Biology 105: Introduction to Genetics Midterm EXAM Part1 Definitions 1 Recessive allele Name Student ID 2 Homologous chromosomes Before starting, write your name on the top of each page Make sure you have all pages You can use the back-side of the pages for scratch, but we will not grade answers written on the back-side of the page. Unsolved fractions are acceptable answers NO PARTIAL POINTS for INCOMPLETE ANSWERS! 3 Inversion loop Part I Definitions 15 points -------------------- Part II Short Answer 20 points -------------------- Part I Multiple Choice 25 points -------------------- Part V Extended Calculation 30 points -------------------- Total -------------------- 4 Non-dysjunction 5 Coupling confirmation 1 2
Part2 Short answers B What problems if any would the hybrid face during mitosis and meiosis. 1 (3 pt) In chickens there are two unlinked genes, R and P that control the phenotype of the comb. For each gene there is a dominant allele (R and P) and a recessive allele (r and p). The genes produce the following genotype/phenotype relationships: R_P_ = walnut comb R_pp = rose comb rrp_ = pea comb rrpp = single comb. (_ indicates that either allele is present) A true breeding pea comb is crossed with a true breeding rose comb. A What is the genotype of the pea comb and rose comb parents? B What will be the proportions of the different genotypes and phenotypes in the F1? C If R and P had been tightly linked such that no recombination occurred, how would this have changed your answer to part B 4 (4pt) Human cells normally have 46 chromosomes (44 autosomes and 2 sex chromosomes). For each of the following stages state the number of DNA molecules present in a human cell Metaphase of mitosis G1 phase of mitosis Metaphase of meiosis I Telophase of meiosis II 5 (4 pt) The blood groups A, B, AB and O are determined by multiple alleles- A, B and o. A and B are dominant over o and A and B are codominant to one another. What type of blood type would the children be expected to have from the following crosses: ABxoo ABxAA 2 (2pt) Recombination frequency between two genes far apart on the same chromosome is around 50%. So too is recombination frequency between two unlinked genes. How would you differentiate between the two cases? 3 (2pt) The red fox has 17 pairs of long chromosomes. The arctic fox has 26 pairs of shorter chromosomes. A What do you expect to be the chromosome number in the somatic tissue of the F1 hybrid? ABxBo AoxBo 6 (2pt) Two genes T and S code for height and skin tone in vampires. (T=tall, t=short; S=smooth, s=wrinkled). A heterozygous TtSs individual mates with a homozygous ttss individual. They have progeny: 41 tall wrinkled; 39 short smooth; 11 tall smooth; 9 short wrinkled. Are the two genes linked or on separate chromosomes. If linked what is the distance between these two genes. 3 4
7 A human disease afflicted a family as shown in the accompanying pedigree. Part4 Multiple choice: 1 A human disease afflicted a family as shown in the accompanying pedigree. Deduce the most likely mode of inheritance of this single gene trait. Cousin numbered 2 married cousin numbered 4 What is the genotype of cousin 2 and cousin4 What percentage of their male progeny will be normal What percentage of their female progeny will be normal 8 You are given one red-eyed male fly. The genotype of this fly is not known. Red is an autosomal dominant trait. In the lab you have true breeding red-eyed and white-eyed flies. Diagram one cross you would need to do and the expected results you would get in order to unambiguously determine the genotype of the red-eyed fly. 9 The karyotype of a child is shown below. What event during meiosis can best explain the child s observed karyotype. In which parent did this event occur. The most likely mode of inheritance is A X-linked dominant B X-linked recessive C Y-linked dominant D Y-linked recessive E autosomal dominant 2 Normal mitosis takes place in a diploid yeast cell of the genotype A/a B/b. Which genotype is most likely to be present in the daughter cells A A B B a b C A/A B/B D A/a B/b E a/a b/b 3 Inversions result in A the loss of chromosomal material B the gain of chromosomal material C rearrangements involving two different chromosomes D rearrangements involving only a single chromosome E none of the above 4 Dosage compensation in humans is accomplished by the following means: A. Two-fold expression of X-linked genes in males B. Inactivation of an X chromosome in females C. One-half fold expression of genes on the two X chromosomes in females D. One-half fold expression of genes on the two X chromosomes in males 5 6
