AaBbCc x AaBbcc How would you characterize this kind of relationship between the genes: A-B- (gray); A-bb (yellow); aaB- (black); aabb (cream) The CC and Cc genotypes allow color according to the expression of the A and B alleles. However, the cc genotype results in albino rats regardless of the A and B alleles present
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- The following genotypes of two independently assorting autosomal genes determine coat color in rats:A-B- (gray); A-bb (yellow); aaB-(black); aabb (cream)A third gene pair on a separate autosome determines whetherany color will be produced. The CC and Cc genotypes allow coloraccording to the expression of the A and B alleles. However,the cc genotype results in albino rats regardless of the A and Balleles present. Determine the F1 phenotypic ratio of the followingcrosses: (a)AAbbCC * aaBBcc; (b) AaBBcc * AABbcc;(c) AaBbCc * AaBbcc.In rats, the following genotypes of two independently assorting autosomal genes determine coat color: A-B- (gray) A-bb (yellow) aaB- (black) aabb (cream) A third gene pair on a separate autosome determines whether or not any color will be produced. The CC and Cc genotypes allow color according to the expression of the A and B alleles. However, the cc genotype results in albino rats regardless of the A and B alleles present. Determine the F1 phenotypic ratio of the following crosses: (a) AAbbCC * aaBBcc (b) AaBBCC * AABbcc (c) AaBbCc * AaBbcc (d) AaBBCc * AaBBCc (e) AABbCc * AABbccMultiple crosses were made between true-breeding lines of black and yellow Labrador retrievers. All the F1 progeny were yellow. When these progeny were intercrossed, they produced an F2 consisting of 121 yellow, 9 black and 30 chocolate. What epistatic ratio and what kind of epistasis is approximated in the F2? Propose a biochemical pathway for coat color in Labrador retrievers based on the type of epistasis. Correlate each genotype with the phenotype that would occur in your pathway. Also show the frequency of each genotype. A-B- A-bb aaB- aabb
- Construct a map for the genes d,e,f. Assume that: d and e = 3%; e and f = 5%. Give 2arrangements of the genes/maps.In rats, the following genotypes of two independently assorting autosomal genes determinecoat color:A–B– (gray)A–bb (yellow)aaB– (black)aabb (cream)A third gene pair on a separate autosome determines whether or not any color will be produced.The CC and Cc genotypes allow color according to the expression of the A and B alleles.However, the cc genotype results in albino rats regardless of the A and B alleles present.Determine the F1 phenotypic ratio of the following crosses: (1) AaBBCc × AaBBCc (2) AABbCc × AABbcc (3) AaBbCc × AaBbcc5 f There are two genes that determine the coat colour expression in dogs: eumelanin and merle. These genes are located on two separate chromosomes. For the eumelanin gene, black coat colour (E) is dominant over red coat colour (e). The merle gene controls the degree to which these coat colours are expressed through incomplete dominance. The following table describes the merle gene expression. Genotype MM Mm mm Phenotype White AAY Half colour (Grey or light red) Full colour (Black or Red) 1. If a dog breeder wants to determine the unknown genotype of a dog, what colour of dog must she use as a testcross? 2. A white dog was testcrossed, and all of the puppies had light red coats. What is the genotype of the white dog? Use a Punnett square to show your work. Hint - When a trait is controlled by two genes, an individual's genotype consists of 4 letters. = = = B I 123 ||| 14 E E I ос C GO * ●● ● ●0 00 00 00 ✓ C U up X₂ x²
- . The human IGF2 gene is autosomal and maternallyimprinted. Copies of the gene received from themother are not expressed, but copies received fromthe father are expressed. You have found two allelesof this gene that encode two different forms of theIGF2 protein distinguishable by gel electrophoresis.One allele encodes a 60K (Kilodalton) blood protein;the other allele encodes a 50K blood protein. In ananalysis of blood proteins from a couple named Billand Joan, you find only the 60K protein in Joan’sblood and only the 50K protein in Bill’s blood. Youthen look at their children: Jill is producing only the50K protein, while Bill Jr. is producing only the 60Kprotein.a. With these data alone, what can you say about theIGF2 genotype of Bill Sr. and Joan?b. Bill Jr. and a woman named Sara have two children, Pat and Tim. Pat produces only the 60K protein and Tim produces only the 50K protein. Withthe accumulated data, what can you now say aboutthe genotypes of Joan and Bill Sr.?Star eye A peculiar eye condition known as "star” is manifested as a dominant gene in Drosophila. Its recessive allele R* produces the normal eye of wild type. The expression of R can be suppressed by the dominant allele of another locus, Ru-R. Ru-R*, as the recessive allele of the said locus, has no inhibitory effect on R*. When a normal-eyed male of genotype Ru-R Ru-R RR is crossed to a homozygous wild-type female of genotype Ru-R* Ru-R* R*R*, what phenotypic ratio is expected in the F2?In a certain plant, leaf size is determined by fourgenes whose alleles assort independently and actadditively. Thus, alleles A, B, C, and D each add4 cm to leaf length and alleles A′, B′, C′, andD′ each add 2 cm to leaf length. Therefore,an AA BB CC DD plant has leaves 32 cm longand an A′A′ B′B′ C′C′ D′D′ plant has leaves16 cm long.a. If true-breeding plants with leaves 32 cm longare crossed to true-breeding plants with leaves16 cm long, the F1 will have leaves 24 cm longand the genotype AA′ BB′ CC′ DD′. Listall possible leaf lengths and their expectedfrequencies in the F2 generation produced fromthese F1 plants.
- A recessive epistasis cross gives in the F2 a 9:3:4 phenotypic ratio. What is the expected phenotypic ratio when the F1 (BbEe) is crossed with the double recessive homozygote (bbee)? Keeping in mind ee masks the expression of B.Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?In Drosophila, a female fly is heterozygous for three mutations, Bareyes (B), miniature wings (m), and ebony body (e). Note that Bar isa dominant mutation. The fly is crossed to a male with normal eyes,miniature wings, and ebony body. The results of the cross are asfollows.111 miniature29 wild type117 Bar26 Bar, mimatue101 Bar, ebony31 Bar, miniature, ebony35 ebony115 miniature, ebonyInterpret the results of this cross. If you conclude that linkage isinvolved between any of the genes, determine that map distance(sbetween them.