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- Which of the following would be the correct Haworth projection of a-D-Talose (linear form is in the left side)? * CHO CH,OH CHOH OH HO HO -H- он OH OH HO OH OH OH OH C. CH,OH CH,OH H- он он OH OH CHOH OH OH он O Dcomplete steps of haworth projection of D-tagatoseAaBbCcDdEe AaBbCcDc AaBbCcDdEe AaBb( AaBbccDdEe AaBbCcDdEe AaBbCcDd. x, x A I v Av Normal No Spacing Heading 1 Heading 2 Title Subtitle Subtle Emp In tomatoes, red fruit (R) is dominant over yellow fruit (r) and tall plant (T) is dominant over short plant (t). A plant that is heterozygous for both traits is crossed with a plant that is homozygous recessive for both traits. What is the genotype and phenotype of the offspring? Saguaro cacti are very tall cylindrical plants that usually have two L-shaped arms, one on each side. Suppose you live in southern Arizona where the Saguaro cactus is common, and you happen to have one growing in your yard. Your Saguaro has two arms, but one is longer than the other. Now, assume that arm length in these cacti i controlled by a single gene with arms of the same length (A) being dominant to arms of different lengths (a). What is the genotype of your cactus? Could one of the parents of your cactus have had a phenotype with arms actus
- Abigail Griffith- Ph... KAbigail Griffith - DN... 100% Genetics Glencoe Reading Essentials.pdf ements, people se, but there a dominant eople with os. They Think it Over 3. Explain How can scientists determine if achondroplasia developed from a new mutation? a have ant, parents pre, when plasia, the ace -A pedigree . In the tal line are oldest to Picture This ions-I for numbers are 4. Evaluate Circle the carriers in the second tionposition position (5' end) position (3' end) UUU Phe - F UUC Phe UAU Tyr Y. UAC Tyr UAA Stop UGU Cys UGC Cys UGA Stop UGG Trp UCU Ser UCC Ser U UUA Leu UUG Leu UCA Ser UCG Ser UAG Stop W 輔 CAU His CGU Arg CGC Arg CGA Arg CGG Arg CUU Leu CCU Pro CUC Leu CCC Pro P. CCA Pro CÁC His CỦA Leu CAA Gln CUG Leu CCG Pro CAG GIn AUU lle ACU Thr AAU Asn AGU Ser AUC lle ACC Thr -T ACA Thr AAC Asn_ AGC Ser AGA Arg AGG Arg A AUA lle AAA Lys EK AAG Lys AUG Met ACG Thr GUU Val GGU Gly GGC Gly GGA Gly GGG Gly GAU Asp D GAC Asp GCU Ala GUC Val -V GUA Val GCC Ala GCA Ala GCG Ala G - A GAA Glu FE GAG Glu GUG Val Nonpolar Polar Basic Acidic Stop codon Table 4.1 Biology: How Life Works, Third Edition © 2019 W. H. Freeman and Company Consider the DNA sequence–3'-CACTGGTAGCGA–5', • What is the sequence of the messenger RNA sequence that is transcribed from this DNA sequence? (Indicate the 5' and 3' ends of the RNA molecule.) • Consult the genetic code that is shown above. What is the amino acid sequence of the…I didnt get answer for this Can you explain in a few sentences how the change in the structure of the gene affects the function of the corn plant?
