ANSWERS TO ASSIGNED PROBLEMS FALL 1999
************** Chapter 3 **************
1a) WW (black) x ww (white) gives all Ww (black) x sib gives 3/4 W- (black) 1/4 (ww) white
1b) all white
1c) cross #1 is WW x W- and cross #2 is Ww x Ww
2a) both parents are Cc
2b) the man is probably CC, the woman is cc
2c) the man is Cc and the woman cc
2d) expect 4 normal and 4 albino
6) Checkered must be dominant to plain according to cross #b, so Ck > ck
cross a= CkCk x Ck-; cross b= CkCk x ckck; cross c= ckck x ckck; cross d= CkCk x ckck
cross e= Ckck x ckck; cross f=Ckck x Ckck; cross g= CkCk x Ckck
9a) Ww YY x Ww YY 9b) ww Yy x Ww Yy
9c) Ww Yy x Ww Yy 9d) Ww Yy x ww yy
10) cross D
13) Homologous chromosomes carry the same set of genes, non-homologs carry different genes
15) not done in correct notation because it is too much trouble
15a) gray long GG LL x ebony short gg ll gives all gray long Gg Ll which interbred give
9 gray long G-L- : 3 gray short G- ll : 3 ebony long gg L- : 1 ebony short gg ll
15b) gray short GG ll x ebony long gg LL same as in "a"
15c)gray long GG LL x gray short GG ll gives all gray long GG Ll which interbrad give
3 gray long GG L- : 1 gray short GG ll
18)The parental cross was yellow YY x green yy to give all yellow Yy. When interbred, the F1
produced 1/4 true-breeding yellow YY, 2/4 heterozygous yellow Yy than when selfed give 3:1
ratios, and 1/4 true-breeding green yy.
21) Thalassemia shows incomplete dominance.
26) The phenotype shows with orange symbols must be recessive as proven by I-3 x I-4
producing II-6. Individuals I-1, II-4, and II-5 must be heterozygous Aa as must be all individuals
in generation IV. Individual III-5 must be homozygous AA.
27) Myopia is inherited as a recessive phenotype. I-3, I-4, II-1, II-2, II-3, II-4 must be
heterozygous.
28e) These 2 pedigrees are compatible with either dominance or recessivity.
32) IMPORTANT a) 1/6 b) 1/6 x 1/6 = 1/36 probability that the first throw will be 3 times the probability that the second throw will be a 6 c) 2/36 or 1/18 (prob. of a 3 on die #1 and a 6 on #2=1/6) plus (prob. of a 6 on #1 and a 3 on #2=1/6) Here the event can occur in either of 2 ways.
32d) Here probability is conditional. Only three numbers are odd and one of these is a "5". So
prob. is 1/3.
34) The probability that a parent is heterozygous is 2/3 (normal offspring of two heterozygotes).
The chance that two heterozygotes have a homozygous child is 1/4. 2/3 x 2/3 x 1/4 = 1/9.
38) answer is in textbook with clear pedigree
************* CHAPTER 4 ***************
2. Co-dominance applies to gene products at the molecular level where each allele controls the
production of about half the product in the cell. Hemoglobin variants and ABO antigens are good
examples. Incomplete dominance occurs at the organismal level when the heterozygote's
phenotype is intermediate between the two homozygotes.
3. Platinum is a dominant visible and recessive lethal. Expect 2 platinum : 1 normal.
4. Short-tail (really the t gene in mice) is a dominant visible, recessive lethal. Short x normal is
really Tt x tt. Tt x Tt give 2:1 phenotypic ratio because TT dies in utero.
5. Showing ONLY SUPERSCRIPTS!! Not showing base I, genotypes are AA, AO, AB, BB, BO,
OO. A and B alleles are co-dominant (both antigens are detectable). O is recessive to both.
6. The mother must be AO because her B father did not give her B. The father must be BO
because his mother was O. AO x BO gives 1 AB : 1AO : 1 BO : 1 OO, also four phenotypes.
8a) Mating is AB Ss x OO Ss gives 1/2 AO and 1/2 BO, 3/4 secretors and 1/4 non-secretors.
Thus ration is 3/8 Type A secretors 3/8 Type B secretors 1/8 Type A non-sec 1/8 Type B non-sec
8b) Each of the above 4 phenotypes would be subdivided into 1/4 hh which appear to be Type O
non-secretors and 3/4 unchanged. So the ratio would be 9/32 Type A secretors 9/32 Type B
secretors 3/32 Type A non-sec 3/32 Type B non-sec and 8/32 Type O non-secretor (really
caused by Bombay phenotype).
10. Again, this notation is not proper and should have superscripts. Himalayan allele= Ch
Chincilla allele= Cn Full color=C albino=c. Beware!
10a) Himalayan x Himalayan Ch c x Ch c gives 3/4 Ch- himalayan and 1/4 cc albino Use cc as
next parent. Full x albino must be C Cn x c c to give 1/2 full 1/2 Chinchilla. Use Cn c as parent.
Expected result is 1/2 chinchilla and 1/2 albino.
