) You are performing a dihybrid cross with a strain of flies that is ebony (e) and a strain that is pink (p). The results of the F2 are shown below. Use x² test to help you determine if e and p are linked (Note: refer to Table I for chi-square table). Number 1827 Phenotype + p 424 386 298 e + ep Genetic map is the relative positions of genetic markers on a chromosome. Explain the rationale behind genetic mapping.
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- The genotypes below that give viable animals for an mdm2+/- p53+/- cross are the following (p53 on chr 17 and mdm2 on chr 12) are:Consider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PC was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PC products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?A RFLP is discovered that is linked to the gene for Duchenne’s muscular dystrophy (DMD). DMD is an X-linked, recessive trait. The RFLP is 2 map units from the gene for DMD. Consider the following pedigree and Southern blot using a probe that hybridizes to the RFLP. Which band/s is/are associated with DMD? What is the genotype for individuals 3 and 4? (Remember, this is an X linked disease, so use X’s and Y’s to denote). Individual 9 married a man who does NOT have muscular dystrophy, and she is pregnant. DMD is an X-linked trait. What is the probability for their child to have DMD? An amniocentesis is performed and it is determined that 9’s child in utero has only a 10 kb band that hybridizes to the same probe used above. What can you say about the child now?
- n corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines Using the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspringConsider a maize plant: Genotype C/cm ; Ac/Ac+ where cm is an unstable colorless allele caused by Ds insertion. What phenotypic ratios would be produced and in what proportions when this plant is crossed with a mutant c/c Ac+/Ac+? Assume that the Ac and c loci are unlinked, that the chromosome-breakage frequency is negligible, and the C allele encodes pigment production.In corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines Using the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm of each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspring
- Consider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PCR was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PCR products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?n corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines sing the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm of each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspringIn fruit flies, you are mapping three genes in a three point cross. The mutants are hairy body (h), sepia colored eyes (se) and female sterility (g). You cross a heterozygous parent with a homozygous recessive parent and obtain the following results: Type Number h se g. 5 + se + 450 + se g 27 ++g_ h se + + + + h + g. h + + TOTAL is the gene in the middle and the distance in map units between se and g is Oh; 16.4 se; 7.1 Oh; 7.1 70 82 7 327 32 1000 se; 16.4
- What would justify the following ratio appearing after phenotyping the outcome of a crossing trial: 8.9: 2.9: 3.2:1? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a b C d Obviously this represents independent assortment based on crossing dihybrid heterozygotes. Obviously this represents gene linkage based on test crossing a dihybrid heterozygote. Obviously this represents the results of a trihybrid test cross. Obviously this represents independent assortment based on crossing monohybrid heterozygotes.An individual has the following genotype. Gene loci (A) and (B) are 15 m.u. apart. What are the correct frequencies of some of the gametes that car be made by this individual? Bl a O A. Ab = 7.5%; AB = 42.5% B. ab = 25%; aB = 50% O C. AB = 7.5%; aB = 42.5% O D. aB = 15%; Ab = F0% E. aB = 70%; Ab = 15% Reset Selection OMark for Review What's This?Consider three genes L, U, and W, for which the count of F2 phenotypes after a 3-point cross is as follows: Phenotype F2 count: L U w 19 L u W 1 l u W 21 L U W 33 l U W 274 l u w 41 l U w 2 L u w 259 Which of the following statements about genes L, U, and W are TRUE? (may be more than one correct ans) A. L, U, and W are each on a different chromosome B. Only U and L are on the same chromosome C. Only U and W are on the same chromosome D. Only W and L are on the same chromosome E. L, U, and W are all on the same chromosome