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Question: beyond mendel 65 exercise c gene linkage and independent assortment...

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Beyond Mendel 65 Exercise C: Gene linkage and independent assortment 17. Based on Morgans understanding of gene linkage, construct a Punnett square of a grey bodied, red-eyed female (ex hypothesis that he tested. with a yellow-bodied, white-eyed male (xy.This is the 18. What is the expected genotypic frequency of Xerr according to the gene linkage hypothesis? 19. what is the expected phenotypic frequency of rx-x xy according to the gene linkage hypothesis? Morgans results did not support the hypothesis of gene linkage. For example, he expected to find relatively equal proportions of grey-bodied/white-eyed males (xery) to grey-bodied/red-eyed males (Xy). Do you see that in your Punnett square? In Table 2 are his results for males of the xoxarxXry cross. While the phenotypes representing the genotypes xery and xry were nearly equally proportionate as predicted. He did not expect to see yellow bodied males at all. But how? Genotype Phentoypes Number grey body/red eyes 4292 grey body/white eyes 4605 yellow body/red eyes 86 yellow body/white eyes 45 Table 2. Male results of Morgans sex-linked dihybrid cross Again, the inner working of meiosis are responsible for Morgans results. He suggested that whille genes are usually on same chromosome and remain linked during meiosis, but not always. Sometimes alleles on the same chromosomes dont always stay together. Morgan proposed that during meiosis paired homologous chromosomes can swap segments of chromosomes between them. He referred to these recombined chromosomes (part paternal-part maternal) as recombinant chromosomes Further work by Morgan and colleagues supported this hypothesis, and they found that homologous chromosomes do, in fact, recombine during prophase I of meiosis (Fig. 4). Morgan termed this process as crossing over, and it involves the physical exchange of the same segments of non-sister chromatids between homologous chromosomes. www.thebiologyprimer.com
Beyond Mendel Exercise C: Gene linkage and independent assortment E In Fig. 4, label the process of crossing over In Fig. 4, label the recombinant chromosomes. Early Prophase I 20. How did the results of Morgans dihybrid cross violate his gene linkage hypothesis? Late Prophase I Metaphase I Anaphase Telophase and Cytokinesis I Meiosis II Fig. 4. Crossing over mixes segments of Fig. 5. Phenotypes expressed by the I gene. i codes for non-sister chromatids generating recombinant chromosomes a base polysaccharide, while Pand P code for the base plus a sugar Exercise D: One gene, multiple alleles Different alleles emerge via mutations during DNA replication, and they occur randomly. T possible for a gene to have more than one allele, and some do. Human blood of multiple alleles. A gene known as l is responsible for the production to a glycoprotein found in the cellular membrane of red blood cells (Fig. 5). The recessive base polysaccharide. The alleles and l code for the same base polysaccharide plus an added to the end. codes for one sugar, while codes for a different sugar herefore, it is type is an excellent example of a specific polysaccharide attached allele i codes for a sugar www.thebiologyprimer.com
Beyond Mendel 67 Exercise D: One gene, multiple alleles Codominance Genotype Blood type In human ABO blood type there are three blood type phenotypes possible: pt, P, , i, P, and W (Table 3). Understanding these blood types is important in blood transfusions, and it is due to the acceptance of the glycoproteins on the blood cellular membranes coded by the gene. Similar to Mendelian genetics, f is recessive to both Pand P. But what happens when both alleles are activated. So blood cells of the genotype produce different glycoproteins with both sugars. This double expression is known Type A Type AB both μ and p alleles are in a gene? As it turns out Type O Table 3. ABO Blood types 21. What is the expected ratio of blood types in children from a type AB mother and a type O father? Blood type is autosomal (not sex-linked). Exercise E: Incomplete dominance Mendel discovered that when a hybrid is created in the F, generation, one phenotype is present (dominant) while the other is absent (recessive). In flower color in four-oclocks, this is not the case. If you cross a pure line purple flower (F Fp) with a pure line white flower (FwFwl you wont get either purple or white flower in the Fi generation. You get pink flowers! If you cross the hybrids of the Fi generation, you will get the purple and white flowers back, but you also get about twice as many pink plants as either purple or white. In the case of a heterozygote, both alleles are coding for the phenotype not just one. purple white F all FFw (pink) 22. Which inheritance hypothesis that Mendel tested does incomplete dominance support? 1 F F lpurple): 2 FFw(pinkl: 1 Fwfw (white) Fig.6.Incomplete dominance in four-o clocks. www.thebiologyprimer.com
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