Reciprocal crosses were performed using two inbred strains of mice, AKR and PWD, that have different alleles of many polymorphic loci. In each of the two crosses, placental tissue was isolated whose origin was strictly from the fetus (this can be separated by dissection from placental tissue originating from the mother). RNA was prepared from the fetal placental tissue and then subjected to deep sequencing (that is, RNA-Seq). Because of the polymorphisms, investigators could compare the number of reads of mRNAs for specific genes that were transcribed from maternal or paternal alleles, as shown in the following figure. (The x-axis shows the percentage of reads for the given mRNA that correspond to the AKR allele of that gene.)
a. | Which of the genes (A, B, or C) is maternally imprinted? Which is paternally imprinted? Which is not imprinted? |
b. | Why was it important to perform reciprocal crosses to determine whether any of the genes were imprinted? |
c. | Using the same type of diagram that indicates the percentage of AKR alleles, diagram the expected results for these same three genes if a female F mouse from the cross on the left (that is, a daughter of a cross between an AKR female and a PWD male) was then crossed to a PWD male. Describe the two possible outcomes for each gene. |
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Genetics: From Genes to Genomes
- ISSR is generally a dominant STS DNA marker. Nonetheless, with validated experimental evidence (e.g. laboratory and population genetics data), the marker can be used in codominance marker genotyping. Briefly explain each case below: a) Codominant marker targets specific locus and reveals allelic variations in that locus among DNA samples. b) Dominant marker: primers can complement other repeat sequences or in multiple loci thereby non-specificity in sampled genomes.arrow_forwardGenomic DNA from a family where sickle-cell disease is known to be hereditary, is digested with the restriction enzyme MstII and run in a Southern Blot. The blot is hybridised with two different 0.6 kb probes, both probes (indicated in red in the diagram below) are specific for the β-globin gene (indicated as grey arrow on the diagram below). The normal wild-type βA allele contains an MstII restriction site indicated with the asterisk (*) in the diagram below; in the mutated sickle-cell βS allele this restriction site has been lost. What size bands would you expect to see on the Southern blots using probe 1 and probe 2 for an individual with sickle cell disease (have 2 βS alleles)? Probe 1 Probe 2 (a) 0.6kb 0.6kb and 1.2kb (b) 0.6kb and 1.8kb 0.6kb, 1.2kb and 1.8kb (c) 1.2kb 0.6kb (d) 1.8kb 1.8kb a. (a) b. (b) c. (c) d. (d)arrow_forwardNot all inherited traits are determined by nuclear genes (i.e., genes located in the cell nucleus) that are expressed during the life of an individual. In particular, maternal effect genes and mitochondrial DNA are notable exceptions. With these ideas in mind, let’s consider the cloning of a sheep (e.g., Dolly). A. With regard to maternal effect genes, is the phenotype of such a cloned animal determined by the animal that donated the enucleatedegg or by the animal that donated the somatic cell nucleus? Explain.arrow_forward
- What is a recombinant vector? How is a recombinant vector constructed? Explain how X-Gal is used in a method of identifying recombinant vectors that contain segments of chromosomal DNA.arrow_forwardThe DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non-shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. Given the information above I calculate the level of penetrance seen in image B to be "Blank" 1 percent.arrow_forwardThe DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non- shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. ANSWER: Given the information above I calculate the level of penetrance seen in image B to be Blank 1 percent. A KEY Homozygous Homozygous Heterozygous Heterozygous Wild Type Male Female Male Female Male Note: Completely red symbol denotes an individual exhibiting the phenotype of interest CI || III IV V 1/4 1/2 1/2 1/2 1/2 Wild Type Female 1/4 1/2 Affected Known carrier Affected female Normal female Affected male Normal male D ●●●arrow_forward
- Compared to the normal A allele, the disease-causing allele in sickle cell anemia (S allele) is missing an MstII restriction site. On a Southern blot of genomic DNA cut with MstII and hybridized with the probe shown on the diagram below, a person with sickle anemia, carrying two S alleles, will show Choose an answer below: a single band at 1.1 kb. a single band at 1.3 kb. a single band at 0.2 kb. one band at 0.2 and one at 1.3 kb. one band at 1.1 and one at 1.3 kb.arrow_forwardTransgenic tobacco plants were obtained in which the vector Ti plasmid was designed to insert the gene of interest plus an adjacent kanamycin-resistance gene. The inheritance of chromosomal insertion was followed by testing progeny for kanamycin resistance. Two plants typified the results obtained generally. When plant 1 was backcrossed with wild-type tobacco, 50 percent of the progeny were kanamycin resistant and 50 percent were sensitive. When plant 2 was backcrossed with the wild type, 75 percent of the progeny were kanamycin resistant and 25 percent were sensitive. What must have been the difference between the two transgenic plants? What would you predict about the situation regarding the gene of interest?arrow_forwardA small-scale pedigree study involving 15 families with a total of 50 children was performed to determine whether a SNP in a candidate gene is linked to a disease. The result showed that among 22 children who developed the disease, 19 belonged to non-recombinants while 3 recombinants. For the remaining 28 children, 21 were non-recombinants and 7 belonged to recombinants. A Lod (Z) score can be calculated and used to determine whether this SNP is linked to the disease. By using a very stringent threshold, which of the following conclusions is correct? O A.Z is approximately 4.03, indicating a linkage O B. Z=0.25, indicating no linkage OCZ=0.2, indicating no linkage O D.Z is approximately 4.19, indicating a linkagearrow_forward
- DNA from 100 unrelated individuals from one population of Chinook salmon were amplified at a single microsatellite locus, and run on an agarose gel. The results from gel electrophoresis shows three different fragment lengths (i.e. alleles/band positions) corresponding to 3 alleles; Allele F (250bp), Allele R (180bp), and Allele Y (100bp). The numbers at the top of each lane is the number of fish observed with that particular genotype or banding pattern in the population. Note that a single band means a homozygote for that allele (band size). a) Calculate the allele frequency for Allele Y. (b) Calculate the genotype frequencies for the following genotypes: FF, FR, and RY. (c) What is the expected number of Chinook salmon with homozygous genotype for allele Y in the study population? (d)What is the name of the statistical test that you could conduct to test whether this population of Chinook salmon is in Hardy Weinberg Equilibriumarrow_forwardA 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?arrow_forwardYou are using the restriction enzyme HAEIII to digest different samples of the taster gene isolated from cheek cells of different people and amplified by PCR. When viewing the bands on the electrophoresis gel, one would expect that a taster (homozygote) would have---------band(s), whereas a carrier (heterozygote) would show--------band(s), and a non-taster would show------band(s).arrow_forward
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