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- In cats, the genotype AA produces tabby fur color; Aa is also a tabby, and aa is black. Another gene at a different locus is epistatic to the gene for fur color. When present in its dominant W form (WW or Ww), this gene blocks the formation of fur color and all the offspring are white; ww individuals develop normal fur color. What fur colors, and in what proportions, would you expect from the cross AaWw Aa Ww?1. How many different types of gametes can be produced by an individual with the genotype Yy Ww Pp ee Bb? 2. If an individual with the genotype Yy Ww Pp ee Bb is crossed with another individual with the same genotype, what proportion of offspring will be homozygous recessive at all genes? 3. If an individual with genotype Yy Ww Pp ee Bb is crossed with an individual with the same genotype, what proportion of offspring will have the dominant phenotype for the four heterozygous loci?2. 9, 1 pts Cystic fibrosis is a disease which is caused by a defective CFTR gene. This form of the gene is recessive. Let "N"=normal CFTR gene and "n" = abnormal CFTR gene. What is the genotype of individual 1 in generation I? 1. %3D 2 3 II 2 3. 5. O NN O Nn O Either NN or Nn
- 2. Let's assume you did a dihybrid cross of homozygous dpy-5 hermaphrodites and homozygous him-5 males. You let the F1 hermaphrodites to self-fertilize and collected the F2 progeny. What are the ratios of the following genotypes among all the F2 progeny? In the table below, "dpy-5" or "him-5" indicates the mutant allele and the plus sign "+" indicates wildtype allele. The slash “/” separates two alleles of the same locus. The semicolon ";" separates different chromosomes. Use words or any type of diagrams that we learned in the class to show your work. Genotype dpy-5/dpy-5; him-5/him-5 dpy-5/+; him-5/him-5 +/+; him-5/him-5 dpy-5/dpy-5; him-5/+ dpy-5/+; him-5/+ +/+; him-5/+ dpy-5/dpy-5; +/+ dpy-5/+; ++ +/+; +/+; Ratio15. Albinism in corn plants is caused by a recessive lethal gene that results in death before maturity. What will the adult phenotypic ratio be for the F1 generation of heterozygous parents? 16. Huntington's chorea is a dominant lethal in humans. The disease does not appear until later in life, so that afflicted individuals may already have produced children. What are the F1 genotypic and phenotypic ratios of parents who are homozygous dominant and heterozygous?2. Having a "Widow's Peak" is a recessive trait. If both parents are heterozygous for the "Widows Peak gene. A) What are the genotypes and phenotypes of each parent? B) What are the chances that their children will have a "Widow's Peak"? (MA CAM) M 7 Chance that Nidars Pa their children ulihave the 3. Pattern baldness is a recessive sex-linked trait. Jake has a full head of hair. His wife Jeannie is heterozygous for the pattern baldness gene. A) What is Jeannie's phenotype? Why? B) What are the chances that their children will have pattern baldness? S
- 7. Below is a pedigree for a family in which some members have achondroplastic dwarfism. This trais is inherited as an autosomal dominant. Individuals shaded on the pedigree have achondroplastic dwarfism. The homozygous dominant condition usually results in death in-utero. Complete the chart indicated relationship and whether both, one or neither has achondroplastic dwarfism.1. How many asci contain a parental order of ascospores? 2. How many asci contain ascospores representing crossing over? 3. How many total segregants were analyzed? 4. What is the distance from the spore color gene to the centromere?Part 1 Monohybrid crosses 1. The following lists represent genotypes at imaginary genetic loci. In each case there may be one or more correct answers – list ALL correct answers a. Homozygous dominant Mm NN oo pp Qq RR b. Homozygous recessive Mm NN oo pp Qq RR c. Genotypes in which dominant allele must show Mm NN oo pp Qq RR d. Genotypes in which recessive allele must show Mm NN oo pp Qq RR
- 5. SpongeBob SquarePants recently met SpongeSusie Roundpants at a dance. SpongeBob is heterozygous for his square shape, but SpongeSusie is round. Create a Punnett square to show the possibilities that would result if SpongeBob and SpongeSusie had children. 1. List the possible genotypes of the offspring: Shape: Square (S), round (S). 2. List the possible phenotypes of the offspring: A Genotype for SpongeBob:SS B Genotype for SpongeSusie SS C. Complete the Punnett Square 3. What are the chances of having an offspring with square shape: S S out of or %. 4. What are the chances of having an offspring with round shape: out of or %. 5. What is the ratio of square to round shape? SI10. What phenotypes are expected and in what proportion from each of the following n crosses? (Assume D, G and Ware dominant over d, g and w, respectively.) 31 (a) DdGGww x DDGgWw how bon (b) ddggWw x DdGgww HD: (c) DdGgWw x DdggWw (d) DdGgWW x DdGgWW baildhalos 6 am bunjid-roloos (d) nomow sof bemam nam lemona (0) namnow beild-oloss of bemem nem tarono [b) int euogyson M 11. Radishes can be long, round or oval in shape. They may be red, white or purple in color. A long red variety was crossed with a round white variety. In the F₁ generation, all were oval purple. In the F2 generation, there were 8 long red, 15 long purple, 17 oval red, 31 oval purple, 7 long white, 15 oval white, 8 round red, 16 round purple and 9 round white. Explain the genetic basis of these results. hot tmsidorss endohes barlastis bed terftet ben hemouth bra (Jisi edohss srillo sonsmedni Snem ar to gaylonog ori ai fer msil) ov fa cisiv lemon diw be (!) 12. In shorthorn cattle, the allele for red coat color (R) is…11. Below is a pedigree chart of an autosomal recessive disorder. Answer the following questions using the correct genetic terminology (do not just write letters like “Ee”). A. What is the genotype of individual 1 in generation II? B: What is the genotype of individual 2 in generation I?C: Is it true that individuals 6, 7, 8, 9, and 10 in generation III all have the same genotype? Why or why not?