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Unformatted text preview: 3 Independent Assortment of Genes BASIC PROBLEMS 1. a. The expected phenotypic ratio from the self cross of A/a ; 3/1) is 9 A/— ; 8/—
3 A/— ; b/b
3 a/a ;B/w
1 61/61 ; b/b b. The expected genotypic ratio from the self cross of A/a ; B/b is 1 A/A ; 3/3
2 A/A ;B/b
1 A/A ; 19/12
2 A/a ; 8/3
4 A/a ;B/b
2 A/a ; b/b
1 a/a ;B/B
2 a/a ;B/b
1 0/61 ; b/b c. and d. The expected phenotypic and genotypic ratios from the test cross of A/a ;B/b is
1 A/a ;B/b
1 A/a ; b/b
1 a/a ;B/b
1 (1/61 ; b/b 2. The resulting cells will have the identical genotype as the original cell: A/a ;
B/b. 38 Chapter Three 3. The general formula for the number of different male/female centromeric
combinations possible is 2", where n = number of different chromosome pairs. In this case, 25 = 32. 4. Because the DNA levels vary six—fold, the range covers cells that are haploid
(spores or cells of the gametophyte stage) to cells that are triploid (the
endosperm) and dividing (after DNA has replicated but prior to cell division).
The following cells would fit the DNA measurements: 0.7
1.4 2.1
2.8
4.2 haploid cells
diploid cells in G1 or haploid cells after S but prior to cell division diploid cells after S but prior to cell division
triploid cells after S but prior to cell division Parent cell /\/ a
w
3% Chromosome duplication triploid cells of the endosperm
a t /a+7 z/
:82? :81): “‘°‘l’ Daughter cells Segregation A sporophyte of A/a ;
following proportions: 1/4A ;B 1/4 A ; b 1/4 a ; B 1/4 a ; 19
Random fertilization of the spores from the above gametophytes can occur
4 x 4 = 16 possible ways. Four of these combinations (A ; B x a ; b, a ; b xA
; B, A ; b x a ; B, a ; B xA ; b) will result in the desired A/a ;B/b sporophyte
genotype. Therefore 1/4 of next generation should be of this genotype. B/b genotype will produce gametophytes in the. 7. Mitosis produces cells with the same starting genotype: A/a ;B/b ; C/c. 9. 10. 11. P
Transient diploid F1 ad— ; a x ad+ ; a
adﬂad‘ ; a/a 1/4 ad+ ; a, white
1/4 ad— ; a, purple
1/4 ad+ ; a, white
1/4 ad— ; a, purple Chapter Three 39 The cross is female Xd/Xd ; p/p x male XD/Y ; P/P where P = dominant allele
for pink and d = recessive allele for dwarf. F1 F2 1/2 XD/Xd : P/p (pink female)
1/2 Xd/Y ;P/p (dwarf, pink male) 1/16 xD/xd : P/P (pink female) 1/8 XD/Xd : P/p (pink female) 1/16 XD/Xd : p/p (wild type female)
1/16 Xd/Xd : P/P (dwarf, pink female)
1/8 Xd/Xd : P/p (dwarf, pink female)
1/16 Xd/Xd : p/p (dwarf female) 1/16 XD/Y ; P/P (pink male) 1/8 XD/Y ; P/p (pink male) 1/16 XD/Y ; p/p (wild type male) 1/16 Xd/Y ; P/P (dwarf, pink male)
1/8 Xd/Y ; P/p (dwarf, pink male)
1/16 Xd/Y ; p/p (dwarf male) His children will have to inherit the satellitecontaining 4 (probability = 1/2), the
abnormallystaining 7 (probability = 1/2), and the Y chromosome (probability 2 1/2). To get all three, the probability is (1/2)(1/2)(1/2) = 1/8. The parental set of centromeres can match either parent, which means there are
two ways to satisfy the problem. For any one pair, the probability of a centromere from one parent going into a specific gamete is 1/2. For n pairs, the
probability of all the centromeres being from one parent is (1/2)”. Therefore, the total probability of having a haploid complement of centromeres from either
parent is 2(1/2)" = (1/2)”‘1. 40 12. 13. Chapter Three Dear Monk Mendel: I have recently read your most engrossing manuscript detailing the results of
your most wise experiments with garden peas. I salute both your curiosity and
your ingenuity in conducting said experiments, thereby opening up for scientific
exploration an entire new area of our Maker’s universe. Dear Sir, your findings
are extraordinary! While I do not pretend to compare myself to you in any fashion, I beg to bring to
your attention certain findings I have made with the aid of that most fascinating
and revealing instrument, the microscope. I have been turning my attention to
the smallest of worlds with an instrument that I myself have built, and I have
noticed some structures that may parallel in behavior the factors which you have
postulated in the pea. I have worked with grasshoppers, however, not your garden peas. Although you
are a man of the cloth, you are also a man of science, and Ipray that you will not
be offended when I state that I have specifically studied the reproductive organs
of male grasshoppers. Indeed, I did not limit myself to studying the organs
themselves; instead, I also studied the smaller units that make up the male
organs and have beheld structures most amazing within them. These structures are contained within numerous small bags within the male
organs. Each bag has a number of these structures, which are long and threadlike
at some times and short and compact at other times. They come together in the
middle of a bag, and then they appear to divide equally. Shortly thereafter, the
bag itself divides, and what looks like half of the threadlike structures goes into
