EBK CAMPBELL BIOLOGY IN FOCUS
2nd Edition
ISBN: 8220101459299
Author: Reece
Publisher: PEARSON
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Chapter 12.3, Problem 2CC
For each type of offspring of the testcross in Figure 12.9, explain the relationship between its
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In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows:
genotype
number
sn ct
15
sn ct+
34
sn+ ct
33
sn+ct+
18
What is the map distance between sn and ct?
Draw a Punnett square for the dihybrid cross described below (it is the same story as given for question 8, above) and use it to fill in
the blanks correctly in the text that follows. NOTE: Please type in whole numbers, no symbols.
There are two known alleles of gene occupying a specific locus in the X chromosome. The gene in question codes for a transcription factor
involved in digit development. The mutant allele is dominant and gives rise to an additional but non-functioning little finger (polydactyly) on
both hands.
A couple have had their DNA sequenced at the region of interest, the male exhibits polydactyly because of the mutation, the female is
homozygous wild type at the same locus and therefore has the wild type phenotype. Both have green eyes. In this story; eye colour shows a
monogenic autosomal inheritance pattern and the allele for brown eyes shows incomplete dominance with that for blue eyes, the
heterozygote phenotype is green eyes.
The genes for eye colour and…
The image shows a pair of homologous chromosomes from a single parent before gamete production. M1 and M2 are maternal chromosomes, while P1 and P2 are paternal chromosomes. Two traits are shown: D represents seed color (D – green, d – yellow), while F represents flower color (F – purple, f – white). These two traits follow the patterns of basic Mendelian genetics.
During crossing-over between the M2 F allele and the P1 f allele, a mutation occurred and the portion of P1 did not reattach to the chromosome.
Which of the following explains what would happen to the proportion of white flowers in a population resulting from this mutation?
A - There would be an increase in the proportion of white flowers because the f allele is distributed to more gametes.
B - There would be a decrease in the proportion of white flowers because the f allele is not distributed to as many gametes.
C - There would be an increase in the proportion of white flowers because the f allele would not be masked by the…
Chapter 12 Solutions
EBK CAMPBELL BIOLOGY IN FOCUS
Ch. 12.1 - Which one of Mendels laws relates to the...Ch. 12.1 - MAKE CONNECTIONS Review the description of meiosis...Ch. 12.1 - WHAT IF? Propose a possible reason that the first...Ch. 12.2 - A white-eyed female Drosophila is mated with a...Ch. 12.2 - Neither Tim nor Rhoda has Duchenne muscular...Ch. 12.2 - MAKE CONNECTIONS Consider what you learned about...Ch. 12.3 - When two genes are located on the same chromosome,...Ch. 12.3 - For each type of offspring of the testcross in...Ch. 12.3 - WHAT IF? Genes A, B, and C are located on the same...Ch. 12.4 - About 5% of individuals with Down syndrome have a...
Ch. 12.4 - WHAT IF? The ABO blood type locus has been mapped...Ch. 12.4 - MAKE CONNECTIONS The gene that is activated on the...Ch. 12.4 - Women born with an extra X chromosome (XXX) are...Ch. 12 - A man with hemophilia (a recessive, sex-linked...Ch. 12 - Pseudohypertrophic muscular dystrophy is an...Ch. 12 - A space probe discovers a planet inhabited by...Ch. 12 - Using the information from problem 3, scientists...Ch. 12 - A man with red-green color blindness (a recessive,...Ch. 12 - You design Drosophila crosses to provide...Ch. 12 - A wild-type fruit fly (heterozygous for gray body...Ch. 12 - Assume that genes A and B are 50 map units apart...Ch. 12 - Prob. 9TYUCh. 12 - SCIENTIFIC INQUIRY DRAW IT Assume you are mapping...Ch. 12 - FOCUS ON EVOLUTION Crossing over is thought to be...Ch. 12 - FOCUS ON INFORMATION The continuity of life is...Ch. 12 - SYNTHESIZE YOUR KNOWLEDGE Butterflies have an X-Y...
