A research paper published in the summer of 2012 presented a method to obtain the whole-genome sequence of a fetus without any invasive procedure such as amniocentesis that could on rare occasions cause miscarriage. This new technique is based on the fact that some fetal cells leak into the mother’s bloodstream and then break down, releasing their DNA. Assume that exactly 10% of the DNA fragments in the mother’s blood serum come from the fetus, while the remaining 90% of the DNA fragments in the serum come from the mother’s genome.
The investigators collected cell-free DNA from a pregnant woman’s bloodstream and subjected it to an advanced high-throughput sequencing method. The table at the end of this problem looks at seven unlinked loci; the number of reads of particular alleles (identified by Greek letters) are shown. You should assume for the sake of simplicity that all numerical differences are statistically significant (even though actual data are never this clean).
a. | Determine whether each locus is autosomal, X-linked, or Y-linked. |
b. | Describe the diploid genomes of the mother and fetus by using Greek letters for the alleles, or a dash (–) if no Greek letter is appropriate. |
c. | Is the fetus male or female? |
d. | At an eighth locus, 1500 reads of a single type of sequence were found. Provide a possible explanation for this result, being as specific as possible. |
Want to see the full answer?
Check out a sample textbook solutionChapter 11 Solutions
Genetics: From Genes to Genomes
- In contrast with the genomic manipulations of animals and plants described in this chapter, human genetherapy is directed specifically at altering the genomes of somatic cells rather than germ-line cells.Why couldn’t or wouldn’t medical scientists try to alter the genome of human germ-line cells?arrow_forwardthis is what i have said about this image so far, what else can be said aswell including the raw count column. " Interpreting the results of an RNA-Seq analysis is pivotal in understanding the underlying genetic mechanisms of diseases such as breast cancer. In this analysis, Figure 1 provides comprehensive data on differentially expressed genes associated with breast cancer. By delving into the provided information, we can gain valuable insights into the molecular landscape of this disease. First focus is on the gene with the highest fold change, EYA4, situated on chromosome 6. With a staggering fold change of 3604.4176, EYA4 exhibits an unprecedented level of overexpression in cancerous cells compared to normal cells. This profound alteration suggests a pivotal role for EYA4 in breast cancer pathogenesis. The log2 fold change of 11.81555 further emphasizes the magnitude of this difference in gene expression. Statistical significance is evident, with an exceptionally low p-value of…arrow_forwardSee the attachment and answer the following parts of the question: A) If the binturong genome is 2.87 x 109 base pairs, and the "highly repetitive DNA" fraction is composed entirely of copies of sequence 5'ATGGTCC3' and its complement, how many copies of this sequence are present in the binturong genome? B) Briefly explain, in your own words, why the fraction of the binturong DNA fragments that reannealed relatively slowly took so much longer to renature than the other DNA fragments. C) If you took more of the same randomly generated 1000 bp fragments of binturong DNA (the same sample that you used in the equilibrium density gradient centrifugation experiment described in part a and the C0t curve described in part b of this question) and used them as a sample in agarose gel electrophoresis, how many bands would you expect to find in the gel when you turned off the current and stained the gel with ethidium bromide? Briefly explain why you would predict that number of bands.arrow_forward
- https://www.khanacademy.org/science/biology/biotech-dna-technology/dna-cloning-tutorial/a/overview-dna-cloning That is the link for the example ^arrow_forwardwhat is the answer of (a)?arrow_forwardAll the cells of one organism share the same genome. However, during development, some cells develop into skin cells while others develop into muscle cells. Briefly explain how the same genetic instructions can result in two different cell types in the same organism.arrow_forward
