Page 1 of 7. Name: A. Preliminary Assessment #3. You may need a calculator for numbers 2&3.

Transcription

1 Page 1 of 7 Name: A Preliminary Assessment #3 You may need a calculator for numbers 2&3. You may bring one 3 inch by 5 inch card or paper with anything handwritten on it (front and back). You have also been provided with some resources (separate sheets). Do not write on the backs of pages. If you run out of space, the second-to-last page in the packet has a blank page for you to put your answers on. If you put anything there that you want us to grade, please clearly indicate that on the page with the question. Use ink DO NOT WRITE ON THE BACKS OF PAGES

2 Page 2 of 7 1) In a species of vines, they are either branched or branchless. The majority of the vines are branchless and are homozygous recessive (two non-functional copies) for genes a and b (aabb). A small fraction of vines have a functional allele that codes for the enzyme Anuliate. You breed these together for a long time until you have some true-breeding plants with branches (genotype AAbb). One day you cross one of your true-breeding (AAbb) plants with another plant that has no branches and a dominant copy of a gene of unknown function which codes for a working copy of the enzyme Bughery (genotype aabb). Half of the offspring have branches and half of the offspring do not. a) Draw a Punnett square for this cross and explain how these genes work together. Include in your explanation what the function of the Bughery enzyme is. (10 points) Bughery blocks expression/function of Anuliate (4 points) Punnett square and phenotypes (6 points) ab ab Ab AaBb Aabb Ab AaBb Aabb Phenotype: 1:1 branchless (AaBb) and with branches (Aabb) b) You take one of the branchless offspring from part (a, above) and cross it back to one of the common, all recessive, branchless plants (aabb). What phentoypes and in what ratios would you expect for the offspring? (13 points) AB Ab ab ab aabb AaBb Aabb aabb aabb phenotype ratio: 1:3 branches (Aabb) to no branches (AaBb, aabb, aabb) (10pts) Explaining the phenotypes (3pts): AaBb: B blocks A so no branch aabb: recessive for branches (regardless of dominant allele of repressor B) aabb: recessive no branches Aabb: branches, heterozygous for branches and no dominant B allele to block the function

3 Page 3 of 7 2) In house flies, an uncommon, dominant gene that codes for an enzyme that disrupts gene development and creates wingless flies is on the X chromosome. Also on the X chromosome is an uncommon, dominant gene that codes for an enzyme that converts the normal black eye pigment to yellow. The two genes are 4 map units apart. a) A male fly with yellow eyes and no wings is mated to a normal female (with black eyes and wings, who is not a carrier for either trait). What are the phenotypes (and in what ratio) would you expect the female offspring of this cross to be? Explain in reference to meiosis and probabilities of generating gametes with different allele combinations. (15 points) Phenotypes: female offspring should have yellow pigment/no wings; now heterozygous dominant for the traits, inherited dominant/rare alleles from father (8pts) Male fly = only one X chromosome that carries the dominant alleles; gives these to ALL female offspring (male fruit flies do not have crossing over) 4 map units=4% recombination=4cm => very close on chromosome/linked and during meiosis I in prophase I the probabilities of generating gametes with different allele Meiosis I/ prophase I = cross over with 4% chance of recombination BUT no in father no recombination between X and Y) (7 pts) b) If you pick a female born in part a (above) and mate her with a normal male (black eyes with normal wings), what phenotypes (and in what ratio) would you expect the male offspring of this cross to be? Explain in reference to meiosis and probabilities of generating gametes with different allele combinations. (15 points) In this case the males born from the cross will have no functional genes on the Y chromosome from their father and so their phenotype will depend 100% on mom s eggs. The female is heterozygous for both genes. 96% of her eggs will be non-recombinant, meaning 48% YW and 48% yw. 4% will be recombinant: 2% Yw and 2% yw. So of the sons born from this heterozygous female, 48% should be yellow eyed and wingless, 48% should be black eyed and have wings, 2% will be yellow eyed with wings, and 2% will be black eyed and wingless. NOTES: -2 points for having 8% recombinants -8 points for discussing recombination but not saying that it can happen in the offspring.

4 Page 4 of 7 3) a) A population of 10,000 gorillas is homozygous recessive for the inability to taste phenolthiolcarbamide (all tt). 1,000 new gorillas arrive in the population from another area. The 1,000 new gorillas are homozygous dominant for the ability to taste (all TT). There is no further gene flow, no mutation, no natural selection, and random mating (no sexual selection). If you come back to this population after many generations and it still has 10,000 gorillas, what fraction of each genotype would you expect to find? Explain. (15 points) p = 1/11 q = 10/11 No drift. (4 points) It is required to explicitly state this in this part to distinguish from other situation No other selection/evolution, so HW holds (4 pts) Expected ratios (7 points) Expected TT = p 2 = 0.83% or about 83 gorillas Expected Tt = 2pq = 16.53% or about 1653 gorillas Expected tt = q 2 = 82.64% or about 8264 gorillas NOTE: the only time you would see 1:2:1 ratio of genotypes is if 1:1 ratio of alleles, that is not the case here (or with any of the Homework or practice assessment questions on this topic) b) A population of 20 gorillas is homozygous recessive for the inability to taste phenolthiolcarbamide (all tt). 2 new gorillas arrive in the population from another area. The 2 new gorillas are homozygous dominant for the ability to taste (all TT). There is no further gene flow, no mutation, no natural selection, and random mating (no sexual selection). If you come back to this population after many generations and it still has 20 gorillas, what fraction of each genotype would you expect to find? Explain. (15 points) This smaller population will be subject to genetic drift (5 points) and no other forces (3 points) 7 point for either saying 100% of one allele or the other.

5 Page 5 of 7 4) In the pedigree below, males are represented by squares, females are represented by circles, open symbols represent individuals who do not have a rare trait that we are following and closed circles represent individuals who do have that trait. Explain how it is possible (even if very unlikely) that this pedigree could be for an autosomal recessive allele. Then explain why it is much more likely that the trait is an autosomal dominant allele. (10 points) Recessive is possible but would require the following to be carriers: Male in generation 1 (2 pts) 2 females and 2 males marrying in in generation 2 (2 pts) This is unlikely for a rare trait (2 points) Dominant can be explained just by the one affected gene from the female in generation 1 passing down to all kids (4 pts)

6 Page 6 of 7 5) If a system is Positive Inducible (2 points): a) A loss-of-function mutation to the regulator gene that keeps the regulator from binding to the operator would cause the system to be (choose one, 1 pt): b) A loss-of-function mutation to the structural gene would cause the system to be (choose one, 1 pt): 6) If a system is Negative Repressible (2 points): a) A loss-of-function mutation to the regulator gene that keeps the regulator from binding to the operator would cause the system to be (choose one, 1 pt): b) A loss-of-function mutation to the structural gene would cause the system to be (choose one, 1 pt): 7) You are studying the color of duck feathers. You have one true-breeding strain of ducks that have active alleles at the two genes (on two different chromosomes) that control feather color (AABB) and are also all white. Another strain of ducks have non-functional (recessive) alleles for both genes (aabb) and are all brown. You cross one duck from each population together. They produce F1 offspring which are mated to each other and produce an F2 generation with a ratio of 12 white : 3 yellow : 1 brown. What is the most likely type of genetic interaction. Explain. (3 points) The most likely interaction is that one of the functional enzymes (or genes) inhibits the production of the brown phenotype causing white to appear (2 points) The other gene codes for an enzyme that converts the brown pigment to yellow (1 point)

7 Page 7 of 7 Blank page. You may use this for scratch work. If you need to continue a question on this page, please make a note by that question so we know to come looking for your answer here.