10b) Cross 1 cc x Cn c produces cc albino and Cross 2 C - x cc produces full Cc. Cross the progeny and expect 1/2 full 1/2 albino.
10c) Cross 1 CnCh x cc produces Himalayan Chc. Cross 2 CCh x cc produces Himalayan Chc.
Cross and expect 3/4 Ch - himalayan and 1/4 Ca Ca albino.
11a) Ck Ca x Cd c gives 1/4 Ck Cd 1/4 Ck Ca 1/4 Cd Ca 1/4 Ca Ca genotypic ratio. The
phenotypic ratio is 1/4 cream 1/2 sepia 1/4 albino.
11b) Sepia parent must be Ck Ca but cream parent could be either Cd Cd or Cd Ca. So the cross
could produce either 1/2 sepia 1/2 cream OR 1/4 cream 1/2 sepia 1/4 albino.
11c) Tedious! Here neither parent's genotype can be known for sure. Sepia might be Ck Ch, Ck
Ca, Ck Ck. The cream parent might be Cd Cd or Cd Ca. So 6 different crosses might occur. All
are solved like the ones above.
11d) The sepia parent must be Ck Ca. The cream parent might be Cd Cd or Cd Ca.
12. The parental cross is RR PP TT x rr pp tt. F1 is all pink, personate, tall Rr Pp Tt. The F2
plants will have 12 different phenotypes and 27 different genotypes. To have a pink, personate,
tall phenotype the probability is (2/4 RR) x (3/4 P-) x (3/4 T-) = 18/64.
15. If a phenotype is not mentioned, you should assume that the individual is wild-type (homo-
zygous unless told otherwise). So write out the full genotype for each fly.
a) b+b+ st+st+ x bb stst gives all heterozygous wildtype b+b st+st interbred and get 9/16 wild,
3/16 brown, 3/16 scarlet, 1/16 white.
b) b+b+ st+st+ x b+b+ stst gives all b+b+ st+st wildtype interbred to give 3/4 wild 1/4 scarlet
c) bb st+st+ x bb stst gives all brown bb st+st interbred to give 3/4 brown 1/4 white.
16. It is much better to call aa=non-agouti because it is a PATTERN gene and other genes
determine hue. I prefer to call cc white or albino.
16a) 9/16 agouti 3/16 black 4/16 colorless, an epistatic ratio.
16b) female #1= AACc female #2 = AaCC female #3 = AaCc
20a) The 9:3:3:1 ratio is a give-away. This is two genes contributing to unique phenotypes.
A-B- = green, aaBB = brown, AAbb = gray, (does not matter which is A and which is B), aabb is
blue
20b) Because the F2 is a dihybrid ratio, the cross must have been AaBb x AaBb. Because the
KNOWN parent was gray AAbb, the other parent must have been aaBB brown.
22) Exactly the same as 20. A-B- = yellow, aaBB = blue, AAbb = red, aabb = mauve
22b) red x mauve = AAbb x aabb gives all red. F2 is 3/4 red 1/4 mauve
23) 1=c unique; 2=d unique; 3=b; 4=e unique; 5=a unique; so only one set of answers
29) This combines unique phenotypes and X-linkage so beware. Also, answers can be expressed
2 ways.
29a) F1 has X+Xv bw+bw wild females and XvY bw+bw vermillion males.
F2 has either 1) Females 3/8 normal, 1/8 brown, 3/8 vermillion, 1/8 white
and
Males same ratio (note each sex sums to 1)
OR
2) 3/16 normal females, 3/16 normal males, 1/16 brown females, 1/16 brown males, 3/16 vermillion females, 3/16 vermillion males, 1/16 white females,
1/16 white males (note each sex sums to 1/2)
29b) F1 all normal X+Xv bw+bw females and X+Y bw+bw males
F2 6/16 normal females, 2/16 brown females, 3/16 males normal, 1/16 males brown, 3/16
males vermillion, 1/16 males white (different # phenotypes in 2 sexes).
29c)F1 females normal X+Xv bw+ bw and males vermillion XvY Bw+bw
F2 has 3 female normal : 1 female brown : 3 female vermillion : 1 female white. Same ratio in
males.
30) White is X-linked and epistatic to sepia.
30a) Assume female is XwXw se+se+ and male is X+Y se se. F1 has all wild females XwX+
se+se, all XwY se+se white males. F2 each sex is 3/16 normal, 4/16 white, 1/16 sepia
30b) all F1 are normal
F2 has 3/16 wild males, 4/16white males, 1/16 sepia males, 6/16 wild females, 2/16 sepia females.
31a and b) I believe that the answer is wrong/incomplete in the old solutions manual, maybe in new also.
pedigree #1 exclude any dominant - II-3
pedigree #2 exclude any dominant and X-linked recessive - II-5
pedigree #3 exclude X-linked dominant - II-8
31c) II-1 must be Aa or AA or XA Xa ; II-6 must be AA or Aa ; II-9 must be Aa or aa
33) The cross is female HH x male hh. Only hh males can be cock-feathered, all other genotypes are hen feathered. Therefore all F1 Hh birds are hen-feathered. In the F2 all females are hen-
feathered and 1/4 of males (the proportion of hh) are cock-feathered. PLEASE NOTE!! Because
this is an autosomal gene, the answer would be the same if you had considered the reciprocal
cross of female hh (a hen from a flock of true-breeding cock-feathered males) with male hen-feathered HH. This is IMPORTANT.
