Ch.4 Problems

#7) A male of blood type AB could not father a type O child, whose mother was blood type A (ignoring the possibility that each parent was heterozygous for the very rare Bombay allele).  His child would have to receive either the A or B allele and therefore he could not produce a type O child.  Because the O allele is recessive to the A and B alleles, a male of blood type A, B, or O could father such a child but on the basis of blood type alone, paternity can only be excluded; one cannot use such information to prove paternity.  

15) Looking at figure 4-10 sure helps with this one!  Make sure you designate the cross properly.  I'll just do the first one - that should be enough (if your having trouble) to figure out the other two.  wild type  x  white.  The cross is designate thus:  P₁  bw⁺ bw⁺, st⁺  st⁺ (wild type) x  bw bw, st st (white)  The F₁ would all be phenotypically wild type and have the genotype:  F₁  bw⁺ bw, st⁺ st    Crossing this fly with others like it would generate a typical dihybrid 9 (wild type) : 3 (brown) : 3 (scarlet) : 1 (white) ratio - as commonly seen with two independently assorting genes.  Deal with each gene pair separately.   bw⁺ bw  x  bw⁺ bw will generate a 3/4 bw⁺/-- : 1/4 bw bw ratio in the F₂  the st⁺ st  x  st⁺ st will similarly generate a 3/4 st⁺/-- : 1/4 st st  ratio.  Use the 'forked-line' method or a Punnet square to combine these independent probabilities.  (Forked lines aren't easy for me to do in HTML but here goes...)

17) It helps to draw a pathway from the information given:

Colorless precursor -------> cyanidin (red color) -------> purple

Crossing two purple plants gives purple, red, and colorless plants in about a 9 : 3 : 4 ratio.  The three phenotypes and the ratio should tip you off that this is not a single gene with a dominant and recessive allele at work.  This sure looks like epistasis - two or more independently assorting genes acting on a single trait.  The ratio strongly suggests two gene pairs, each with a dominant and recessive allele.  Imagine a gene A and its recessive allele a that produce an enzyme that turn the colorless precursor into the red pigment.  A second gene B and its recessive allele b, turn the red color into a purple one, but can't work on the colorless precursor.  In a standard Aa Bb (purple) x Aa Bb (purple) dihybrid cross.  The following phenotypic ratios would result:

9/16 A/--, B/--  (purple)  3/16 A/--, bb  (red)   3/16 aa, B/--  (colorless) + 1/16  aa, bb (colorless) = 4/16 colorless.  for an overall ratio of 9 purple : 3 red : 4 colorless

25) From reading the question, it should be obvious that the mother is a carrier for color blindness.  Since both parents are type A and produce a son who is type 0, it should also be obvious that they are both heterozygous for the O allele (I^A I^O).  (If these two conclusions are not obvious to you, then re-read the chapter and/or get in to see me soon!)  The question is: what is the probability that their next child will be female, have normal vision and be blood type O?  There is always a 50% chance of having a female.  Since it is the mother who is the carrier of the colorblindness allele, the daughter will inherit a normal (dominant) allele for this trait from her father and therefore have normal vision no matter which allele she receives from her mom.  That is, a 100% chance of having normal vision.  Finally, she will have a 25% chance of inheriting an I^O allele from each parent and being blood type 0.  Combining these probabilities: (1/2 chance of being female) x (1/1 chance normal vision) x 1/4 type 0) = 1/8 probability. 

31) This is a fun one, eh?  It is important to realize that this is the same trait see in three different families.  From looking at the first two pedigrees, one can rule out any dominant mode of inheritance.  In each case an affected child is born to parents who don't have show the trait.  If it were dominantly inherited, you would see it in the parents.  That leaves two possibilities: recessive, sex-linked or recessive, autosomal.  If you don't know what autosomal means, you should;  look it up!  It can't be recessive and sex-linked because of individual II-4 (her father I-3 would have been affected).  That leaves recessive, autosomal.  The mom in the third pedigree (individual I-6) must have been a carrier.  Individual II-1 = AA or Aa,   II-6 = AA or Aa, II-9 = Aa (has to be heterozygous because of dad).

43) Autosomal dominant, sex-limited inheritance.  (the easiest thing to do is to get the Nature article referenced in the text:  Shenker et al, Nature vol 365, 14 October, 1993).