Campbell Essential Biology (7th Edition)
7th Edition
ISBN: 9780134765037
Author: Eric J. Simon, Jean L. Dickey, Jane B. Reece
Publisher: PEARSON
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Textbook Question
Chapter 9, Problem 11SQ
How could you determine the genotype of one of the brown F2 mice in problem 10? How would you known whether a brown mouse is homozygous? Heterozygous?
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In Drosophila, a fully heterozygous female with the X-linked recessive genes a, b, and c (not necessarily in that order on
the chromosome) was mated to a male that was genetically a, b, c (not necessarily in that order on the chromosome).
The offspring occurred in the following phenotypic ratios:
Phenotypes:
Numbers:
What is the cis/trans arrangement in the heterozygous parent?
Wild
426
а, с, b
428
Which gene is in the middle?
a
23
c, b
22
If you added 23, 22, 3, and 2, it would give you the map distance between
genes
C
49
b, a
46
What calculation would you make to determine if interference was
occurring? (you don't have to complete the calculation)
b.
C, a
Total
1000
3.
In 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 P1 cross?
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As stated in Solved Problem 2, recessive mutation in certain mice called waltzers causes them to execute bizarre steps. W. H. Gates crossed waltzers with pure-breeding normal mice and found, among several hundred normal progeny, a single waltzing female mouse. This mouse was mated with a waltzing male, and her offspring were waltzers. When mated with a homozygous normal male, all her progeny were normal. Some of these normal males and females were intercrossed, and, unexpectedly, none of their progeny were waltzers. T. S. Painter examined the chromosomes of some of Gates’s waltzing mice that showed a breeding behavior similar to that of the original, unusual waltzing female. He found that these mice had the normal number of 40 chromosomes. In the unusual waltzers, however, one member of a chromosome pair was abnormally short. Interpret these observations as completely as possible, both genetically and cytologically.
Chapter 9 Solutions
Campbell Essential Biology (7th Edition)
Ch. 9 - The genetic makeup of an organism is called its...Ch. 9 - Which of Mendels laws is represented by each...Ch. 9 - Edward was found to be heterozygous (Ss) for the...Ch. 9 - Whether an allele is dominant or recessive depends...Ch. 9 - Prob. 5SQCh. 9 - Prob. 6SQCh. 9 - Prob. 7SQCh. 9 - Prob. 8SQCh. 9 - Adult height in people is at least partially...Ch. 9 - A purebred brown mouse is repeatedly mated with a...
Ch. 9 - How could you determine the genotype of one of the...Ch. 9 - Tim and Jan have freckles (a dominant trait), but...Ch. 9 - Incomplete dominance is seen in the inheritance of...Ch. 9 - Why was Henry VIII wrong to blame his wives for...Ch. 9 - Both parents of a boy arc phenotypically normal,...Ch. 9 - Heather was surprised to discover that she...Ch. 9 - Prob. 17SQCh. 9 - Prob. 18IMTCh. 9 - Prob. 19IMTCh. 9 - For each pair of your homologous chromosomes, one...Ch. 9 - In 1981, a stray cat with unusual curled-back ears...Ch. 9 - Interpreting Data As shown in the Punnett square...Ch. 9 - There are now nearly 200 recognized breeds of dog,...Ch. 9 - Gregor Mendel never saw a gene, yet he concluded...Ch. 9 - Prob. 25BS
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- Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?arrow_forwardHuman sex chromosomes are XX for females and XY for males. a. With respect to an X-linked gene, how many different types of gametes can a male produce? b. If a female is homozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele? c. If a female is heterozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele?arrow_forwardIndividuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were counted, with the following results: 474 Aabb, 480 aaBb, 20 AaBb, and 26 aabb. What type of cross is it? Are these loci linked? What are the two parental classes and the two recombinant classes of offspring? What is the percentage of recombination between these two loci? How many map units apart are they?arrow_forward
- In Drosophila flies, the allele b gives a black body, and the allele b+ gives brown, the wild-type phenotype. The allele wx of a separate gene gives waxy wings, and wx+ gives non-waxy. The allele cn of a third gene gives cinnabar eyes, and cn+ gives red. A female heterozygous for these three genes is testcrossed, and 745 progenies are produced which are phenotypically classified as follows: (see image) Make a linkage map of the three genes. Compute for interference and explain what the derived value means. Show complete solutions to support your answersarrow_forwardIn a cross between a white-eyed female (ww) and a red-eyed male (w+Y), nearly all the progeny were either red-eyed females (w+w) or white-eyed males (wY). However, about 1 in every 2000 F1 flies had an "exceptional phenotype" and was either a white-eyed female or red-eyed male. How did Bridges explain this unexpected result? A) Crossing over B) Incomplete cytokinesis C) Incorrect synapsis D) Nondisjunction E) Pseudoautosomal regionarrow_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 believe that the gene you're investigating, eyes absent, is an autosomal recessive mutant allele F7 ge in Drosophila. You set up a cross between a true-breeding mutant male and a true-breeding wildtype female to obtain the F1 generation. You then cross the F1 flies. In the F2 offspring, you observe the following: 237 wild-type males 180 wild-type females 70 mutant males 83 mutant females Based on these findings, is the mutant allele autosomal recessive? How do you know?arrow_forwardYou 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_forwardFemale Drosophila heterozygous for three recessive mutations e (ebony body), st (scarlet eyes), and ss (spineless bristles) were testcrossed, and the following progeny were obtained: Phenotype Number wild-type 67 ebony 8 ebony, scarlet 68 ebony, spineless 347 ebony, scarlet, spineless 78 scarlet 368 scarlet, spineless 10 spineless 54 (a) What indicates that the genes are linked? (b) What was the genotype of the original heterozygous females? (c) What is the order of the genes? (d) What is the map distance between e and st? (e) Between e and ss? (f) What is the coefficient of coincidence? (g) Diagram the crosses in this experiment.arrow_forward
- In Drosophila, the dominant Bar mutation (B, chromosome X, position 57) results in thin bar- shaped eyes, while the recessive singed (sn, chromosome X, position 21) results burnt looking bristles. True breeding, wild type females are mated with true breeding males with Bar eyes and singed bristles. 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_forwardIN DROSOPHILA, AN X-LINKED RECESSIVE MUTATION, Xm CAUSES MINIATURE WINGS. LIST THE F₂ PHENOTYPIC RATIOS IF: A MINIATURE-WINGED FEMALE IS CROSSED WITH A NORMAL MALE AND A MINIATURE-WINGED MALE IS ● ● CROSSED WITH A NORMAL FEMALE. WHAT WOULD THE PHENOTYPIC RATIO FROM (A) BE IF THE MINIATURE- WINGED GENE WERE AUTOSOMAL? ASSUME IN ALL CASES THAT THE P1 INDIVIDUALS ARE TRUE-BREEDING.arrow_forwardIn Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. 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 =1arrow_forward
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