During the study of your genetics course, you are learning the different developmental stages of a unique Tsetse fly, named defsis-timen, and you discover differences in their developmental pattern i.e. unusual and normal...... Rest of the question is in pictureAnswer part b and c And in part c its unlinked genes not unlined b) Kindly explain: what type of cross is cross 2?c) If the genes for both the characteristics are unlined, what would be the result of cross 2 interms of phenotypic classes and in what ratios? After careful experimentation; you have observed that unusual developmental pattern phenotypeis recessive. Your friend is also working on another aspect of same experimental model andstudying a changed pattern that produces short halters. The short legs phenotype is recessive tothe normal halters phenotype. You label the short halters allele as (hal) and the normal haltersallele as (+). You set up the following crosses:Unusual pattern, long haltersNormal pattern, short haltersCross 1:unu+hal (n)F1:Normal pattern, Long haltersCross 2:Normal pattern, long haltersUnusual pattern, short halters

Crossing In Mendelian genetics, crossing is the technique of mating two plant of same species with…

Answered: Answer part b and c And in part c its…

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Answer part b and c And in part c its unlinked genes not unlined

BIOLOGY:CONCEPTS+APPL.(LOOSELEAF)

10th Edition

ISBN:9781305967359

Author:STARR

Publisher:STARR

Chapter13: Patterns In Inherited Traits

Section: Chapter Questions

Problem 6GP: In sweet pea plant, an allele for purple flowers (P) is dominant when paired with a recessive allele...

