A complementation analysis was performed using Drosophila which normally have black wings. Six mutants for clear wings were discovered. Themutants were bred in a complementation test to produce the following:Mutant24#11112111111411Where a "0" (zero) means clear wing offspring and a "1" (one) means wildtype (black wing offspring). How many genes can you identify that affectthe production of wing colour in Drosophila?O a. 3O b.4О с. 5O d. 6O e. none of the above3.5

Mutant A species that have mutated gene in his genome.

A complementation analysis was performed using Drosophila which normally have black wings. Six mutants for clear wings were discovered. The mutants were bred in a complementation test to produce the following: Mutant 1 2 3 4 # 1 1 1 1 2 1 1 1 3 1 1 1 4 1 1 6 Where a "0" (zero) means clear wing offspring and a "1" (one) means wildtype (black wing offspring). How many genes can you identify that affect the production of wing colour in Drosophila? O a. 3 O b.4 O c. 5 O d.6 O e. none of the above 5

A complementation analysis was performed using Drosophila which normally have black wings. Six mutants for clear wings were discovered. The mutants were bred in a complementation test to produce the following: Mutant 1 2 3 4 # 1 1 1 1 2 1 1 1 3 1 1 1 4 1 1 6 Where a "0" (zero) means clear wing offspring and a "1" (one) means wildtype (black wing offspring). How many genes can you identify that affect the production of wing colour in Drosophila? O a. 3 O b.4 O c. 5 O d.6 O e. none of the above 5

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter15: Genomes And Genomics

Section: Chapter Questions

Problem 1QP: The gene controlling ABO blood type and the gene underlying nail-patella syndrome are said to show...

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You are given a Drosophila female that looks wild-type but is heterozygous for mutations intan body (t), miniature wings (m), and white eyes (w). You test cross this female with a tanbodied, miniature winged, and white-eyed homozygous mutant male, and you obtain thefollowing 1400 progeny: Phenotype : number+ + + : 608t m w : 516+ m w : 2t + + : 6+ m + : 39t + w : 46+ + w : 81t m + : 102 Calculate the distance between each pair t-m, m-w, and t-w only using the number ofrecombinants between them (i.e. ignoring the gene in the middle). Draw a linear map with thedistances between genes.
In Drosophila, a heterozygous female for the X-linkedrecessive traits a, b, and c was crossed to a male that phenotypically expressed a, b, and c. The offspring occurred inthe following phenotypic ratios.+ b c 460a + + 450a b c 32+ + + 38a + c 11+ b + 9 No other phenotypes were observed.(a) What is the genotypic arrangement of the alleles ofthese genes on the X chromosome of the female?
In Drosophila, a heterozygous female for the X-linkedrecessive traits a, b, and c was crossed to a male that phenotypically expressed a, b, and c. The offspring occurred inthe following phenotypic ratios.+ b c 460a + + 450a b c 32+ + + 38a + c 11+ b + 9 No other phenotypes were observed.(a) What progeny phenotypes are missing? Why?
Consider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PC was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PC products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?
In Drosophila, a female fly is heterozygous for three mutations, Bareyes (B), miniature wings (m), and ebony body (e). Note that Bar isa dominant mutation. The fly is crossed to a male with normal eyes,miniature wings, and ebony body. The results of the cross are asfollows.111 miniature29 wild type117 Bar26 Bar, mimatue101 Bar, ebony31 Bar, miniature, ebony35 ebony115 miniature, ebonyInterpret the results of this cross. If you conclude that linkage isinvolved between any of the genes, determine that map distance(sbetween them.
Females of wild-type Strain A and males of mutant Strain B, as well as females of mutant Strain B and males of wild-type Strain A, make reciprocal crosses. Explain why reciprocal crosses are needed in genetics experiments involving Drosophila fruit flies.
Multiple crosses were made between true-breeding lines of black and yellow Labrador retrievers. All the F1 progeny were yellow. When these progeny were intercrossed, they produced an F2 consisting of 121 yellow, 9 black and 30 chocolate. What epistatic ratio and what kind of epistasis is approximated in the F2? Propose a biochemical pathway for coat color in Labrador retrievers based on the type of epistasis. Correlate each genotype with the phenotype that would occur in your pathway. Also show the frequency of each genotype. A-B- A-bb aaB- aabb
Cinnabar eyes (cn) and reduced bristles (rd) are linked autosomal recesive characters in Drosophila fruit flies. A homozygous wild-type female was crossed to a reduced, cinnabar male, and the F1 females were then crossed with cinnabar reduced males to obtain the F2. Of the 200 F2 offspring obtained, 86 were wild type, 13 were cinnabar, 17 were reduced and 84 were reduced and cinnabar.. What is the map distance between the cn and rd alleles? 8,4 mu 15 mu 17 mu 30 mu 13 mu
You are mapping three linked loci in Drosophila melanogaster (the common laboratory fruit fly). You cross flies that are triply mutant for apricot (pale eyes), bristle (extra bristles) and clipped (notched wings) to wild-type flies. The F+ flies are wild-type in appearance. You then backcross the F+ females to pure-breeding (apricot, bristle, clipped) males and score the phenotypes of 1000 F progeny for all three loci. Here are the results: 359 wild-type 361 apricot, bristle, clipped 89 bristle, clipped 91 apricot 42 apricot, bristle 38 clipped 9 apricot, clipped 11 bristle Using these data, first determine what gametes from the F; trihybrid produced each of the eight F2 categories. Note that apricot = aa (recessive to wild-type A); bristle = bb (recessive to wild-type B); and clipped = cc (recessive to wild-type C). Then determine if each gamete is recombinant (R) or nonrecombinant (R) for each pair of alleles (that is, for each genetic interval). Complete the table by dragging the…
Male Drosophila expressing the autosomal recessivemutations sc (scute), ec (echinus), cv (crossveinless),and b (black) were crossed to phenotypically wildtype females, and the 3288 progeny listed wereobtained. (Only mutant traits are noted.)653 black, scute, echinus, crossveinless670 scute, echinus, crossveinless675 wild type655 black71 black, scute73 scute73 black, echinus, crossveinless74 echinus, crossveinless87 black, scute, echinus84 scute, echinus86 black, crossveinless83 crossveinless1 black, scute, crossveinless1 scute, crossveinless1 black, echinus1 echinusa. Diagram the genotype of the female parent.b. Map these loci.c. Do the data provide evidence of interference?Justify your answer with numbers.
In a cross in Drosophila, a female heterozygous for the autosomallylinked genes a, b, c, d, and e (abcde/ + + + + +) was testcrossedwith a male homozygous for all recessive alleles (abcde/abcde).Even though the distance between each of the loci was at least3 map units, only four phenotypes were recovered, yielding thefollowing data: Phenotype No. of Flies+ + + + + 440a b c d e 460+ + + + e 48a b c d + 52 Total = 1000 Why are many expected crossover phenotypes missing? Can anyof these loci be mapped from the data given here? If so, determinemap distances.
In Drosophila, the wildtype eye color is black. In the laboratory, you screened for mutants than when homozygous results to different eye phenotypes. You found two mutants 1) red eye (re) and 2) white eye (we). You performed a complementation test and the resulting eye phenotype was gray. From the results, what conclusion can you make? A. There was complementation, thus the two mutations are alleles of different genes. B. There was no complementation, thus the two mutations are alleles of different genes. C. The two mutations failed to complement, thus they are alleles of different genes. D. The two mutations failed to complement, thus they are alleles of the same genes. E. The two mutations complemented, hence they are likely controlled by different genes.