- A method for detecting methylated CpGs involvesthe use of a chemical called bisulfite, which convertscytosine to uracil but leaves methylated cytosine untouched. You want to know whether a particularCpG dinucleotide at one location in the genome ismethylated on one or both strands in a tissue sample.The genomic sequence containing this CpG is:5’...TCCATCGCTGCA…3’. You take genomicDNA from the sample tissue, treat it exhaustivelywith bisulfite, and then use flanking primers toPCR-amplify the region including this CpGdinucleotide. You then want to Sanger sequence(see Fig. 9.7) the amplified PCR product. a. After you treat genomic DNA with bisulfite, the twoDNA strands will melt into single strands. Why?b. Your answer to part (a) introduces a potential complication, because if you do not account for this result of bisulfite treatment, the PCR primers willnot amplify the DNA. What special considerationswould be necessary when you design your PCRprimers for this experiment? Could one pair…arrow_forwarda) Bioinformatics is an interdisciplinary field that integrates computer science with mathematics and statistics to solve biological questions. Many bioinformatics tools for gene prediction, homology modelling and such are available free online. (i) How can online tools such as BLAST and FASTA assist in our genomics research? Is the sequence below in FASTA format? Justify your answer. >gi 129295|sp|P01013 | OVAX_CHICK GENE X PROTEIN (OVALBUMIN-RELATED) QIKDLLVSSSTDLDTTLVLVNAIYFKGMWKTAFNAEDTREMPFHVTKQESKPVQMMCMNNSFNVATLPAE KMKILELPFASGDLSMLVLLPDEVSDLERIEKTINFEKLTEWTNPNTMEKRRVKVYLPQMKIEEKYNLTS VLMALGMTDLFIPSANLTGISSAESLKISQAVHGAFMELSEDGIEMAGSTGVIEDIKHSPESEQFRADHP (ii) FLFLIKHNPTNTIVYFGRYWSParrow_forwardFriedreich ataxia (FRDA) is an autosomal recessive, neurodegenerative disease that causes a lack of voluntary coordination of muscle movements. Affected individuals are homozygous for an unusually large number (expansion) of repeats of a trinucleotide sequence (GAA) in the first intron of the X25 gene. Unaffected individuals typically have between 7 and 38 repeats of the trinucleotide (GAAGAAGAAGAA…). FRDA patients have anywhere from 66 to over 1,700 repeats. To understand how the GAA trinucleotide expansion leads to FRDA, researchers looked at X25 gene expression by extracting RNA from affected and unaffected patients and doing a northern blot analysis (see the figure below): In panel “a,” the researchers used a probe to detect X25 mRNA. In panel “b,” the researchers used a probe on a duplicate of the original blot to detect human GAPDH mRNA (GAPDH is an enzyme involved in glycolysis). The sample labeled “YR” is mRNA from yeast cells that was used as a control. Explain…arrow_forward
- Woolly mammoths have been extinct for about 4,000 years, but we often find their well-preserved remains in Siberian permafrost. Research groups are now planning to use SCNT to resurrect these huge elephant-like mammals. No mammoth eggs have been recovered yet, so elephant eggs would be used instead. An elephant would also be the surrogate mother for the resulting embryo. The researchers may try a modified SCNT technique used to clone a mouse that had been dead and frozen for 16 years. Ice crystals that form during freezing break up cell membranes, so cells from the frozen mouse were in bad shape. Their DNA was transferred into donor mouse eggs, and cells from the resulting embryos were fused with undifferentiated mouse cells. Four healthy clones were born from the hybrid embryos. What are some of the pros and cons of cloning an extinct animal?arrow_forwardAnnotation of the human genome sequence reveals a discrepancy between the number of protein-coding genes and the number of predicted proteins actually expressed by the genome. Proteomic analysis indicates that human cells are capable of synthesizing more than 100,000 different proteins and perhaps three times this number. What is the discrepancy, and how can it be reconciled?arrow_forwardPeople who carry a theoretical genetic disorder (called B-disease) can be identified from a 2kb DNA sequence. People who carry this genetic disorder have a single nucleotide polymorphism that results in a change of GTATCC to GGATCC, a site that only occurs once at nucleotide number 750 in this DNA sequence. Answer the following questions based on the information provided. (a) How can you develop a simple molecular test to identify the genetic disorder?r B-dif w. (41 (b) If you have carried out the molecular test (based on the information above) on a 100 individual and found that 24 were healthy (BB) and 26 were carriers (bb); 1) What is the ratio of heterozygous? 2) Show how can you identify the three types from the agarose gel (H focaiarrow_forward
- Biology Today and Tomorrow without Physiology (Mi...BiologyISBN:9781305117396Author:Cecie Starr, Christine Evers, Lisa StarrPublisher:Cengage Learning