1. The western coyote (Canis lupus latrans) can interbreed with the gray wolf (Canis lupus lupus) to produce healthy and fertile F, hybrids, but these hybrids are usually unable to attract mates (either among themselves, or backcrossed to the western coyote or the gray wolf). This reproductive isolation: A. is a post-zygotic barrier known as behavioral F, inferiority B. is a post-zygotic barrier known as F2 hybrid breakdown C. is a post-zygotic barrier known as F, hybrid sterility D. is a post-zygotic barrier known as F, hybrid inviability E. is a post-zygotic barrier known as ecological F, inferiority

1. The western coyote (Canis lupus latrans) can interbreed with the gray wolf (Canis lupus lupus) to produce healthy and fertile F, hybrids, but these hybrids are usually unable to attract mates (either among themselves, or backcrossed to the western coyote or the gray wolf). This reproductive isolation: A. is a post-zygotic barrier known as behavioral F, inferiority B. is a post-zygotic barrier known as F2 hybrid breakdown C. is a post-zygotic barrier known as F, hybrid sterility D. is a post-zygotic barrier known as F, hybrid inviability E. is a post-zygotic barrier known as ecological F, inferiority

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter12: Mendel, Genes, And Inheritance

Section: Chapter Questions

Problem 4TYK: What genotypes, and in what frequencies, will be present in the offspring from the following...

