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In rabbits, albinism is an autosomal recessive conditioncaused by the absence of the pigment melanin from skin and fur. Pigmentation is a dominant wild-type trait. Three purebreedingstrains of albino rabbits, identified as strains
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- 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_forwardConsider the following pedigree. Solid symbols represent individuals affected by the trait. Assume complete penetrance and non-variable expressivity. II 3 4 III 1 2 3 5 6 a) what is the mode of inheritance of this trait? b) Does the ratio of affected to unaffected offspring in generation III-1 to 1II-4 match the expected ratio for this mode of inheritance? Explain your answer in terms of the expected ratio versus the ratio observed. Give a reason for your answer. No mark is assigned for yes or no)arrow_forward) Cats also have a locus (gene) for a character called agouti. In cats with agouti fur, pigment is unevenly distributed (see Fig. 1). The agouti allele is dominant to the allele for regular fur. A male cat who long haired and is heterozygous for agouti fur is crossed with a female who has regular, non-agouti fur and is homozygous for short fur. a) What are the genotypes of these cats? b) What are all of the possible allelic combinations in the gametes from each of these cats? c)What are all of the possible phenotypic combinations of these characters in their kittens and in what ratios?arrow_forward
- 1)se; 12 cM 2)h; 12 cM 3)g; 8 cM 4)se; 8 cMarrow_forwardThe DNA of every individual in the pedigree shown below has been sequenced at the causative locus. All the non-shaded individuals are wild type apart from III.1. III.1 has been proven to have the causative mutation for this autosomal dominant condition, but they exhibit no symptoms. Based on this small pedigree, what is the level of penetrance for the condition? Please give your answer as a WHOLE percentage, give the number only, no percentage symbol. Answer: The level of penetrance for the condition shown in the pedigree below is Blank 1 percent. 1:1 1:2 Il:1 I1:2 I1:3 Il:4 I1:5 I1:6 II:1 I:2 III:3 III:4 III:3 III:6 III:7 III:8 III:9 III:10 III:11 III12 II:13 III:14 IV:1 | IV:2 IV:3 IV:4 IV:5 IV:6 IV:7 IV:8 IV:9 IV:10 IV:11 IV:12 IV:13 IV:14 IV:15 IV:16 IV:17 IV:18 IV:19 V:1 V:2 V:3 V:4 V:5 V:6 V:7 V:8 V:9 V:10 V:11 V:12arrow_forwardThe pedigree below shows that inheritance of a disease that is caused by a late onset, dominant, autosomal mutation that is rare, but only 50% penetrant. The gene that is mutated in the disease is linked at a distance of 10 cm to a microsatellite marker that has alleles numbered 1, 2, and 3. The marker alleles detected in each individual are indicated below. What is the probability that individual A will develop the disease? Explain using an illustration of this occurs.arrow_forward
- A 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_forward10 cM separates two hypothetical autosomal human genes. The dominant alleles are have complete penetrance and will result to Crossed eyes (e+) and short thumbs (th+). Four children are born to a normal guy and cross-eyed, small-thumbed woman. Two of the children have short thumbs and the other two have crossed eyes. She is carrying her fifth child. What is the probability that this fifth child will be cross-eyed and have short thumbs?arrow_forwardIn goats, a beard is produced by an autosomal allele that is dominant in males and recessive in females. We’ll use the symbol Bb for the beard allele and B+ for the beardless allele. Another independently assorting autosomal allele that produces a black coat (W) is dominant over the allele for white coat (w). Give the phenotypes and their expected proportions for the following crosses. a. B+ Bb Ww male × B+ Bb Ww female b. B+ Bb Ww male × B+ Bb ww female c. B+ B+ Ww male × Bb Bb Ww female d. B+ Bb Ww male × Bb Bb ww femalearrow_forward
- The following is a linkage map of chromosome 5 for three genes in tomato: (see image) The cross between the triple heterozygote (Lf J W/ lf j w) and a triple homozygous recessive produced 500 progeny. Assume that there is no interference in the Lf-W region. Give the expected number of individuals for each of the following progeny types and show complete solutions.a. with crossover in the Lf-J and J-W regionsb. with crossover in the Lf-J regionc. with crossover in the J-W regiond. without crossover in the Lf-W regionarrow_forwardThe wild-type (normal) fruit fly, Drosophila melanogaster, has straight wings and long bristles. Mutant strains have been isolated that have either curled wings or short bristles. The genes representing these two mutant traits are located on separate chromosomes. Carefully examine the data from the following five crosses shown below (running across both columns). (a) Identify each mutation as either dominant or recessive. In each case, indicate which crosses support your answer. (b) Assign gene symbols and, for each cross, determine the genotypes of the parents.arrow_forwardA female animal with genotype A/a ⋅ B/b is crossed with a double-recessive male (a/a ⋅ b/b). Their progeny include 442 A/a ⋅ B/b, 458 a/a ⋅ b/b, 46 A/a ⋅ b/b, and 54 a/a ⋅ B/b. Explain these results.arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning