A prokaryotic endogenote of genotype his+ leu+ met+ trp+ ala+ undergoes conjugat
ID: 173016 • Letter: A
Question
A prokaryotic endogenote of genotype his+ leu+ met+ trp+ ala+ undergoes conjugation with and now requires leucine supplementation to live. Which of the following answers could explain this outcome? a.) The exogenote must have contained the allele 'leu-'. b) An uneven number of crossovers took place. c.) An even number of cross-overs occurred. d.) The e) All recombined genome of the endogenote turned it into an mutant. e) All the answers are correct are except for B. Of the following statements, four are 'fake news' and only one is correct. Which one is TRUE? a.) Recombination involves DNA strand breakage and invasion of the homologous non-sister chromosome; whether or not the sister chromatids exchange DNA with one another is irrelevant since their genetic information will be identical. b.) By default, a human male's sperm will always carry a Y chromosome. c.) An organism with 44% thymine in its DNA will have 6% cytosine since C and T are both single-ring structures. d.) By its very nature, an epistatic gene represents a lethal allele since it overrides the effects of another gene. e.) A mitochondrial disorder will appear in all of the offspring of an affected father. What is the middle gene in the following outcome from a 3-point test cross? a) b is the middle gene b) y is the middle gene c) z is the middle gene d) is the middle gene e) I wish knew.Explanation / Answer
Answer:
2). a). Recombination involves DNA stands breakage and invasion of the homologous non-sister chromosomes, whether, or not the siste chromatids exchange DNA with one another is irrelevant since their genetic information will be identical
3). y is the middle gene
Explanation:
Total progeny=1854
Parental combinations are more so, the parental combinations are z + y & + b +
1. If single cross over (SCO) occurs between z & + and + & b
Normal order= z -------- + & + ----- b
After cross over= z ----- b & + ------ +
z ----- b recombinants are 119+307= 426
+ ------ + recombinants are 112+310= 422
Total recombinants = 848
RF = (848/1854)*100 =45.74%
2. If single cross over (SCO) occurs between + & y and b & +
Normal order= + --------- y & b --------- +
After cross over= + --------- + & b ------ y
+ --------- + recombinants are 310+3= 313
b ------ y recombinants are 307+5=312
Total recombinants = 625
RF = (625/1854)*100 = 33.71%
3. If single cross over (SCO) occurs between z & y and + & +
Normal order= z --------- y & +------+
After cross over= z --------- + & +---------y
z --------- + recombinants are 119+3= 122
+--------- y recombinants are 112+5= 117
Total recombinants = 239
RF = (239/1854)*100 = 12.89%
% RF = Map unit distance
The order of gene is -----
z----12.89 m.u.-----y-------33.71 m.u.------+
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