Concept explainers
You have four guinea pigs for a genetic study. One male and one female are from a strain that is pure-breeding for short brown fur. A second male and female are from a strain that is pure-breeding for long white fur. You are asked to perform two different experiments to test the proposal that short fur is dominant to long fur and that brown is dominant to white. You may use any of the four original pure-breed-ing guinea pigs or any of their offspring in experimental matings. Design two different experiments (crossing dif-ferent animals and using different combinations of pheno-types) to test the dominance relationships of alleles for fur length and color, and make predictions for each cross based on the proposed relationships. Anticipate that the litter size will be 12 for each mating and that female guinea pigs can produce three litters in their lifetime.
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Genetic Analysis: An Integrated Approach (2nd Edition)
- Pedigree Analysis Is a Basic Method in Human Genetics Using the pedigree provided, answer the following questions. a. Is the proband male or female? b. Is the grandfather of the proband affected? c. How many siblings does the proband have, and where is he or she in the birth order?arrow_forwardBelow is a partially filled in complementation table. Please answer the following questions. 1 2 3 4 LO 5 6 1 2 I 3 4 5 4 + + 5 6 a. If you were to perform a complementation test with mutants 1 and 2, what would be the result? [Select] b. If you were to perform a complementation test with mutants 5 and 6, what would be the result? [Select] c. If you were to perform a complementation test with mutants 3 and 4, what would be the result? [Select] d. There is one complementation group that you can determine from the information above. If you performed a complementation test with mutant 1 and 6 and there was growth, would you say that the mutants were part of that complementation group? [Select] e. What is the minimum number of complementation groups that could exist based on the information above? [Select] f. What is the maximum number of complementation groups that could exist based on the information above? [Select]arrow_forwardA couple enters your genetic counseling clinic for some family planning advice. The woman’s father was color blind, but her own vision is normal. The man has no family history of color blindness. Neither the man nor woman have any known history of hemophilia, but their first child (a boy) has hemophilia. They ask you to calculate the chance that their nextchild will be affected by one or both conditions. You remember from your genetics training that these are both X-linked recessive conditions and that they are closely linked: in fact, their genetic loci are separated by only 10cM! During the interview with this couple, you draw the following pedigree to represent their information. Given what you know, determine for this couple what chance they have of each of the following (in the table).arrow_forward
- please make sure to read the question (THERE ARE OTHER VARIATIONS OF THE QUESTION ON DIFFERNT WEBSITES THAT ARE DIFFERENT THAN MINE)arrow_forwardWe are interested in detremining the genetic map of three different house finch traits: beak color (dark vs light), song length (short vs long) and patch size (reduced vs normal). We have done a large experiment involving the usual PO crossed, with the F1 test crossed. The following are the data for the offspring from the test cross: Beak Shade Patch Size Song Length Number Observed Dark Normal Long 67 Light Normal Long 3 Light Normal Short 498 Dark Normal Short 43 Dark Reduced Short 13 Light Reduced Short 80 Light Reduced Long 50 Dark Reduced Long 389 Choose the map below that best fits the data Beak Shade- 14.26CM- Patch Size- 9.54CM-Song Length O Patch Size-9.54cM- Beak Shade- 14.26CM-Song Length O Patch Size- 14.26cM- Beak Shade- 9.54CM-Song Length O Beak Shade-9.54cM-Patch Size- 14.26CM-Song Length O Patch Size-9.54CM- Song Length- 14.26CM-Beak Shade O Patch Size- 14.26cM-Song Length- 9.54cM-Beak Shadearrow_forwardI need a lot of help here :/arrow_forward
- 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 2 4 # 1 1 1 1 2 1 1 1 1 1 1 4 1 1 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 О с. 5 O d. 6 O e. none of the above 3. 5arrow_forward1)se; 12 cM 2)h; 12 cM 3)g; 8 cM 4)se; 8 cMarrow_forwardThe table below shows the progeny of a test cross of a heterozygote (HhFfGg). Each row shows the number of progeny that inherited each combination of alleles from the heterozygous parent. You've already determined the gene order (G is in the middle) and labeled each type of progeny. Using this information, what is the recombination frequency between H and G? Alleles: Type: HGF hgf HgF h Gf Hgf h GF HGf hg F O 0.543 0.302 O 0.241 O 0.23 0.139 Parental Parental DCO DCO SCO (H-G) SCO (H-G) SCO (G-F) SCO (G-F) Count: 124 119 15 18 51 55 38 40 O The correct answer is not available.arrow_forward
- A purple snow pea plant known to be hybrid for three linked genes (a, b, and d) is testcrossed. The progeny phenotypes and frequencies are: +a+ 86 b+d 56 +ad 78 b++ 49 +++ 12 bad 6 ba+ 515 ++d 460 a) Draw the chromosomes of the trihybrid and tester lined mentioned above. b) Give a map of the gene order with distance.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_forwardConsider 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 PCR 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 PCR 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?arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning