of 16 have brown hair be and floppy ears?
1. 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 F, female crossed with a resistant male were all sensitive. I 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 Big pink circle Small half-blue-half-pink circle Small blue circle Small pink circle - male cytoplasm - female cytoplasm F, nucleus male nucleus -female nucleus
12) You are studying two flies that you found in your lab one has bristly hair the other has dumpy wings. You cross these two mutants together and all the resultant flies are wild type. You then cross these animals to a double mutant and collect the following offspring. Phenotype Observed Normal 92 Bristly Hair 122 Dumpy Wings 129 Bristly Hair and Dumpy Wings 97 Which are your dominant alleles? Check all that apply O Bristly Hair O Normal Wings O Dumpy Wings O Normal hair 13) Assuming that these traits are sorting in a Mendelian way, and given that you collected 440 flies from your cross, how many flies do you expect that have just bristly hair _(1). ;bristly hair and dumpy wings_(2). 1) 2) 14) Given the numbers you just calculated for expected and the observed listed earlier, what is the chi-squared value for this cross? Numerical answer
. In a particular kind of ornamental flower, the wildtype flower color is deep purple, and the plants aretrue-breeding. In one true-breeding mutant stock, theflowers have a reduced pigmentation, resulting in alavender color. In a different true-breeding mutantstock, the flowers have no pigmentation and are thuswhite. When a lavender-flowered plant from the firstmutant stock was crossed to a white-flowered plantfrom the second mutant stock, all the F1 plants hadpurple flowers. The F1 plants were then allowed toself-fertilize to produce an F2 generation. The 277 F2plants were 157 purple : 71 white : 49 lavender. a. Explain how flower color is inherited. Is this traitcontrolled by the alleles of a single gene?b. What kinds of progeny would be produced if lavender F2 plants were allowed to self-fertilize?
1. 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 F, 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 Big pink circle Small half-blue-half-pink circle - F, nucleus Small blue circle - male cytoplasm - female cytoplasm - male nucleus - female nucleus Small pink circle
8. In guinea pigs, the allele for black fur (B) is dominant over the allele for white fur (b). Similarly, the allele for straight hair (H) is dominant over the allele for having rough hair (h). Pure breeding rough haired guinea pigs with black fur were crossed with pure breeding straight haired guinea pigs with white fur. a. State the genotype and the phenotype of the F1 individuals produced as a result of this cross.b. Two F1 offspring were mated together. Calculate the expected ratio of phenotypes in the F2 generation.c. Show the completed Punnett square below.