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BIO 304 Lecture 8

BIO 304 Lecture 8. A cross is made between AaBb and aabb plants and the offspring include approximately equal numbers of plants with the genotypes AaBb , Aabb , aaBb , and aabb . These results are consistent with which of the following? 1 . complete linkage

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BIO 304 Lecture 8

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  1. BIO 304 Lecture 8

  2. A cross is made between AaBb and aabb plants and the offspring include approximately equal numbers of plants with the genotypes AaBb, Aabb, aaBb, and aabb. These results are consistent with which of the following? 1. complete linkage ✓ 2. independent assortment 3. interference 4. co-dominance 5. X-linkage

  3. A cross is made between AaBb and aabb plants. If the A and B genes demonstrate complete linkage, the offspring will be expected to have which of the following genotypes? ✓ 1. AaBb and aabb only 2. AaBb, aabb, Aabb, and aaBb 3. aaBb and Aabb only 4. AABB, aabb, and AaBb 5. AABB, aabb, AaBb, aaBb, and Aabb

  4. The phenomenon in which one crossover decreases the likelihood of crossovers in nearby regions is called ______________. 1. first division segregation 2. chiasmata ✓ 3. interference 4. reciprocal genetic exchange 5. coefficient of coincidence

  5. In the cross AaBbCc X aabbcc, one of the gametes produced by the heterozygous parent will be genotype AbC. What is the genotype of the reciprocal gamete? 1. ABC 2. abc 3. AbC ✓ 4. aBc 5. ABc

  6. During recombination, double crossovers would be expected to occur 1. More frequently than single crossovers 2. At the same frequency as single crossovers ✓ 3. Less frequently than single crossovers 4. At exactly 1/2 the frequency of single crossovers

  7. In Neurospora, the linear arrangement of the eight ____________ reflects the sequence of their formation during meiosis. 1. chiasmata 2. tetrads 3. parental gametes ✓ 4. ascospores 5. chromosomes

  8. Variable expressivity piebald spotting in beagles (SP allele) Drosophila eyeless gene

  9. Temperature-sensitive mutations

  10. Linkage vs. independent assortment

  11. Example of a cross demonstrating linkage: 1st gene: eye color2nd gene: wing length pr+ = red vg+ = normal pr = purple vg = vestigial purple eye, vestigial wing X red eye, normal wing P: (pure breeding) F1: red eye, normal wing F1 test cross: red eye, normal wing X purple eye, vestigial wing red eye, normal wing 1339 purple eye, vestigial wing 1195 red eye, vestigial wing 151 purple eye, normal wing 154 F2:

  12. Try at home: What would the parental and recombinant phenotypes be in the F2 generation if the first cross was the following: P: (pure breeding) red eye, vestigial wing X purple eye, normal wing F1: red eye, normal wing F1 test cross: red eye, normal wing X purple eye, vestigial wing F2:

  13. Allele designations for linked genes: Calculating recombination frequency (RF):

  14. Recombination frequency is proportional to the distance between two genes genes close together: few recombinants genes far apart: many recombinants

  15. Map units are based on RF

  16. Single crossovers and double crossovers

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