Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 2, Problem 7EQ
The
43 resistant to strain 22 but sensitive to strain 24
9 sensitive to strain 22 and strain 24
32 sensitive to strain 22 but resistant to strain 24
110 resistant to strain 22 and strain 24
Explain the inheritance pattern for flax resistance and sensitivity to M. lini strains.
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The fungus Melampsora lini causes a disease known as flax rust.Different strains of M. lini cause varying degrees of the rust disease.Conversely, different strains of flax are resistant or sensitive to thevarious varieties of rust. The Bombay variety of flax is resistant toM. lini-strain 22 but sensitive to M. lini-strain 24. A strain of flaxcalled 770B is just the opposite; it is resistant to strain 24 but sensitive to strain 22. When 770B was crossed to Bombay, all F1 individuals were resistant to both strain 22 and strain 24. When F1individuals were self-fertilized, the following data were obtained:43 resistant to strain 22 but sensitive to strain 24 Explain the inheritance pattern for flax resistance and sensitivityto M. lini strains.9 sensitive to strain 22 and strain 2432 sensitive to strain 22 but resistant to strain 24110 resistant to strain 22 and strain 24
Wild-type strains of the haploid fungus Neurospora canmake their own tryptophan. An abnormal allele td renders the fungus incapable of making its own tryptophan.An individual of genotype td grows only when its medium supplies tryptophan. The allele su assorts independently of td; its only known effect is to suppress the tdphenotype. Therefore, strains carrying both td and su donot require tryptophan for growth.a. If a td ; su strain is crossed with a genotypically wildtype strain, what genotypes are expected in the progenyand in what proportions?b. What will be the ratio of tryptophan-dependent totryptophan-independent progeny in the cross of part a?
From a cross between e+ f+ g+ and e− f − g− strains ofNeurospora, recombination between these linkedgenes resulted in a few octads containing the followingordered set of spores:e+ f+ g+e+ f+ g+e+ f − g+e+ f − g+e− f − g−e− f − g−e− f − g−e− f − g−a. Where was recombination initiated?b. Did crossing-over occur between genes e and g?Explain.c. Why do you end up with 2 f+ : 6 f − but 4 e+: 4 e−and 4g+: 4g−?d. Could you characterize these unusual octads as MIor MII for any of the three genes involved?Explain.
Chapter 2 Solutions
Genetics: Analysis and Principles
Ch. 2.1 - 1. Experimental advantages of using pea plants...Ch. 2.1 - The term cross refers to an experiment in which a....Ch. 2.1 - 3. To avoid self-fertilization in his pea plants,...Ch. 2.2 - Prob. 1COMQCh. 2.2 - Prob. 2COMQCh. 2.3 - A pea plant has the genotype rrYy. How many...Ch. 2.3 - A cross is made between a pea plant that is RrYy...Ch. 2.3 - Prob. 3COMQCh. 2.4 - Which of the following would not be observed in a...Ch. 2.4 - Prob. 2COMQ
Ch. 2.5 - A cross is made between AABbCcDd and AaBbccdd...Ch. 2.5 - Prob. 2COMQCh. 2.5 - Prob. 3COMQCh. 2 - 1. Why did Mendel’s work refute the idea of...Ch. 2 - 2. What is the difference between...Ch. 2 - 3. Describe the difference between genotype and...Ch. 2 - 4. With regard to genotypes, what is a...Ch. 2 - 5. How can you determine whether an organism is...Ch. 2 - In your own words, describe Mendels law of...Ch. 2 - Based on genes in pea plants that we have...Ch. 2 - Prob. 8CONQCh. 2 - Do you know the genotype of an individual with a...Ch. 2 - 10. A cross is made between a pea plant that has...Ch. 2 - Prob. 11CONQCh. 2 - 12. Describe the significance of nonparentals with...Ch. 2 - For the following pedigrees, describe what you...Ch. 2 - Ectrodactyly, also known as lobster claw syndrome,...Ch. 2 - Identical twins are produced from the same sperm...Ch. 2 - In cocker spaniels, solid coat color is dominant...Ch. 2 - A cross was made between a white male dog and two...Ch. 2 - 18. In humans, the allele for brown eye color (B)...Ch. 2 - Albinism, a condition characterized by a partial...Ch. 2 - A true-breeding tall plant was crossed to a dwarf...Ch. 2 - 21. For pea plants with the following genotypes,...Ch. 2 - 22. An individual has the genotypeand makes an...Ch. 2 - 23. In people with maple syrup urine disease, the...Ch. 2 - Prob. 24CONQCh. 