strains of Drosophila that have either normal legs or abnormally short legs and you are studying the gene responsible. You know that normal legs are dominant to short legs. You come across a misplaced fly with normal legs, but you are not sure of his genetic background and you want to keep him in your experiments. (Without doing a molecular analysis), How could you figure out whether he was heterozygous or homozygous for the leg
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You are working in the lab with strains of Drosophila that have either normal legs or abnormally short legs and you are studying the gene responsible. You know that normal legs are dominant to short legs. You come across a misplaced fly with normal legs, but you are not sure of his genetic background and you want to keep him in your experiments.
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(Without doing a molecular analysis), How could you figure out whether he was heterozygous or homozygous for the leg gene that you are studying? (Describe what you would do and how the results would answer the question.)
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What is the procedure you described above called?
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- This research exploited the principles of Mendelian genetics combined with the MCR technique to genetically manipulate fruit flies. Now, suppose you do a similar experiment. You perform two crosses: one that allows for Mendelian patterns of inheritance, and another that uses MCR with CRISPR-Cas9 to genetically edit the eye color of fruit flies. In fruit flies, the gene for eye color is located on the X chromosome. Red eyes (wild type, w+) are dominant to white eyes (mutant, w-). Let’s see how well you can predict genotype and phenotype ratios resulting from the two crosses. Genetic cross 1 (without MCR): Cross between a white-eyed male and a homozygous red-eyed female. Genetic cross 2 (with MCR): Cross between an MCR-element (carries the CRISPR-Cas9 construct) white-eyed male and a homozygous red-eyed female. Keep in mind, this technology converts a heterozygous genotype to a homozygous recessive genotype. Drag the labels to the table to fill in the genotypic and phenotypic ratios.The genetic map was based on crosses in Drosophila involving the three sex-linked genes a, b and c. “a” gives red eyes, “b” gives normal wings and “c” gives black body. The recombination frequencies between these genes are as follows; a and b is 23.8, b and c is 2.6 and a and c is 28.1, respectively. Could you draw a basic genetic map based on distance between these genes using dots to show distance(s)? Could you make one fundamental comment using these distances based on genetic linkage?A geneticist is using a three‑point testcross to map three linked Drosophila recessive mutations called a, b, and c, where a is associated with anomalous gait, b is associated with buckled wings, and c is associated with curved bristles. She first crosses homozygous anomalous, buckled flies to homozygous curved flies. Next, she testcrosses the F1 progeny to anomalous, buckled, curved flies. She obtains 1000 progeny distributed as shown. From this data, calculate the map distance between b and c. progeny curved 277 anomalous, buckled 283 wild type 5 anamalous, buckled, curved 5 buckled, curved 133 anomalous 127 anomalous, curved 86 buckled 84
- Eye colors are passed down through generations, but sometimes genetic variations can lead to surprising results in eye colors (dark eye color, still rocks!?) Some examples are found below. Is there truth behind the following? Provide explanations for your answer. Two parents with blue eyes cannot have a brown-eyed child and vice-versa. Eyes with more than one color stems from parents with two different eye colors. People with violet eyes are mutants. People with red or pinkish eye color can have dark-eyed parents.The allele b gives Drosophila flies a black body and b+ gives brown, the wild-type phenotype. The allele wx of a separate gene gives waxy wings and wx+ gives non-waxy, the wild-type phenotype. The allele cn of a third gene gives cinnabar eyes and cn+ gives red, the wild-type phenotype. A female heterozygous for these three genes is testcrossed, and 1000 progeny are classified with the following phenotypes. 382 cinnabar 379 black, waxy 69 waxy, cinnabar 67 black 48 waxy 44 black, cinnabar 5 wild type 6 black, waxy, cinnabar Based on this data, what is the correct map of these genes in terms of order and distance?Two different strains of Drosophila, strain A and strain B, each has a recessive mutation that results in abnormally bright red eye color. (Wild type flies have brownish red eye color). When a homozygous strain A fly is crossed with a homozygous B fly, all of the progeny have the dominant wild type eye color. The wild type-eyed progeny were allowed to breed among themselves to produce the F2 generation. The F2 generation consisted of 92 wild type and 74 bright red-eyed flies. Write the genotype(s) of the flies in each generation. Use a low dash (e.g. A_ B_) to indicate genotypes that could be either homozygous or heterozygous) a) parental strain A b) parental strain B c) wild type progeny (F1) d) wild type F2 e) bright-eyed F2
- When you are visiting the local pet store you start talking to them about their Ceti alpha-5 eel colony. They tell you that some eels have red antennas, some have white antennas, and some have pink antennae. Whenever you cross an eel with red antennae to an eel with white antennae you only recover eels with pink antennae. If you assume antennae color is under the control of one gene and you cross two eels with pink antennae what would you predict would be the genotypes and phenotypes of the progeny? How would you explain these results? (Describe why the expected results from ‘a’ would happen.)Part A: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be AB? Enter your answer as a decimal to three places (for example: 0.120). Part B: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be Ab? Part C: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be aB? Part D: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a signal plant will be ab?In the lab, you discover two yeast haploid mutants that cannot produce arginine. You cross them together and the resulting diploid produces arginine. If producing arginine is the wild type phenotype, what can you definitively conclude? A.The haploid strains have identical mutations in the same genes. B.The haploid strains have identical mutations in different genes. C.The haploid strains have mutations in different genes. D.The haploid strains have mutations in the same gene. E.The haploid strains must belong to the complementation group encoding the first enzyme in the biosynthetic pathway.
- Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?You are a gene hunter, trying to find the genetic basis for a rare inherited disease. Examination of six pedigrees of families affected by the disease provides inconsistent results. For two of the families, the disease is co-inherited with markers on chromosome 7. For the other four families, the disease is co-inherited with markers on chromosome 12. Explain how this difference might have arisen.Often geneticists want to change one allele in an outcrossing organism while keeping the rest of the genome the same. For example, they might wish to take a specially designed stock of flies and alter the eye color from red to white. Suppose that the white-eye allele is dominant, meaning that flies with one or two white-eye alleles will have white eyes. One procedure used is to take a white-eyed fly and cross it with the red-eyed stock. The whiteeyed offspring are then considered to be the first generation, and are crossed with the red-eyed stock. Their white-eyed offspring are considered to be the second generation, and are again crossed with the red-eyed stock, and so forth. The special red-eyed stock is homozygous for the desirable allele A at some other locus, but the white-eyed fly is homozygous for the inferior a allele at that locus. What fraction of flies will have the a allele (at the second locus) after two generations