2. What is the probability that, in the family described above, their first two children will have dentinogenesis imperfecta? *

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1. Dentinogenesis imperfecta is a rare, autosomal, dominantly inherited disease of the teeth which occurs about one in 8. The teeth are somewhat brown in color and the crowns wear down rapidly. Assume that a male with dentinogenesis imperfecta (assume he is heterozygous), marries a woman with normal teeth. What is the probability that their first child will have dentinogenesis imperfecta? *.5.25 2. What is the probability that, in the family described above, their first two children will have dentinogenesis imperfecta?.5 *.25 3. What is the probability that, in the family described above, their first child will be a girl with dentinogenesis imperfecta?.5 *.25 4. You are growing pea plants in your a garden and some pea plants have solid leaves, while others have striped leaves. You conduct five different crosses (a

through e, above) and obtain the results given in the table. Define gene symbols and give the possible genotypes of the parents of each cross. Using upper case letters for dominant alleles and lower case letters for recessive alleles, the parental genotypes in cross a are most likely to be: AA x AA AA x Aa AA x aa Aa x Aa *Aa x aa aa x aa 5. The parental genotypes in cross d (above) are most likely to be: AA x AA AA x Aa AA x aa *Aa x Aa Aa x aa aa x aa 6. In Drosophila melanogaster, vestigial (short) wings (vg) are caused by a recessive gene which independently assorts with a gene pair which influences body hair. Homozygotes for the hairy allele (hh) have hairy bodies. A cross is made between a fly with normal wings and a hairy body and a fly with vestigial wings and a normal body. The phenotypically normal F1 flies were crossed among each other and 124 F2 flies were reared. What phenotypes would you expect in the F2 and in what actual numbers (not ratio) would you expect to find them? 1728 wild type: 256 vestigial and hairy 512 wild type: 512 vestigial and hairy 256 wild type : 256 hairy : 256 vestigial : 256 vestigial and hairy 32 wild type : 32 hairy : 32 vestigial : 64 vestigial and hairy *576 wild type : 192 hairy : 192 vestigial : 64 vestigial and hairy 7. With incomplete dominance, a likely ratio resulting from a monohybrid cross would be 9:3:3:1 3:1 1:2:2:4 *1:2:1 3:3

8. A condition in which one gene can mask the expression of a different gene is called additive alleles dominance recessiveness *epistasis codominance 9. Typical phenotypic ratios resulting from epistatic interactions in dihybrid crosses would be *9:3:4, 9:7 3:1, 1:1 1:1:1:1, 1:4:6:4:1 9:3:3:1, 1:2:1 1:2:2:4:1:2:1:2:1 1. A mutant gene which produces brown eyes (bw) is located on chromosome #2 of Drosophila melanogaster whereas a mutant gene producing bright red eyes, scarlet (st), is located on chromosome #3. Phenotypically wild type flies (with dull red eyes), whose mothers had brown eyes and whose fathers had scarlet eyes were mated. The 8 offspring possessed the following phenotypes: wild type (dull red), white, scarlet (bright red), and brown. Most of the 8 offspring had wild type eyes while those with white eyes were the least frequent. How many white-eyed flies would you expect in the F2 generation? *5 6 75 2 4 11. The actual numbers in each phenotypic class in the above cross were: 445 wild type : 155 scarlet : 14 brown : 6 white. Use a chi-square test to test the ssumption that the F2 phenotypic ratio is due to an epistatic interaction between the brown and scarlet loci. What values do you get for chi-square, the degrees of freedom, and the P value of the test. Hint: use a chi-square table from your lecture notes, slides or text book. Chi-square=.4; df=1; P is greater than.9

Chi-square=.16; df=1, P is greater than.5 and less than.9 Chi-square=2.89; df=1; P is greater than.5 and less than.2 Chi-square=.4; df=3; P is greater than.9 Chi-square=.16; df=3; P is greater than.9 *Chi-square=2.89; df=3; P is greater than.2 and less than.5 12. Many of the color varieties of summer squash are determined by interacting loci: and genotype with AA or Aa produces white fruit, aabb or aabb produce yellow fruit, and aabb produces green fruit. Assume that two fully heterozygous plants are crossed. Give the phenotypic ratio of their offspring 9 white: 4 yellow: 3 green 9 white: 3 yellow: 4 grren 9 white : 6 yellow : 1 green 1 white : 3 yellow : 3 green *12 white : 3 yellow : 1 green 13. A form of hemophilia are determined by genes on the X chromosome in humans. Assume that a phenotypically normal woman whose father had hemophilia is married to a normal man. What is the probability that their first daughter will have hemophilia? *.25.5.75 14. In a Drosophila experiment a cross was made between homozygous wild type females and yellow-bodied males. All of the resulting F1's were phenotypically wild type. However, adult flies of the F2 generation (resulting from matings of the F1's) had the following characteristics: all females were wild type, half the males were wild type and half the males were yellow. Is the yellow locus autosomal or X-linked or Y-linked? Autosomal *X-linked Y-linked 15. Is the mutant gene for yellow body in Drosophila recessive or dominant?

*Recessive Dominant Neither 16. Glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as an X- linked recessive gene in humans. A woman who does not have the disease, but whose father suffered from G6PD, marries a normal man. What proportion of their sons is expected to be G6PD?.25 *.5.75 17. If the husband in the above question was not normal but was also G6PD deficient, what proportion of their sons is expected to be G6PD.25 *.5.75 18. The pattern of inheritance of both Fragile-X Syndrome and Huntington Disease can be characterized by the term Sex-linked X-linked Y-linked Maternally inherited *Genetic aniticipation 19. A tortoiseshell female cat (with patches black and patches of orange fur) mates with a black male cat. The kittens are expected in what phenotypic ratio? Females 1% black; males 5% black and 5% orange Females 1% orange; males 5% black and 5% orange Females 1% black; males 5% black and 5% orange

Females 5% black and 5% orange; males 5% black and 5% orange *Females 5% black and 5% tortoiseshell; males 5% black and 5% orange Females 5% black and 5% tortoiseshell; males 5% black and 5% tortoiseshell 2. How many Barr bodies would one expect to see in cells of Turner syndrome females (karyotypically 45,X) and Klinefelter syndrome males (karyotypically 47,XXY)?, *,1.2 1, 1,1,2 21. In humans, the genetic basis for determining the sex "male" is accomplished by the presence of. a balance between the number of X chromosomes and the number of haploid sets of autosomes multiple alleles scattered throughout the autosomes one X chromosome one Y chromosome *a particular portion of the Y chromosome 22. Assuming a normal number of autosomes, what would be the sexes of an XXY mouse and an XXY Drosophila, ZW butterfly Female, female, female Female, male, male *Male, female, female Male, female, male Male, male, male