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1. A male lion (Panthera leo) crossed with a female tiger (Panthera tigris) produces F, hybrid cubs which are called ligers, but these F, ligers (especially males) are typically infertile (as predicted by the British geneticist J.B.S. Haldane). This type of reproductive isolation: A. is a post-zygotic barrier known as behavioral F1 inferiority B. is a post-zygotic barrier known as F2 hybrid breakdown C. is a post-zygotic barrier known as F, hybrid sterility D. is a post-zygotic barrier known as F1 hybrid inviability E. is a post-zygotic barrier known as ecological F, inferiority
2. In the following types of matings, the phenotypes of the parents are listed together with the frequencies of phenotypes occurring among their offspring. Indicate the genotype of each parent (you may need to use testcrossesl). Parents Offspring I B:0 을 A: 을B 을 AB : B: A: 0 를 AB : 를 A a. Вx В Ь. Ох АВ Ox AB Bx A d. Bx A C. TV Sex Lin ko
2. A sample of 400 individuals is found to contain: 32 A1A1: 36 A1 A2 : 60 A1A3 57 A2A2 : 90 A2A3 : 125 A3A3 a) Find the gene frequencies after one generation of selfing ww b) Find the percent heterozygosity after one generation of selfing. w w c) Find the frequencies of each heterozygous type.
What genotypes, and in what frequencies, will be present in the offspring from the following matings? a. AABBaaBB b. AaBbaabb c. AaBBAABb d. AaBbAaBb
3. In sesame plants, the one-pod condition (P) is dominant to the three-pod condition (p), and normal leaf (L) is dominant to wrinkled leaf (I). Pod type and leaf type are inherited independently. Determine the genotypes for the two parents for all possible matings producing the following offspring: 318 one-pod normal, 98 one-pod wrinkled b. а. 323 three-pod normal, 106 three-pod wrinkled 401 one-pod normal d. C. 150 one-pod normal, 147 one-pod wrinkled, 51 three-pod normal, 48 three-pod wrinkled 223 one-pod normal, 72 one-pod wrinkled, 76 three-pod normal, 27 е. three-pod wrinkled
3. The chestnut coat color of horses is due to a recessive gene, while the dominant allele results in black. The pacing gait is due to a recessive gene, whereas the dominant allele results in the trotting gait. Show the types of offspring that cou result from a cross of a black trotter (male), heterozygous for both genes, with a chestnut pacer (female). ld a. Possible genotypes for Black coat:Chestnut coat: b. Possible genotypes for Trotting gait:Pacing gairi Male possible gametes: Female possible gametes: c. Male genotype d. Female genotype: e. f. Genotypic Ratio: g. Phenotypic Ratio:
7. Here are the results of a crossing trial with fruit flies (D. melanogaster) Parents: Ebony bodied, white-eyed 9 fly x Normal-bodied, red-eyed a fly • Both parents were from true-breeding (homozygous) lines • Ebony body means the body colour is black instead of orange F1: Males All white-eyed, normal bodied Females Red eyed, normal bodied F2: (200 offspring in total produced) Male Females Normal body, red eye Normal body, white eye Ebony body, red eye Ebony body, white eye Total 38 36 37 39 12 11 13 14 100 100 Explain the pattern of inheritance of each trait body colour and eye colour Write the genotype of the parents, F1 offspring and F2 offspring. (ii)
1. In a particular bird species, the tips of the feathers are tinged with blue, white, or silver. The BB genotype calls for blue-tipped feathers. The WW genotype calls for white-tipped feathers. The BW genotype calls for an intermediate with silver-tipped feathers. A mating occurs between a blue-tipped bird and a white-tipped bird. What types of offspring would be expected? Give the genotypes and phenotypes of the possible offspring and the probability of each. 2. The bird species described in question 5 has an ornamental plumage on its head. The plumage feathers are either flexible or stiff. The allele for flexible plumage is dominant over the allele for stiff plumage. A male who is heterozygous for plumage and has silver-tipped wings mates with a female who has stiff plumage with silver-tipped wings. What types of offspring would be expected from this cross? Give the probability of each. 3. In some cats, the gene for tail length shows incomplete dominance. When a cat with a…
4. When the F1 bees were backcrossed to the inbred resistant line produced 4 types of offspring in equal numbers: ¼ if the back-cross offspring showed the complete hygienic behavior pattern¼ of the back-cross offspring showed none of the elements of hygienic behavior¼ of the back-cross offspring would uncap the cells of the diseased larvae but these workers would not remove the diseased larvae¼ of the back-cross offspring would remove the diseased larvae (once the experimenter had uncapped the cells of the diseased larvae) a. If the resistant behavioral pattern is due to two gene loci with two alleles each, can this account for the 4 phenotypes shown in the back-cross? How? b. If you use UURR (for the non-resistant inbred line) and uurr (for the resistant inbred line), what are the phenotypic and genotypic ratios for the F2 cross between these lines (i.e., F1 x F1)? Show your work: c. Given the genetic model given in part d, do the backcross offspring in part c (from an F1 x resistant…
17. Three recessive traits in aardwolf (a real animal) are stripe-less (s), dwarf (d), and fluffy mane (f). A stripe-less and fluffy female aardwolf mates with a dwarf male aardwolf to produce an F1 that is all WT (has stripes, normal size, and non-fluffy mane). When an F1 is test crossed to a fully recessive aardwolf, the following F2 are produced: Stripe-less, fluffy, normal sized 89 Striped, non-fluffy, dwarf 92 Stripe-less, non-fluffy, dwarf 1 Striped, fluffy, normal sized 2 11 13 19 20
7. In a certain animal, black fur is dominant to white fur. Determine the possible offspring from crosses between: black fur (B), white fur (b) a. Homozygous black X white Genotype: Phenotype: Genotypic Ratio: Phenotypic Ratio: b. Heterozygous black X heterozygous black Genotype: Phenotype: Genotypic Ratio: Phenotypic Ratio:
7. In fowl, genotype rrpp gives single comb, R_P_ walnut comb, rrP_ pea comb, and R_pp rose comb. a. If single-combed birds are crossed with birds of a true-breeding walnut-combed strain, predict the proportions of comb types that will appear in the F₁ and F₂. b. What is the probable genotypes of parent fowls in a walnut X rose mating that would produce progenies in this proportion: 3 rose: 3walnut: 1pea: 1 single?