Campbell Biology In Focus
2nd Edition
ISBN: 9780134203072
Author: Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, Jane B. Reece
Publisher: PEARSON
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Chapter 11.3, Problem 1CC
Incomplete dominance and epistasis are both terms that define genetic relationships. What is the most basic distinction between these terms?
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Incomplete dominance and epistasis are both terms thatdefine genetic relationships. What is the most basicdistinction between these terms?
L, R, P and D are four genes located on the same chromosome in a diploid
organism. Map distances between different pairs of genes were determined and
the following results obtained. Based on these data, what is the order of the
genes on the chromosome?
D-R: 35 CM
P-D: 5 CM
L-R: 10 CM
D-P: 30 CM
D-L 25 CM
P-R: 40 CM
Note that only one orientation of the order is given as an answer choice. For
example: ABCD has the same order as DCBA, but only ABCD would be listed
as a possible answer. So be sure to check both orientations when comparing
your gene order to the possible answer choices.
D-L-P-R
P-D-L-R
P-R-D-L
L-P-R-D
R-L-D-P
In the fruit fly, recessive mutations in either of two independently assorting genes, brown and purple, prevent the synthesis of red pigment in the eyes. Thus, homozygotes for either of these mutations have brownish-purple eyes. However, heterozygotes for both of these mutations have dark red, that is, wild-type eyes. If such double heterozygotes are intercrossed, what kinds of progeny will be produced, and in what proportions?
Chapter 11 Solutions
Campbell Biology In Focus
Ch. 11.1 - DRAW IT Pea plants heterozygous for flower...Ch. 11.1 - List all gametes that could be made by a pea plant...Ch. 11.1 - Prob. 3CCCh. 11.2 - For any gene with a dominant allele A and...Ch. 11.2 - Two organisms, with genotypes BbDD and BBDd, are...Ch. 11.2 - Prob. 3CCCh. 11.3 - Incomplete dominance and epistasis are both terms...Ch. 11.3 - Prob. 2CCCh. 11.3 - WHAT IF? A rooster with gray feathers and a hen of...Ch. 11.4 - Beth and Tom each have a sibling with cystic...
Ch. 11.4 - MAKE CONNECTIONS In Table 11.1, note the...Ch. 11 - DRAW IT Two pea plants heterozygous for the...Ch. 11 - A man with type A blood marries a woman with type...Ch. 11 - A man has six fingers on each hand and six toes on...Ch. 11 - Prob. 4TYUCh. 11 - Flower position, stem length, and seed shape are...Ch. 11 - Prob. 6TYUCh. 11 - Prob. 7TYUCh. 11 - Prob. 8TYUCh. 11 - Prob. 9TYUCh. 11 - Prob. 10TYUCh. 11 - In tigers, a recessive allele that is pleiotropic...Ch. 11 - Prob. 12TYUCh. 11 - Imagine that you are a genetic counselor, and a...Ch. 11 - Prob. 15TYUCh. 11 - Prob. 16TYUCh. 11 - Prob. 17TYU
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- Individuals 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_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_forwardWhat some ways that incomplete penetrance can occur? Give at least two possible ways on why it occurs.arrow_forward
- 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? X X ++ e + + + O+ X + X + ■ + X + + + 3+ X X X X + + Y Y cu cu cu + cu cu J e e e e e (D e + cu cu (Darrow_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_forwardIn a breed of cattle, rough coat is dominant over smooth, and black coat is codominant to red. What do the statement The loci of the two pair of genes concerned are on the same pair of chromosomes (assuming no crossing over)mean?arrow_forward
- A standard three-point mapping is conducted for recessive mutations in autosomal genes purple eye (pr), curved wing ( c) and black body (b). Their wild type alleles are also used for genetic mapping. An F1 Drosophila female heterozygous for purple eye (pr), curved wing (c) and black (b) is crossed to a triply homozygous mutant male. The observed numbers and phenotypes of the offspring are as follows: 360 pr c b 380 pr+ c+ b+ 104 pr c+ b 96 pr+ c b+ 30 pr c b+ 20 pr+ c+ b 6 pr c+ b+ 4 pr+ c b PROVIDE THE FOLLOWING: A) State the order of genes on this chromosome. B) Calculate map distances between the gene pairs: pr-c, pr-b, c-b. Show calculations, state the number of map units and which gene pairs they refer to.arrow_forwardA researcher crosses mice with brown eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes: F1: 6 apricot, short : 30 brown, long : 15 brown, short : 9 apricot, long You know the genes encoding these traits are autosomal, completely dominant and assort independently. You want to use a chi-square test to analyse these results. a) Making use of the appropriate genetic convention for naming alleles, give the genotype of the male parent in this cross. b) What is your null hypothesis for the chi-square test? c) Give the expected number of individuals in the "brown, long" class. d) You obtain a value of 3.47 for the chi-square test. What conclusion can you make from the results of the chi-square test? P df 0.995 0.975 0.9 0.5 0.1 0.05* 0.025 0.01 0.005 1 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 2 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 4 0.207 0.484 1.064 3.357…arrow_forwardA phenotypically abnormal individual has a phenotypically normalfather with an inversion on one copy of chromosome 7 and a phenotypicallynormal mother without any changes in chromosomestructure. The orders of genes along the two copies of chromosome7 in the father are as follows: R T D M centromere P U X Z C (normal chromosome 7)R T D U P centromere M X Z C (inverted chromosome 7) The phenotypically abnormal offspring has a chromosome 7 withthe following order of genes: R T D M centromere P U D T R Using a sketch, explain how this chromosome was formed. In youranswer, explain where the crossover occurred (i.e., between whichtwo genes).arrow_forward
- In an autotetraploid organism, the gene controlling coat color is very near the centromere of the chromosome carrying it. The allele Y for the yellow coat is dominant to y for blue coat. A homozygous yellow autotetraploid strain is crossed with a homozygous blue autotetraploid strain. What types and proportions of gametes would the F1 be expected to produce if the Y locus were 40 or more map units from the centromere? Derive the expected F2 phenotypic ratio. Please explain coherently. Thank you!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_forwardIn corn, the genes v (virescent seedlings), pr (red aleurone), and bm (brown midrib) are all on chromosome 5, but not necessarily in the order given. The cross: v+ pr bm/ v pr+ bm+ with v pr bm/ v pr bm produces 1000 progeny with the following phenotypes: v+ pr bm 226 v pr+ bm+ 229 v+ pr bm+ 153 v pr+ bm 185 v+ pr+ bm 59 v pr bm+ 71 v+ pr+ bm+ 36 v pr bm 41 What is the gene order, and the (b) genetic map of these three genes?arrow_forward
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