2. The homozygous mutant a, an allele of the albinism gene in the mouse, has a white coat and dark eyes. The homozygous mutant p has pink eyes. Mice of genotype a+/+p are testcrossed with mice homozygous for the two mutations. The progeny obtained are shown below. Calculate the recombination frequency and map distance between the two genes. Phenotype of Progeny White coat, dark eyes White coat, pink eyes Dark coat, white eyes Dark coat, pink eyes Number 240 31 34 474

2. The homozygous mutant a, an allele of the albinism gene in the mouse, has a white coat and dark eyes. The homozygous mutant p has pink eyes. Mice of genotype a+/+p are testcrossed with mice homozygous for the two mutations. The progeny obtained are shown below. Calculate the recombination frequency and map distance between the two genes. Phenotype of Progeny White coat, dark eyes White coat, pink eyes Dark coat, white eyes Dark coat, pink eyes Number 240 31 34 474

Name: 2. The homozygous mutant a, an allele of the albinism gene in the mou
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Description: 2. The homozygous mutant a, an allele of the albinism gene in the mouse, has awhite coat and dark eyes. The homozygous mutant p has pink eyes. Mice ofgenotype a+/+p are testcrossed with mice homozygous for the two mutations.The progeny obtained are

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter13: Modern Understandings Of Inheritance

Section: Chapter Questions

Problem 2VCQ: Figure 13.4 Which of the following statements is true? Recombination of the body color and red/...

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11. Shown below are the offspring of the testcross qr/q +r+ X q r/q r. Which groups are in the recombinant class? q+r and qr+ q+r+ and qr+ qr and qr+ q+r+ and qr
5. A strain of Neurospora with genotype A B is crossed with a strain with the genotype a b. Half the progeny are A B and half are a b. Explain how this is possible?
1) By decomposing the cross, MmNn X Mmnn, into single gene outcomes, the possibility of obtaining a double recessive outcome for the ‘N’ gene is: a) 50%! b) 100! c) 0%! d) 25%! 2) A 16-square Punnett is time-consuming to draw out. Dr. Thompson can easily solve this problem by decomposing her dihybrid crosses into separate monohybrid Punnett squares, which would also be quicker than using a forked line diagram. If so, how many individual monohybrid Punnett squares will she need to draw? a) Four Punnett squares. b) Two Punnet squares. c) One Punnett square. d) Sixteen Punnett squares since there are 16 possible outcomes.
1. a) If you were at risk of a fatal disease such as Huntington’s disease and only the 95% accurate preliminary genetic test was available, would you take the test? After all, 95% of the time the test would tell you whether you would eventually get Huntington’s disease or whether you did not have to worry about getting this disease. b) The linkage distance between the DNA sequence used in the test and the actual Huntington’ disease locus was 5%. In a large sample of the Venezuelan family, what percentage of people would inherit the DNA sequence but not the Huntington’s allele (i.e. be a false positive test for Huntington’s)? What percentage of that large Venezuelan family would not inherit the DNA sequence but would inherit the Huntington’s allele (i.e., be a false negative test for Huntington’s)?
2. A maize plant heterozygous for three genes is crossed in a test cross. Are obtained the following results: Progeny phenotype ABC 304abc 301Abc 35aBC 33ABc 8abC 6AbC 172aBc 169 A. Determine the order and distance of the genesb. Determine the coefficient of coincidence and the interference.
1) Given the map a15b24c (where 15 and 24 are the respective map distance between the loci) with a coefficient of coincidence of 0.5, how many "abc" pheontype individuals would be observed among 1000 progeny of a test cross of the following: a+c/+b+ 2)A,B,C, and D are genes linked on the same chromosome. Given the following recombination frequencies, what is the gene order? A-B 15%, B-C 26%, C-D 30%, A-C 11%,A-D 19%, and B-D 4%.
8. In guinea pigs, the allele for black fur (B) is dominant over the allele for white fur (b). Similarly, the allele for straight hair (H) is dominant over the allele for having rough hair (h). Pure breeding rough haired guinea pigs with black fur were crossed with pure breeding straight haired guinea pigs with white fur. a. State the genotype and the phenotype of the F1 individuals produced as a result of this cross.b. Two F1 offspring were mated together. Calculate the expected ratio of phenotypes in the F2 generation.c. Show the completed Punnett square below.
2.In dogs, there is a hereditary deafness caused by a recessive allele, d. A kennel owner has a male dog that she wants to use for breeding purposes if possible. The dog can hear, but the owner is unsure of the genotype. She does a testcross (crosses it to a homozygous recessive dog), and two of the five offspring are deaf. This means that the male dog A.has the genotype DD B.has the genotype Dd C.is still of unknown genotype since there were offspring of both deaf and hearing phenotypes. D.has the genotype dd
3) A. How are the following proteins involved in cell division: condensing, cohesin, kinetochore complex, tubulin? B. Why does phenotype not always reveal genotype? C. How can Mendel's postulates of segregation and independent assortment be explained by meiosis? D. Explain the genotypic and phenotypic ratios predicted when performing a monohybrid and a dihybrid cross of heterozygotes? E. How can you use the product rule and sum rule to calculate the probability of a particular genotype resulting from crossing parents of specific genotypes?
1) By convention, the recessive trait will determine the abbreviation used to track crosses if it is a mutant condition, since it stands out in contrast to the rest of the population. For example, if pea pods are typically green in color, a mutant condition might result in yellow pea pods; therefore, the lowercase letter ‘y’ would be used to depict the mutant state, yellow, while uppercase ‘Y’ would depict the wildtype (wt) condition, green. Remember that the mutant condition is not necessarily always recessive. 1 A) . If green is dominant wt (Y) and yellow is a recessive mutant condition (y), depict a Yy father mated with a YY mother in the Punnett. 1B) Considering the dominant allele, what colors are the parents in the cross above? Yy = _____________________ YY = _____________________ 1C) What colors are the offspring in the cross above? 1D) What is the phenotypic ratio of the offspring? 1E) What is the genotypic ratio of the offspring? (Remember, the genotypic ratio is based…
4. Define the following terms: A. genotype B. phenotype C. monohybrid D. dihybrid
4) A testcross is used to determine which of the following? a) The phenotype of an individual with a dominant genotype b) The phenotype of an individual with a recessive genotype c)The genotype of an individual with a dominant phenotype d) The genotype of an individual with a recessive phenotype