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Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?
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.
When many families were tested for the ability to tastethe chemical phenylthiocarbamide, the matings weregrouped into three types and the progeny were totaled,with the results shown below:ChildrenNumber NonParents of families Tasters tastersTaster × taster 425 929 130Taster × nontaster 289 483 278Nontaster × nontaster 86 5 218With the assumption that PTC tasting is dominant (P)and nontasting is recessive (p), how can the progenyratios in each of the three types of mating be accountedfor?
A recessive allele in mice results in an unusally long neck. Sometimes, during early embryonic development, the long neck causesthe embryo to die. An experimenter began with a population oftrue-breeding normal mice and true-breeding mice with longnecks. Crosses were made between these two populations to produce an F1 generation of mice with normal necks. The F1 micewere then mated to each other to obtain an F2 generation. For themice that were born alive, the following data were obtained:522 mice with normal necks62 mice with long necksWhat percentage of homozygous mice (that would have had longnecks if they had survived) died during embryonic development?
In Figure 1-6, the students have 1 of 15 different heights,plus there are two height classes (4′11″ and 5′ 0″) forwhich there are no observed students. That is a total of17 height classes. If a single Mendelian gene can account for only two classes of a trait (such as purple orwhite flowers), how many Mendelian genes would beminimally required to explain the observation of 17height classes?
Eye Color is a trait that involves a number of genes. To simplify our example, we can break down the traitinto two phenotypes: dark color and light color eyes. Dark coloration is considered dominant, or isexpressed in a heterozygous condition (when the alleles are different).Q: If you have one light eyed parent and one parent that is homozygous for dark eye color, what are thepotential eye colors of their offspring (young)? To start, what do the Parents’ alleles look like? What istheir Genotype? (Remember, you pick which letter you want to use)Light eye colored parent:Dark eye colored parent:One of the things Mendel designed was a way to provide information about how the genes are passed onfrom parent to offspring. He designed something called a Punnett Square. This square places the parents’alleles outside the box, then uses those alleles to fill in the squares.Let’s start with a reminder of what we know about the parents: Parent Phenotype(appearance) Genotype(alleles) 1 2
In a particular human family, John and his motherboth have brachydactyly (a rare autosomal dominant allele causing short fingers). John’s father hasHuntington disease (another rare autosomal dominant allele). John’s wife is phenotypically normaland is pregnant. Two-thirds of people who inheritthe Huntington (HD) allele show symptoms by age50, and John is 50 and has no symptoms.Brachydactyly is 90% penetrant.a. What are the genotypes of John’s parents?b. What are the possible genotypes for John? Howlikely is John to have each of these genotypes?c. What is the probability the child will express bothbrachydactyly and Huntington disease by age 50 ifthe two genes are unlinked?d. How will your answer to part (c) change if insteadthese two loci are 20 m.u. apart?
A recessive allele in mice results in an unusually long neck. Sometimes,during early embryonic development, the long neck causesthe embryo to die. An experimenter began with a population oftrue-breeding mice with normal necks and true-breeding mice withlong necks. Crosses were made between these two populations toproduce an F1 generation of mice with normal necks. The F1 micewere then mated to each other to obtain an F2 generation. For themice that were born alive, the following data were obtained:522 mice with normal necks62 mice with long necksWhat percentage of homozygous mice (that would have had longnecks if they had survived) died during embryonic development?
Three recessive traits in garden pea plants are as follows: yellowpods are recessive to green pods, bluish green seedlings are recessiveto green seedlings, creeper (a plant that cannot stand up) isrecessive to normal. A true-breeding normal plant with green podsand green seedlings was crossed to a creeper with yellow pods andbluish green seedlings. The F1 plants were then crossed to creeperswith yellow pods and bluish green seedlings. The following resultswere obtained: 2059 green pods, green seedlings, normal151 green pods, green seedlings, creeper281 green pods, bluish green seedlings, normal15 green pods, bluish green seedlings, creeper2041 yellow pods, bluish green seedlings, creeper157 yellow pods, bluish green seedlings, normal282 yellow pods, green seedlings, creeper11 yellow pods, green seedlings, normal Construct a genetic map that indicates the map distances betweenthese three genes?
. A geneticist mapping the genes A, B, C, D, and E makestwo 3-point testcrosses. The first cross of pure lines isA/A ⋅ B/B ⋅ C/C ⋅ D/D ⋅ E/E × a/a ⋅ b/b ⋅ C/C ⋅ d/d ⋅ E/EThe geneticist crosses the F1 with a recessive tester andclassifies the progeny by the gametic contribution ofthe F1:A ⋅ B ⋅ C ⋅ D ⋅ E 316a ⋅ b ⋅ C ⋅ d ⋅ E 314A ⋅ B ⋅ C ⋅ d ⋅ E 31a ⋅ b ⋅ C ⋅ D ⋅ E 39A ⋅ b ⋅ C ⋅ d ⋅ E 130a ⋅ B ⋅ C ⋅ D ⋅ E 140A ⋅ b ⋅ C ⋅ D ⋅ E 17a ⋅ B ⋅ C ⋅ d ⋅ E 131000The second cross of pure lines is A/A • B/B • C/C • D/D •E/E × a/a • B/B • c/c • D/D • e/e.The geneticist crosses the F1 from this cross with arecessive tester and obtainsA ⋅ B ⋅ C ⋅ D ⋅ E 243a ⋅ B ⋅ c ⋅ D ⋅ e 237A ⋅ B ⋅ c ⋅ D ⋅ e 62a ⋅ B ⋅ C ⋅ D ⋅ E 58A ⋅ B ⋅ C ⋅ D ⋅ e 155a ⋅ B ⋅ c ⋅ D ⋅ E 165a ⋅ B ⋅ C ⋅ D ⋅ e 46A ⋅ B ⋅ c ⋅ D ⋅ E 341000The geneticist also knows that genes D and E assortindependently.a. Draw a map of these genes, showing distances inmap units wherever possible.b. Is there any evidence of interference?
In rabbits, black hair is due to a dominant allele B and brown hair to its recessive allele b. Short hair (H) is dominant to long hair (h). Show punnet squares In a cross between a homozygous black, longhaired rabbit and a brown, homozygous shorthaired one, what would the F1 generation look like? If you did not know the genotype of an F1 rabbit, you could determine its genotype by a test cross in which it is crossed with an animal with which phenotype AND genotype? If you carried out this test cross (from 2, using the F1 from 1), what phenotypes and in what ratio would you expect? What phenotypes in what ratio would be expected in the F2 generation of the cross in point 1?
Red eyes is the wild-type phenotype in Drosophila, and severaldifferent genes (with each gene existing in two or more alleles) affect eye color. One allele causes purple eyes, and a different allelecauses vermilion eyes. The purple and vermilion alleles are recessive to red eye color. Two types of crosses provided thefollowing data:Cross 1: Males with vermilion eyes × females with purple eyes354 offspring, all with red eyesCross 2: Males with purple eyes × females with vermilion eyes212 male offspring with vermilion eyes221 female offspring with red eyesExplain the pattern of inheritance based on these results. Whatadditional crosses might you make to confirm your hypothesis?