37) You already know that genotypes are "expressed" differently in males and females" (see
feathering). This implies that genes are regulated differently and one method of regulation is to
methylate/de-methylate DNA. Therefore, in a cross of AA x AA, although the allele from each
parent is identical in DNA sequence, it may not be identical in methylation pattern and may
behave differently. If by any method (deletion of one copy or by uniparental disomy), alleles from
both parents are not present, abnormal phenotypes may result. This is a violation of the very basic
tenet in genetics that genes are preserved unchanged in the germ-line cells. It is also very rare.
Anticipation also violates the tenet. Genes are supposed to be passed on unchanged regardless of
what phenotype "houses" them temporarily. In "triplet expansion" diseases, a stretch of identical
DNA triplets is mis-copied and becomes longer (the reverse has not been observed). The longer
alleles cause increasingly severe phenotypes.
38) As soon as you look at cross H, with three phenotypes in proportions of base 16, you know
you are dealing with a two gene epistatic interaction Where A-B- is black, A-bb is chocolate, and
aa-- is golden. True-breeders must be AABB black, aaBB or aabb golden, AAbb chocolate.
Interpret the crosses as follows (parental genotypes in same order as phenotypes in text.
Cross a is AABB x AABB; Cross b is AABb x AAbb; Cross c is AaBB x Aabb;
Cross d is AABB x aabb or aaBB; Cross e is AaBb x aaBb; Cross f is AaBb x aabb;
Cross g is Aabb x Aabb; Cross h is AaBb x AaBb
***************** Chapter 5 ****************
1) In continuous variation, the phenotypic differences among geneotypes is very small and is
blurred by environmentally induced variation. This inheritance pattern produces results in
breeding experiments that are best interpreted by the statistics of the normal curve, the t-test, and
analysis of variance (i.e. are quantitative). Discontinuous traits cannot be treated with the same
mathematics. They must be analyzed with the stochastic equations of finite math and chi square.
3a & b))As soon as you see that the rarest phenotypes are in proportions of 1/16, you know that
two genes, each with two alleles are involved. The 1/16 is both = (1/4)2 and = (1/2)4. If you use
1/4 as the base fraction, you are estimating the number of genes and the power term, 2 in this
case, means 2 genes. If you are using 1/2, you are estimating alleles and the power term means 4
alleles. Because each gene has 2 alleles, the methods are identical. Number of alleles = (number
of genes) x 2 and number of genes = (number of alleles)÷2.
c) medium red = AaBb or aaBB or AAbb; light red = aaBb or Aabb; med dark = AABb or AaBB
d) P mating is AAbb x aabb so F1 is all Aabb or light red. F2 will be 1/4 white, 1/4 medium, and
1/2 light red.
5) As in question 3, examine the proportion of extreme phenotypes, here 4/1000 which is 1/250
which is very close to 1/256 which happens to be (1/2) to the 8th power. So 8 additive alleles
arranged in 4 genes. Problem is nearly solved. NOTE: I will never give you a number that is
"about" the right fraction. It will be exact.
5a) This is quantitative inheritance.
5b) Four gene pairs (#alleles/2).
5c) This requires new calculation. Max is 36 and min is 12, giving a range of 24 cm. If 8 alleles are involved, the largest plants are AABBCCDD, so each must add 24/8=3 cm.
5d) The parents must have had 4 additive alleles each, in homozygous arrangement, and different from each other. AABBccdd and aabbCCDD fits these requirements. So do AAbbCCdd and aaBBccDD, etc.
5e) Any plant with 2 additive alleles will be 18 cm tall, so AAbbccdd, AaBbccdd etc. The
binomial expansion would give them all. Any plant 33 cm tall must have 7 additive alleles,
AABBCCDd etc.
6a) As soon as you see that the extreme classes are 1/64, you know that (1/2) to the sixth power
is 1/64 and that there are 6 alleles, 3 genes involved. Harvey=6 cm and Erma=30 for a range of
24 cm. You are actually told that each allele adds 4 cm. to a base of 6 cm. Note that 4x6=24,
confirming that there are 6 additive alleles. Harvey is aabbcc and Erma is AABBCC and their F1
are all AaBbCc.
6b) The 18 cm parent could have one of several genotypes. If it was AaBbCc crossed to aabbcc,
then you get 1/8 AaBbCc or 18 cm, 3/8 14 cm (2 additive alleles), 3/8 10 cm (1 additive allele)
and 1/8 6 cm (no additive alleles).
9) Proven characters are fingerprints, height and IQ (yes, very good proof). By analogy to other mammals, max running speed (jumping height etc = athletic performance), weight gain on a given diet, growth rate, personality and temperment (can be fixed at various values in breeds of dogs). and many other characters.