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- You cross a true-breeding yellow-bodied, smooth-winged female fly with a true-breeding red-bodied, crinkle-winged male. The red body phenotype is dominant to the yellow body phenotype and smooth wings are dominant to crinkled wings. Use B or b for body color alleles, and W or w for wing surface alleles.(4 points) a) What are the genotypes of the P generation flies? b) What will be the genotype(s) and phenotype(s) of the F1 offspring? c) You discover that the genes for body color and wing surface are linked. You perform a dihybrid test cross between the F1 flies from part (b) with a true-breeding yellow-bodied, crinkle-winged fly. Use the following results of this cross to determine the recombination frequency (%) between the body color and wing surface genes. (Remember that the recombinants are the ones that do not resemble the parental types from the P generation.) Body Color Wing Surface # of Individuals red smooth 102 yellow smooth 404 red crinkled 396 yellow crinkled…arrow_forwardAssume performing only one cross, which produced 274 flies, which consisted of 193 wild-type flies and 81 white-eyed flies. (a) Based upon this information, what is the most likely genotype of the female parent? (b) How to confirm the above answer with only the result from this one cross?arrow_forwardIn silkmoths (Bombyx mori), red eyes (re) and white-banded wings (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these two genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild-type traits. The F1 have wild-type eyes and wild-type wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are wild-type eyes, wild-type wings red eyes, wild-type wings wild-type eyes, white-banded wings red eyes, white-banded wings a. What phenotypic proportions would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes? b. What is the rate of recombination between the gene for red eyes and the gene for white-banded wings?arrow_forward
- In Drosophila, the allele for red eyes (pt) is wild-type and the allele for purple eyes (p¯) is mutant. The allele for grey body (b+) is wild-type and the allele for black body (b¯) is mutant. Flies with p*p* b*b* genotypes are mated with flies that have p p- b¯b¯ genotypes. A testcross was then performed in which the F1 offspring with p*p¯ b*b¯ genotypes were mated with flies with p p¯ b¯b genotypes. Ten-thousand flies were produced from this test cross. The following results were observed: 4,300 red eye, grey body flies 550 red eye, black body flies 4,500 purple eye, black body flies 650 purple eye, grey body flies Which F2 phenotypes are parental types? Which F2 phenotypes are recombinant types? What is the distance between the gene loci for eye color and body color? Use the equation for Hardy-Weinberg equilibrium for the following questions. p+q = 1 p2 + 2pq + q2 = 1arrow_forwardIn Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn+ and ct+ is crossed to a sn ct male. The F1 flies are interbred. The F2 males are distributed as follows sn ct 36 sn ct+ 13 sn+ ct 12 sn+ ct+ 39 What is the map distance between sn and ct?arrow_forwardIn Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up, while ebony (e, chromosome 3, position 70.7) results in a dark body. True breeding, wild type females are mated with true breeding males with curled wings and ebony bodies. Considering Drosophila notation, which of the following correctly diagrams the F1 cross? X X 3+ cu e + X X e + + + + + cu e + O + ■ 3+ X X X X Y Y + + ■ cu cu cu ' + ■ cu ■ ' + e + e e e e e + cu +arrow_forward
- You are studying ear shape in dogs (with XY se x determination) and cross a true-breeding pointed-ear female to a true-breeding floppy-ear male and collect all pointy-ear male and female offspring. Your colleague suspects that the two phenotypes may be caused by alleles of one X-linked gene. What sort of cross would you do to determine whether this is an X-linked trait? Name it and describe it. What offspring phenotypic ratio would you expect in that cross if the green allele is a dominant X-linked allele? Make sure to label your ratio with all relevant phenotypes. How would the result of your cross (from A) be different if the trait is autosomalarrow_forwardTwo genes, A and B, are 10 map units apart along the same chromosome. A cross was made between AAbb and aaBB individuals to produce AaBb F1 offspring. The F1 offspring were then crossed to aabb individuals to yield an F2 generation. What would be the genotype(s) of F2 offspring that carry recombinant chromosomes? (Note: recombinant chromosomes are the product of crossing over). What percentage of F2 offspring would be Aabb?arrow_forwardA cross was performed using Drosophila melanogaster involving a female known to be heterozygous for both ebony body and sepia eyes and a male known to be homozygous wild type male. The resulting progeny were allowed to mate with one another to produce the data set. Three repetitions of the experiment were conducted. The following data were produced from the crosses. Test these data to determine if they are significantly different from the expected phenotypic ratio. Use the 5% level of significance. Your answer should include the hypothesized cross in genotypes, the Chi-squared value, the critical value and whether you reject or do not reject for each experiment. Wild eye Wild body – 112, Wild eye Ebony body – 40, Sepia eye Wild body – 35, Sepia eye Ebony body – 11arrow_forward
- Two pure-breeding strains of flies are mated, and the F1 are intercrossed. The first strain has curled wings and black bodies. The second strain has straight wings and brown bodies. The F2 progeny are 271 straight wings with brown bodies, 31 curled wings with black bodies, 94 curled wings with brown bodies and 90 straight wings with black bodies. If the F1 were backcrossed to the straight, wing brown bodied parent, what phenotypes would be produced among the progeny? What would be the proportion of each phenotype?arrow_forwardThe allele b gives Drosophila flies a black body and b+ gives brown, the wild-type phenotype. The allele wx of a separate gene gives waxy wings and wx+ gives non-waxy, the wild-type phenotype. The allele cn of a third gene gives cinnabar eyes and cn+ gives red, the wild-type phenotype. A female heterozygous for these three genes is testcrossed, and 1000 progeny are classified with the following phenotypes. 382 cinnabar 379 black, waxy 69 waxy, cinnabar 67 black 48 waxy 44 black, cinnabar 5 wild type 6 black, waxy, cinnabar Based on this data, what is the correct map of these genes in terms of order and distance?arrow_forwardIn Drosophila, the brown mutation (bw, chromosome 2, position 104.5) results in brown eyes, while miniature (min, chromosome X, position 36.1) results in wings that are 2/3 the length of wild type. True breeding, wild type females are mated with true breeding males with brown eyes and miniature wings. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1 =1arrow_forward
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