Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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- what is the parental genotype: female___ x___ male?arrow_forwardning x earning.com/index.cfm?method%3DcResource.dspView&ResourcelD=1070&ClassID=D3856718 5. Which offspring could not arise from the parent cells with the chromosomes shown below? Assume crossovers can occur between chromosome pairs in the parent cells. Allele key Parent 1 Parent 2 Dominant C-Curly wing B-Brown body L-Legs on head R-Red eyes HX Recessive c-normal wing b-black body 1-antenna r-orange eyes D. OA. Offspring A OB. Offspring B Oc.offspring C OD. Offspring D acer B.arrow_forwardConsider the parental cross rY x RR, where r is the trait/disease allele, and R is the wild-type allele. The mode of inheritance is sex-linked recessive. Compute the ratios (Wild-type/non-disease phenotype):(Disease phenotype) for the F1 and F2 generations. Check the syllabus to see what formats are allowable for submission (e.g., pdf, docx, etc). Note: if you use a text editor like Notepad, you must use Courier New font. Also, you must show all work, akin to the work shown in the pdf lecture. Hint: In the F1 generation, individuals can only mate with the opposite gender.arrow_forward
- Which of these predictions would you make for a gene that is haplosufficient and essential for viability? A diploid heterozygous for a null allele and the wild-type allele will be viable. A diploid heterozygous for a null allele and a wild-type allele will be nonviable. A diploid homozygous for a null allele will be nonviable. A null allele would lead to nonviable haploid cells.arrow_forwardA male mouse that is heterozygous for a mutant Igf2 allele (Igf2/lgf2-) is crossed with a female who is homozygous for the wild-type lgf2 allele. What would be the expected outcome of this cross? Multiple Choice O O O O 1/2 the offspring would be normal; 1/2 would be dwarf All offspring would be normal because the paternal allele is expressed All offspring would be dwarf because the maternal allele is silenced All offspring would be dwarf because the paternal allele is silenced 3/4 of the offspring would be normal; 1/4 would be dwarf Aarrow_forwardcalculate the frequency of recombination among male parents and female parents separately.arrow_forward
- Consider the following cross involving three linked genes DeF/dEf x def/def. PHENOTYPES NUMBER OBSERVED DeF 76 def 30 dEf 82 DEF 24 Def 4 DEf 18 dEF 1 deF 15 The gene order is:arrow_forwardPlease explain each steparrow_forwardIn pea plants, the tall allele is dominant to the short allele and the yellow pea allele is dominant to the green pea allele. Complete the Punnett square and questions below using the information and parent cross below. TtYy x TTyy 1. Complete the Punnett squares below. T Y y TT Tt yy TT Tt Yy yy 2. The phenotypes for TTYY and TtYy are correct order to receive credit T-trait listed first then the Y-trait) and (make sure the traits are listed in the 2a. What is the chance of an offspring having these two traits? 3. The phenotype for ttyy is and (make sure the traits are listed in the correct order to receive credit T-trait listed first then the Y-trait) 3a. What is the chance of an offspring having these two traits? :: T :: Y :: y :: TT :: Tt :: tt :: YY :: Yy : yy :: tall :: short : yellow :: green : 0% : 12.5% :: 25% : 37.5% :: 50% :: 56.25% :: 75% :: 100%arrow_forward
- Give a brief analysis on the following chi-square for cross 1(wild type female × white eyes vestigial wings male) ( write a paragraph or more)arrow_forwardThe e-gram (graph to the right of allelic ladder image) and above is from a woman. She has variations 14 and 15 at STR D3S1358. If she has children with a man who has variations 12 and 19 at the same STR, what are the possible combinations of variations that their children would have?arrow_forwardA researcher is interested in whether a disease gene is linked to an RFLP marker. A cross is made between pure breeding blind cavefish and pure breeding sighted cavefish. The lines are pure breeding. The resulting RFLP markers of the F1 offspring and the F2 (from an F1XF1 cross) offspring are shown. If the disease gene is 10 m.u. from the RFLP marker, what proportion of the F2 will have the gel pattern shown above AND be blind? Show your work If the disease gene is 10 m.u. from the RFLP marker, what proportion of the F2 will have the gel pattern above AND NOT be blind? Show your workarrow_forward
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