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Unformatted text preview: Rough guide to your grade on this test (after adding basic score + extra credit together): 88-100 = A, 78-87 = B, 65-77 = C, 40-64 = D, <40 = F. Recall that your final grade in the class will factor in extra credit from clicker responses! Median not yet calculated posted later. 1 BICD100 Midterm 1 1/25/10 KEY 1. Non-disjunction during meiosis results in aneuploid gametes, which either lack a particular chromosome or have two copies of it. If an aneuploid gamete participates in fertilization, the resulting embryo either lacks a chromosome (=monosomic) or has an extra chromosome (=trisomic). Monosomic embryos usually dont survive gestation, but trisomics sometimes do. For example, individuals having 3 copies of chromosome 21 (trisomy 21) have Downs Syndrome (a relatively mild impairment), whereas monosomy 21 individuals do not survive. Which of the following could help to explain the difference between the fate of monosomy 21 and trisomy 21 individuals? Circle all that apply (4 pts): a. Dosage compensation >b. Haploinsufficient genes on chromsome 21 c. Gain-of-function mutations on chromosome 21 d. Synapsis >e. Recessive lethal mutations on chromosome 21 f. Mendels Law of Segregation g. X-inactivation h. Maternal inheritance 2. As discussed in class, flower color in snapdragons exhibits semi-dominant inheritance: progeny of crosses between plants with red flowers and plants with white flowers are all pink. Snapdragon flowers can also have two distinct shapes: personate or peloric (see illustration). The difference between personate and peloric is controlled by a single gene where one allele is completely dominant over the other. Crosses between true-breeding lines were carried out as described below: i. red peloric X white personate all F1 progeny pink, personate ii. red personate X white peloric all F1 progeny pink, personate F1 progeny from cross i were crossed to F1 progeny from cross ii. What phenotypic classes were observed in the F2 generation and in what proportions (8 pts)? Let R and W represent the semi-dominant flower color alleles: RR flowers are red, RW flowers are pink, and WW flowers are white. Let P and p represent the flower shape alleles. The results of crosses i and ii show that personate is dominant to peloric so PP and Pp are personate, pp is peloric. F1 progeny from both cross i and cross ii are Pp RW. When two plants of this genotype are crossed together, the progeny will be red, pink and white. Also, will be personate and peloric. The inheritance of alleles for flower color is independent from the inheritance of alleles for flower shape so: p(red and peloric ) or (white and peloric ) = x = 1/16 of each class p(red and personate ) or (white and personate ) = x = 3/16 of each class p(pink and peloric) = x = 1/8 pink and peloric p(pink and personate) = x = 3/8 pink and personate Note that these add up to 1, as they must since these are all the possible outcomes. Rough guide to your grade on this test (after adding basic score + extra credit together):...
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This note was uploaded on 08/19/2010 for the course BICD bicd 100 taught by Professor Soowal during the Winter '08 term at UCSD.
- Winter '08