(These questions are all from the BIOL EXAM 2) Unit 4
CHAPTER 9/10: (EXAM: 2)
MULTIPLE CHOICE:
Q7 (2009)
Q6 (2012) (Would the 1st child ALSO be a 1 in 4 chance?)
Q26, Q27 (2013)
SHORT-ANSWER:
Q8 c (iii) (2015)
2009 MC, Q7 C
Meisham sows must be mated with British males in order to have enough teats to account for all piglets, ie, up to 18; If it were the other way around, there would only be 12 teats between 11-16 piglets. Without a teat, some piglets would die :/
+ The British already have the smaller sized litters and the Chinese already have the larger sized litters.
2012 MC, Q6 A
Each parent is phenotypically normal, yet together they have a child with phenylketonuria. Thus, they're both heterozygous for the trait. Going by the normal cross ratios, the genotypes would be 1 PP: 2Pp: 1 pp. Since the trait is autosomal recessive, the chance is 1/4 of a child with the trait.
The first child's chance is 1, he/she already has it.
2013 MC, Q26 C
MM RR SS × M’M’ R’R’ S’S’ automatically makes a child MM' RR' SS'. So add the height contributions for all alleles, 5+2+5+2+5+2= 21cm.
2013 MC, Q27 D
The stem tells you that there can only ever be one active X chromosome in each somatic cell. The rest become Barr bodies. In A, there would be 1 Barr body. In B, 3. In C, 0. In D (XXXY), there would be 2 Barr bodies, leaving an X and Y chromosome active. therefore, the answer is D.
2015 SA, Q8ciii
I don't believe this question existssssss + haven't done this exam yet.
Hope this helps!