E. Y-inactivation in males 5 A tall parent and a short parent produce F1 offspring that when self-fertilized produce a range of offspring heights, some taller that the tall P1 and some shorter than the short P2. Which set of parental genotypes could best account for these data A AABBCC x aabbcc B AABB x aabb C AABBcc x aabbcc D AA x aa E BB x bb 6 Which of the following does not occur during meiosisi A recombination B replication of homologous chromosomes C separation of homologous chromosomes D pairing of homologous chromosomes E separation of sister chromatids 7 Genes on the second chromosome of Drosophila are shown below ---------S--------------------Q---------------------P-----------------Y---------------T-- 18 MU 15MU 20 MU 10MU Given the map, in a cross between a SsTt x sstt fly, what percentage of recombinant progeny do you expect A less than 63% B greater than 63% C 63% D 50% E 28% 8 The region of the chromosome to which the spindle attaches is known as the A chromatid B telomere C acromere D centromere E centriole 9 How many types of gametes will a diploid individual of the genotype AaBBCcDDEEFfGG produce A 2 B 4 C 8 D 16 E 32 10 Dosage compensation in Drosophila refers to a phenomenon in which A one set of autosomes becomes inactive in females B one set of autosomes becomes inactive in males C One X chromosome becomes inactive in females D The X chromosome becomes doubly active in males E extra chromosome resulting from a non-dysjunction is expelled from the nucleus 11 In a cross of BbDd x BbDd what proportion of the progeny will have the domainant B allele and the dominant D allele A 1/4 B 9/16 C 3/16 D 1/16 E none of the above 12 The most common form of color-blindness in humans results from an X-linked recessive gene. A phenotypically normal couple have a normal daughter and a son who is color blind. What is the probability that the daughter is heterozygous A 100% B 75% C 50% D 25% E 0% 13 A segment of a chromosome has five genes in the order A-B-C-D-E. Three different individuals have these genes in the following order Individual1 A-B-C-D-C-D-E Individual2 A-B-D-E Individual3 A-B-D-C-E These rearrangements are A translocation, deletion, inversion B duplication, deletion, translocation C duplication, deletion, inversion D duplication, translocation, inversion E mutation, interception, misperception 14 A gene for examania (E) resides on chromosome13. There are two alleles for this gene- E A and E F. An E A E A father and an E A E F mother produce a child with the genotype E A E F E F. This was due to A Non-dysjunction in meiosisi in the father and normal dysjunction in the mother B Non-dysjunction in meiosisii in the father and normal dysjunction in the mother 7 8
C D E Non-dysjunction in meiosisi in the mother and normal dysjunction in the father Non-dysjunction in meiosisii in the mother and normal dysjunction in the father normal dysjunction in mother and normal dysjunction in the father C D E 1:2 ratio 1:3 ratio 3:1 ratio 15 In the pedigree below, the incidence of a common hereditary disease is indicated by darkened symbols. The best interpretation from the pedigree is that the disease is caused by a A B C D E autosomal recessive mutation autosomal dominant mutation X linked dominant Y linked recessive X linked recessive 16 A breeder identifies a dog with blue hair. Through multiple generations he crosses blue dogs with one another to establish a true-breeding line. However with each generation he obtains 25% white dogs, 50% blue dogs and 25% black dogs. He concludes that the trait is behaving as a A incomplete dominant trait. B codominant trait. C Dominant trait D lethal trait E recessive trait 17 A true breeding double dominant strain of corn (AABBCC) is crossed to a true breeding recessive strain (aabbcc). The F1 plants are then test crossed. If the A/a locus is linked to the B/b and C/c loci then we should expect more A Abc and abc gametes B abc and AbC gametes C ABC and abc gametes D ABc and abc gametes E AbC and abc gametes 18 Individuals heterozygous for a dominant eye mutation S have small eyes. The S allele is also associated with recessive lethality. When two heterozygotes are crossed, normal-eyed and small eyed individuals are expected in a A 1:1 ratio B 2:1 ratio 9 10
Part5 Extended calculations Problem 1) An individual heterozygous for three genes (AaBbCc) is test-crossed to an aabbcc individual and the progeny are classified by the gamete contribution from the heterozygous parent as follows A c B 650 a C b 655 a c b 152 A C B 144 a c B 140 A C b 145 A c b 28 a C B 30 Problem2 Wildtype sweet pea plants have red flowers and long pollen. These two traits are dominant to white flowers and short pollen. A true breeding wild-type plant is crossed to a true breeding plant with white flowers and short pollen. The F1 was test crossed and the following progeny were obtained: Wildtype 222 Short pollen 27 White flower 23 White flower short pollen 228 Are the genes linked? If they are linked what is the distance between them? A) Are the three genes linked? Show your reasoning B) Which alleles are present on each of the parental chromosomes? Show your reasoning. If the F1 were selfcrossed instead of testcrossed, what would be the frequency of each of the four phenotypic classes. C) What is the order of these genes on the parental chromosomes? Show your reasoning Expected three DCO classes are D) Indicate map distance between these three genes 11 12