2 - 25. A true-breeding pea plant with round and Page...Ch. 2 - Prob. 26CONQCh. 2 - 27. What are the expected phenotypic ratios from...Ch. 2 - Prob. 28CONQCh. 2 - Prob. 29CONQCh. 2 - A pea plant that is dwarf with green, wrinkled...Ch. 2 - 31. A true-breeding plant with round and green...Ch. 2 - Wooly hair is a rare dominant trait found in...Ch. 2 - Huntington disease is a rare dominant trait that...Ch. 2 - 34. A woman with achondroplasia (a dominant form...Ch. 2 - 1. Describe three advantages of using pea plants...Ch. 2 - Explain the technical differences between a...Ch. 2 - 3. How long did it take Mendel to complete the...Ch. 2 - 4. For all seven characters described in the data...Ch. 2 - From the point of view of crosses and data...Ch. 2 - 6. As in many animals, albino coat color is a...Ch. 2 - 7. The fungus Melampsora lini causes a disease...Ch. 2 - For Mendels data for the experiment in Figure 2.8,...Ch. 2 - 9. Would it be possible to deduce the law of...Ch. 2 - In fruit flies, curved wings are recessive to...Ch. 2 - A recessive allele in mice results in an unusally...Ch. 2 - Prob. 12EQCh. 2 - Prob. 13EQCh. 2 - Prob. 14EQCh. 2 - 15. A cross was made between two strains of plants...Ch. 2 - A cross was made between two pea plants, TtAa and...Ch. 2 - Consider this four-factor cross: TtRryyAaTtRRYyaa,...
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- DNA from a strain of Bacillus subtilis with genotype a+ b+ c+ d+ e+ is used to transform a strain with genotype a− b− c− d− e−. Pairs of genes are checked for cotransformation, and the following results are obtained: Pair of genes Cotransformation Pair of genes Cotransformation a+ and b+ No b+ and d+ No a+ and c+ No b+ and e+ Yes a+ and d+ Yes c+ and d+ No a+ and e+ Yes c+ and e+ Yes b+ and c+ Yes d+ and e+ No On the basis of these results, what is the order of the genes on the bacterial chromosome?arrow_forwardDNA from a strain of Bacillus subtilis with genotype a* b* c* d* e* is used to transform a strain with genotype a b c d e. Pairs of genes are checked for cotransformation, and the following results are obtained: Pair of genes Cotransformation Pair of genes Cotransformatic a* and b* No b* and d* No a* and c* No b* and e* Yes a* and d* Yes c* and d* No a* and e* Yes c* and e* Yes b* and c* Yes d* and e* No On the basis of these results, what is the order of the genes on the bacterial chromosome?arrow_forwardE. coli strain B (permissive) were co-infected with rlla and rllb mutant phage T4. Progeny were plated in a dilution series with the following results: Lawn Dilution Number of plaques Strain B 10-6 10 Strain K 10-4 What is the recombination frequency? Remember that the readout is viability and you cannot see the inviable progeny. Answer:arrow_forward
- The following results are derived from crosses between Neurospora strain xy and strain ++: Tetrad Class 3 4 ху x+ x+ xy ++ ++ ++ +y +y ху +y 25 ++ 3 124 4 (i) Name the ascus type of each class from 1 to 4 as P, NP or T. (ii) Are genes x and y linked? Explain your answer. (iii) If they are linked, determine the map distance between the two genes. If they are unlinked, provide all the information you can about why you draw this conclusion.arrow_forwardA cross was performed between a yeast strain that requires methionine and lysine for growth (met− lys−)and another yeast strain, which is met+ lys+. One hundred asci were dissected, and colonies were grownfrom the four spores in each ascus. Cells from thesecolonies were tested for their ability to grow on petriplates containing either minimal medium (min), min+ lysine (lys), min + methionine (met), or min + lys+ met. The asci could be divided into two groupsbased on this analysis:Group 1: In 89 asci, cells from two of the four spore colonies couldgrow on all four kinds of media, while the other two spore coloniescould grow only on min + lys + met.Group 2: In 11 asci, cells from one of the four spore colonies couldgrow on all four kinds of petri plates. Cells from a second one ofthe four spore colonies could grow