- ocument/d/1J-wo90GpYsd_jQSUBtDQHWisqGvSOteUQYoXaXazyS0/edit uction to cell.. R 1 Summary of Philo... E Petrona Andres Mig.. 2 Translations IXL: Par... IXL - Translations: g.. 1 IXL- meostasis Lab Exercise Tools Add-ons Help Last edit was 2 days ago text Calibri 12 BIU Conclusion: 1. List the changes you observed in the body color and perspiration level in response to? 2. Explain how the changes help the body adjust to maintain equilibrium (homeostasis)? 3. Speculate why a change in body temperature occurs? 4. Name which mechanisms your body uses to maintain a constant body temperature? 5. Explain why an increased breathing rate accompanies exercise? 6. Explain why an increased heart rate accompanies exercise? 7. Write a paragraph about the conclusions you can draw about your body's ability to maintain equilibrium (homeostasis). Be sure to include the answers to the questions above.1_30*_SP23 - General Biology I (for majors)/1 of us page The anticodon sequence created from the following DNA: TACGGGGCTGAGATT F1 Select one: O a. Tyr-Gly-Ala-Glu-lle O b. AUGCCCCGACUCUAA c. UACGGGGCUGAGAUU O d. Met-Pro-Arg-Leu-STOP F2 # 80 F3 $ 000 000 F4 % F5 MacBook Air F6 & r F7 DII F8III. Deletion of C IV. Both I & II 6. Refer to the figure answer the following questions. CLUSTAL W (1.83) multiple sequence alignment Human_AA Oyster AA Corn_AA -MKLEWLLFTIGFCWAQYSSN--TOOGRTSIVHLFEWR-------WVDIALECERYLAPK 50 -QVILNCLLYVvGVVRGGTWSNPTCAPGRHTITHLFEWK- -WSDIAAECERFLGPM 52 MAKHLAAMCRCSLLVLVLLCLGSQLAQSOVLFOGFNWESWKKOGGWYNYLLGRVDDIAAT 60 : : : : *:*. %3: .. O Human_AA Oyster AA Corn AA GFGGVOVSPPNENVAIHNPFRPWWERYOPVSYKLCTRSGNEDEFRNMVTRCNNVGVRIYV 110 GYCGVOISPPNENRIVTSPNRFWWERYOPVSYKLVTRSGNEADLRDMVQRCNKVNVRIYA 112 GATHVWLPPPSHSVAPOGYMPGRLYDLD------ASKYGTHAELKSLTAAFHAKGVKCVA 114 ::*.. : ::.:. :.*: Human_AA Oyster AA Corn AA DAVINHMCGNAVSAGTSSTCGSYFNPGSRDFPAVPYSGWDFNDG-KCKTGSGDIENYNDA 169 DVVINHMTG-AGGSGTG-TGGSHWDGGSLSYPGVPFSSWDFNSGSECSTGDGNIHNYNDP 170 DVVINHRCA---DYKDGRGIYCVFEGG------TPDSRLDWGPDMICSDDTQYSN--GRG 163 :: * ***** Figure: Sequence alignment a) How many different species are used as the source of sequence in this analysis? I. two I. one II. three IV. four b) What…
- Cattle in industrial dairy farms are typically given rBST, or recombinant bovine somatotropin. This animal hormone increases a cow's milk production, but there is concern that such hormones may have unforeseen effects on milk-drinking humans. There is no similar concern about plant hormones in the plant foods we eat. Why not?Maize... • Upload Pending Olatunde , Bisola On me Insert Draw Design Layout References Mailings Review View Help Table Design Layout A Garamond v 12 BIU ab x, x A Styles Editing Dictate Editor Reuse A A Aav 会 T Files • Aav A A Font Paragraph N Styles Voice Editor Reuse Files Yellow:Purple Punnett square: in each cell indicate the and the predicted color. genotype Chi-square test of hypothesis: formulate a hypothesis, count the kernels, test the hypothesis. Expected number (O-E) /E Observed O-E (O-E) Phenotypes number Yellow 161 Purple 39 13 words E English (United States) Text Predictions: On D. Focus 18°F 日i A acer 07 12 lı 17AaBbCcDdE AaBbCcDc AaBbC Normal No Spacing Heading 1 Head Activity 1. 3. Based on the following data which of the following would be considered: a. Recessive? b. Dominant? C. A weak loss-of-function mutation? d. A null mutation? e. A haploinsufficiency ? Data - Gene "B" is required for formation of body segments. A wild type fly had 3 body segments. BB (Homozygous dominant) Wild Type Bb (Heterozygous) bb (Homozygous recessive) Lacks all body segments 1 body segment Mutant 1 Wild Type Mutant 2 Wild Type 2 body segments Mutant 3 Wild Type Wild Type 2 body segments Activity 4 tv Ps X W 000 DII DD II
