females . “ /16 A ) 16384 (C ) overall length E ) banding patterns |A ) autoradiographic pattern 10) type and location of genes |B ) position of

PLEase workout steps for numbers 23 and 24 and explain

females .“ /16A ) 16384(C ) overall lengthE ) banding patterns|A ) autoradiographic pattern10) type and location of genes|B ) position of the centromere(B ) 196Calculate the chi – square value ( X 2 )(C ) 14 D ) 28|25 . ( 3 Pts ) What is the basis for homology among chromosomes ?|A ) 3. 05 X 10 -5 B ) 2.35 X 10-5 C ) 225 D ) 450 E ) 4 . 15 X 10 -5C ) Genetic information is encoded in DNA by the sequence of bases22 . ( 3 Pts ) How is genetic information encoded in a DNA molecule ?|A ) X 2 = 0 . 96 B ) X 2 = 2. 01 C ) X 2 = 3. 01 D ) X 2 = 2.71 E ) X 2 = 1. 97|A ) Genetic information is encoded in DNA by the sequence of amino acids .B ) Genetic information is encoded in DNA by the sequence of bases and amino acidscombinations of chromatids are possible during the early phases of anaphase in meiosis !1 ?23 . ( 3 Pts ) What is the probability that , in an organism with a haploid number of 15 , a sperm will be26 . ( 3 Pts ) Determine the probabilities ( p ) that in a family with four children , two will be males and twoSOULSand a paternal member ( for example : AM AP ( one pair ) or AM AP B M BP ( two pair ) ) . How many differentformed that contains all 15 chromosomes whose centromeres were derived from maternal homologs ?|24 . ( 3 Pts ) Consider a diploid cell that contains 14 pairs of chromosomes . Each pair includes a maternal2[seller