Recitation 5 new

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Arizona State University *

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340

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Biology

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Apr 3, 2024

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Name: Date: 25 An ichthyologist new to the study of electric eels in the Amazon basin decides to investigate the genetic variation in electric discharge patterns among the population. They know the E gene codes for the voltage of the electric discharge (E=high voltage and e=low voltage) where E (high voltage) is dominant to e (low voltage). The scientist also understands that the P gene determines the pattern of the electric organ discharges (P=pulsating and p=constant) where P (pulsating) is dominant to p (constant). The ichthyologist observes the results of a cross between a wild-type heterozygous electric eel with high voltage and pulsating discharge (EePp) with a homozygous recessive eel that has low voltage and constant discharge (eepp). The cross results in numerous new offspring, providing a wealth of data for the researcher to analyze the inheritance patterns of electric discharge in electric eels. 1. There are two groups of progeny with parental phenotypes, please list both parental phenotypes? (1.5pt) The two groups of progeny with parental phenotypes are: High Voltage with Pulsating Electric Discharge (E_P_) Low Voltage with Constant Electric Discharge (eepp) 2. There are two groups of recombinant progeny. Write the genotypes of both recombinant progeny groups in slash notation? (1.5pt) The two groups of recombinant progeny observed are: High Voltage with Constant Electric Discharge (E_pp) Low Voltage with Pulsating Electric Discharge (eePp) 3. Given the information provided in the table above, what is the recombination frequency of the observed cross (round to two decimal places)? You must show your work (2.5 pt) Phenotype Observed High Voltage with Pulsating Electric Discharge 423 High Voltage with Constant Electric Discharge 67 Low Voltage with Pulsating Electric Discharge 53 Low Voltage with Constant Electric Discharge 457 Total 1000
Recombination Frequency (RF) = (Number of Recombinant Progeny / Total Number of Progeny) * 100 RF = [(67 + 53) / 1000] * 100 RF = 12.00% 4. What is the distance between the E and P genes (you must include units) ? (1.5 pt) Since 1% recombination corresponds to 1 map unit (cM), the distance between the E and P genes is 12.00 map units.
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