Question

Questlon 21 3 p In Drosophila, nondisjunction during meiosis in the parent can result in flies with abnormal numbers of sex chromosomes. Flies with a single X and no Y are male, and those with XXY are female. Which of the following events could not give rise to a male with a single X in Drosophila? Nondisjunction at meiosis II in the female parent Nondisjunction at meiosis II in the male parent Nondisjunction at meiosis in the female parent All of these CAN give rise to a male with a single X in Drosophila Nondisjunction at melosis in the male parent Question 22 3 pts Given the following cross: AabbCcDdEe X AaBbCcDdee, what is the probability of getting the recessive phenotype? 1256 OO O126 164

Answer 1

ANSWER 21) all of these cases can give rise to male with single X chromosome

Answer 22)1/256

EXPLANATION

Explanation 21) (Non disjunction) meiosis II 3 mal meiosis I n+1 -8X XX n-1 nondisjunction at meiosi n-1 meiosis II nt) meiosis I mal 88 nondisjunction at melosis II n. n means faile in Nondisjunction Separation of chromosome that is if seperation of homologous Chromosome ( meiosis) or sister chromatid s meiosis T.) I

to separato In base of homologous chromosome fail C meiosis I ero nondisjuncions) tuo lack particulare gamete Chromosome while other Euro gamete slould have extra chromosome In case of sister chromatid fail to separate (meiosis II non- disjunction) then gametes acould be normal one gamelo cuill have extra copy of chromosome and another one would lack a copy of chromosome. tuo

Answer 22 -- V 256 Explanation Aabb Cc D d Ee X Aa Bb Cc D dee G Probability of getting all gecessive that is genotype aabbcc ddee А A A Aa b BIBOBb Tabbb с. е. crclcc E a A a e Selec Eeee DIDDDd al Dalad ad = Ee. aa= e bb=2 / 2 Yy ces Yu tu ee = 3/4 5 LV- Overall X 5 4 x x 4 IN X AT ild s 256 256