In Labrador Retrievers, a single gene controls whether an individual dog will be black or brown. Black labs have at least one copy of the dominant allele (B), while chocolate labs have two copies of the recessive allele (b). Imagine that you have a black lab of unknown genotype. Design a test cross that would enable you to determine its genotype.

In Labrador Retrievers, a single gene controls whether an individual dog will be black or brown. Black labs have at least one copy of the dominant allele (B), while chocolate labs have two copies of the recessive allele (b). Imagine that you have a black lab of unknown genotype. Design a test cross that would enable you to determine its genotype.

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter5: The Inheritance Of Complex Traits

Section: Chapter Questions

Problem 2QP: The text outlines some of the problems Frederick William I encountered in his attempt to breed tall...

See similar textbooks

Similar questions

Among Native Americans, two types of earwax (cerumen) are seen, dry and sticky. A geneticist studied theinheritance of this trait by observing the types of offspring produced by different kinds of matings. Heobserved the following numbers:OffspringParents Number of mating pairs Sticky DrySticky × sticky 10 32 6Sticky × dry 8 21 9Dry × dry 12 0 42a. How is earwax type inherited?b. Why are no 3:1 or 1:1 ratios present in the datashown in the chart?
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?
A mouse from a true-breeding population with normal gait was crossed to a mouse displaying an oddgait called dancing. The F1 animals all showednormal gait.a. If dancing is caused by homozygosity for the recessive allele of a single gene, what proportion ofthe F2 mice should be dancers?b. If mice must be homozygous for recessive allelesof each of two different genes to have the dancing phenotype, what proportion of the F2 shouldbe dancers if the two genes are unlinked?(Assume that all the mice in the populationwith normal gait were homozygous fordominant alleles.)c. When the F2 mice were obtained, 42 normal and8 dancers were seen. Use the chi-square test todetermine if these results better fit the one-genemodel from part (a) or the two-gene model frompart (b).
Albino rabbits (lacking pigment) are homozygous forthe recessive c allele (C allows pigment formation).Rabbits homozygous for the recessive b allele makebrown pigment, while those with at least one copy ofB make black pigment. True-breeding brown rabbitswere crossed to albinos, which were also BB. F1 rabbits, which were all black, were crossed to the doublerecessive (bb cc). The progeny obtained were 34black, 66 brown, and 100 albino.a. What phenotypic proportions would have beenexpected if the b and c loci were unlinked?b. How far apart are the two loci?
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 sheep, the formation of horns is a sex-influenced trait; the allelethat results in horns is dominant in males and recessive in females.Females must be homozygous for the horned allele to have horns.A horned ram was crossed to a polled (unhorned) ewe, and the firstoffspring they produced was a horned ewe. What are the genotypesof the parents?
raccoons may have wide, medium-sized, or narrow bands around their tails. They may also havethe habit of washing all, or some of their food, or do not wash their food at all. a) assign genotypes to the phenotypes mentioned (see attached table) b. What mode of inheritance would most likely be exhibited by these traits if crosses were made? c. If two raccoons with medium-sized tail bands and have the habit of washing some of theirfoods will be crossed, what is the probability of having F1 raccoons with: c.1 wide tail bands that won’t wash any of their food? c.2 the same genotype as the parent raccoons? d. If a raccoon with a wide tail band that washes only some of its food is crossed with a raccoonwith a narrow tail band that doesn’t wash any food, what percentage of their offspring wouldbe medium-tailed and washes all its food? Show COMPLETE cross.
Manx cats have no tail. This is caused by a dominantmutation M, which is embryo lethal when homozygous. An unlinked gene B controls the color of the fur onthe cat's tail. The dominant B allele produces black fur on tails, while the recessive allele b produceswhite fur on tails. A Manx cat that is B/B is crossed to a white-tailed cat. Half of the F1 progeny hadthe Manx phenotype and the other half had normal, black tails.The F1 cats with the Manx phenotype were crossed to each other, producing F2 cats. What is the ratioof the following phenotypes in the F2
A woman who is heterozygous, Bb, has brown eyes; B (brown) isthe dominant allele, and b (blue) is recessive. One of her eyes,however, has a patch of blue color. Give three different explanationsfor how this might have occurred?
On a fox ranch in Wisconsin, a mutation arose that gavea “platinum” coat color. The platinum color proved verypopular with buyers of fox coats, but the breeders couldnot develop a pure-breeding platinum strain. Every timetwo platinums were crossed, some normal foxes appeared in the progeny. For example, the repeated matings of the same pair of platinums produced 82 platinumand 38 normal progeny. All other such matings gavesimilar progeny ratios. State a concise genetic hypothesisthat accounts for these results.
A corn geneticist wants to obtain a corn plant that hasthe three dominant phenotypes: anthocyanin (A), longtassels (L), and dwarf plant (D). In her collection ofpure lines, the only lines that bear these alleles are AALL dd and aa ll DD. She also has the fully recessive lineaa ll dd. She decides to intercross the first two and testcross the resulting hybrid to obtain in the progeny aplant of the desired phenotype (which would have to beAa Ll Dd in this case). She knows that the three genesare linked in the order written, that the distance between the A/a and the L/l loci is 16 m.u., and that thedistance between the L/l and the D/d loci is 24 m.u.a. Draw a diagram of the chromosomes of the parents,the hybrid, and the tester.b. Draw a diagram of the crossover(s) necessary toproduce the desired genotype.c. What percentage of the testcross progeny will be ofthe phenotype that she needs?d. What assumptions did you make (if any)?
In Ayrshire cattle, the spotting of the coat can be either red andwhite or mahogany and white. The mahogany-and-white phenotypeis caused by the allele SM (where the letter S indicates spotting, andthe superscript M stands for mahogany). The red-and-white phenotypeis controlled by the allele SR (where the superscript R standsfor red). The table below shows the relationships between genotypeand phenotype for males and females: Explain the pattern of inheritance.