only on min + lys plates andon min + lys + met plates. Cells from a third of the four sporecolonies could only grow on min + met plates and on min + lys+ met. Cells from the…arrow_forwardNeurospora of genotype a + c are crossed withNeurospora of genotype + b +. (Here, + is shorthandfor the wild-type allele.) The following tetrads areobtained (note that the genotype of the four sporepairs in an ascus are listed, rather than listing alleight spores):a + c a b c + + c + b c a b + a + ca + c a b c a + c a b c a b + a b c+ b + + + + + b + + + + + + c + + ++ b + + + + a b + a + + + + c + b +137 141 26 25 2 3a. In how many cells has meiosis occurred to yieldthese data?b. Give the best genetic map to explain these results.Indicate all relevant genetic distances, both betweengenes and between each gene and the centromere.c. Diagram a meiosis that could give rise to oneof the three tetrads in the class at the far right inthe listarrow_forward
- Which of the pedigree diagrams below is most likely to show a family with Haemophilia A?arrow_forwardWhen two plants belonging to the same genus but different species are crossed, the F1 hybrid is more viable and has more ornate flowers. Unfortunately, this hybrid is sterile and can only be propagated by vegetative cuttings. Explain the sterility of the hybrid and what would have to occur for the sterility of this hybrid to be reversed.arrow_forwardA Neurospora colony at the edge of a plate seemed to be sparse (low density) in comparison with the other colonies on the plate. This colony was thought to be a possible mutant, and so it was removed and crossed with a wild type of the opposite mating type. From this cross, 100 ascospore progeny were obtained. None of the colonies from these ascospores was sparse, all appearing to be normal. What is the simplest explanation of this result? How would you test your explanation? (Note: Neurospora is haploid.)arrow_forward
- Eight mutant bacteriophage strains cannot lyse a certain type of bacteria that can be lysed by wild-type bacteriophages. The mutant strains were allowed to infect the bacteria in a complementation test. A "+" indicates that lysis occurred with coinfection. A "-" indicates that lysis did not occur. OXWTMAO G K Т Т М А А Q C GKWTMAQC +++++ - - - - - - - + - + - + - - + - - - O +++ - + - - + + A cistron is defined by no complementation in the How many genes are controlling lysis in this bacteriophage? (Use a number not a word in the space) configuration.arrow_forwardThe figure below shows the life cycle of the fungus Neurospora. The adult stage of the Neurospora is a multicellular haploid. b) Neurospora has an arginine amino acid synthesis pathway shown below. Suppose I take the strain above that only grows with arginine supplements and cross it to a different mutant Neurospora strain that grows with arginine and citrulline supplements but not with ornithine supplements. Assuming gens A, B, and C are unlinked and there is only one mutation per stain: What percentage of the progeny will grow on ornithine? What percentage on citrulline? What percentage on arginine?arrow_forwardBelow are two pictures of perithecium on microscope slides that have been squished under cover slips so that the asci inside have spread out. In the asci, we can clearly see the tan ascospores (light colored) and wildtype ascospores (dark colored). We will perform our tetrad analysis by counting how many of the asci are non-recombinant and how many of the acsi are recombinant. We can tell which are recombinant and non-recombinant by observing the patterns of the ascospores in the asci. See the patterns of ascospores for non- recombinant and recombinant asci in the table below. Ascospore pattern of non-recombinant asci OOOO OOOOO You will count the total number of non-recombinant and recombinant asci in the two pictures below. You will only count the asci that have a star next to them because these asci can be clearly identified as non-recombinant or recombinant. Select the correct number below after you are done. Total number of non-recombinant asci [Select] Ascospore pattern of…arrow_forward
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