each new bag. Could it be, Sir, that these threadlike structures are the very same
as your factors? I know, of course, that garden peas do not have male organs in
the same way that grasshoppers do, but it seems to me that you found it
necessary to emasculate the garden peas in order to do some crosses, so I do not
think it too farfetched to postulate a similarity between grasshoppers and
garden peas in this respect. Pray, Sir, do not laugh at me and dismiss my thoughts on this subject even
though I have neither your excellent training nor your astounding wisdom in the
Sciences. I remain your humble servant to eternity! The hypothesis is that the organism being tested is a heterozygote and that the
A/a and a/a progeny are of equal viability. The expected values would be that
phenotypes occur with equal frequency. There are two genotypes in each case,
so there is one degree of freedom. x2 = Z (observedexpected)2/expected a. x2 = [(120—110)2 + (100—110)2]/110
= 1.818; p > 0.10, nonsignificant; hypothesis cannot be rejected 14. Chapter Three 41 x2 = [(5000—5200)2 + (5400—5200)2]/5200
= 15.385; p < 0.005, significant; hypothesis must be rejected x2 = [(500—620)2 + (540—520)2]/520
= 1.538; p > 0.10, nonsignificant; hypothesis cannot be rejected x2 = [(50—52)2 + (54—52)2]/52
= 0.154; p > 0.50, nonsignificant; hypothesis cannot be rejected This is simply a matter of counting genotypes; there are nine genotypes in
the Punnett square. Alternatively, you know there are three genotypes
possible per gene, for example R/R, R/r, and r/r, and since both genes assort
independently, there are 3 x 3 = 9 total genotypes. Again, simply count. The genotypes are 1 R/R ; Y/Y 1 r/r ; Y/Y lR/R ;y/y '1 r/r ;y/y
2R/r ; Y/Y 2 r/r ; Y/y 2R/r ;y/y 2 R/R ; Y/y 4 R/r ; Y/y To find a formula for the number of genotypes, first consider the following: Number of Number of Number of
genes genotypes phenotypes
1 3 = 31 2 = 21
2 9 = 32 4 = 22
3 27 = 33 8 = 23 Note that the number of genotypes is 3 raised to some power in each case. In other words, a general formula for the number of genotypes is 3”, where
n equals the number of genes. For allelic relationships that show complete dominance, the number of
phenotypes is 2 raised to some power. The general formula for the number of phenotypes observed is 2", where n equals the number of genes. The round, yellow phenotype is R/— ; Y/—. Two ways to determine the exact
genotype of a specific plant are through selfing or conducting a testcross. With selfing, complete heterozygosity will yield a 9:3:3:1 phenotypic ratio.
Homozygosity at one locus will yield a 3:1 phenotypic ratio, while
homozygosity at both loci will yield only one phenotypic class. With a testcross, complete heterozygosity will yield a 1:1:121 phenotypic
ratio. Homozygosity at one locus will yield a 1:1 phenotypic ratio, while
homozygosity at both loci will yield only one phenotypic class. 42 Chapter Three 15. 16. a. 17. a. tapap g. Assuming independent assortment and simple dominant/recessive
relationships of all genes, the number of genotypic classes expected from
selfing a plant heterozygous for H gene pairs is 3" and the number of
phenotypic classes expected in 2”. The data for both crosses suggest that both A and B mutant plants are
homozygous for a recessive allele. Both F2 crosses give 3:1 normal to mutant ratios of progeny. For example, letA = normal and a = mutant, then P A / A x a / 61
F1 A / a
F2 1 A / A phenotype: normal
2 A / a phenotype: normal
1 a / a phenotype: mutant (no trichomes). The cross is A/A ; b/b X 61/51 ; 8/3 to give the F1 ofA/a ; B/b. This is then
test crossed (crossed to a/a ; b/b) to give 1/4 A/a ;B/b (normal) 1/4 A/a ; b/b (no trichomes)
1/4 a/a ;B/b (no trichomes)
1/4 a/a ; b/b (no trichomes) or 1 normal : 3 no trichomes. C/c ;S/S >< C/c ;S/s There are 3 shortzl long, and 3 darkzl albino. Therefore, each gene is heterozygous in the
parents. There are no albino, and there are 1 longzl short
indicating a testcross for this trait. There are no long, and there are 1 darkzl albino.
All are albino, and there are 3 shortzl long.
All are long, and there are 3 darkzl albino. There are no albino, and there are 3 short:
1 long. There are 3 darkzl albino, and 1 shortzl long. C/C ;S/s x C/— ; s/S C/c;S/S x c/c;S/—
c/c ; 5/3 x c/C ;S/s C/c;S/s x C/c;S/s
C/C;S/S x C/—;S/S C/c ;S/s x C/c ; s/S 18. a. and b. Cross 2 indicates that purple (G) is dominant to green (g), and cross 1 indicates cut (P) is dominant to potato (p). Cross 1 : G/g ; P/p x g/g ; P/p There are 3 cut : l potato, and 1 purple :
1 green. Cross 2 : G/g ; P/p x G/g ; p/p There are 3 purple : 1 green, and 1 cut : 1
potato. 19. a. Chapter Three 43 Cross 3: G/G ;P/p x g/g ; P/p There are no green, and there are 3 cutzl potato. There are no potato, and there are 1 purple:1 green. There are 1 cutzl potato, and there are
1 purplezl green. Cross 4: G/g ; P/P x g/g ;p/p
Cross 5: G/g ;p/p x g/g ; P/p
From cross 6, Bent (B) is dominant to normal (b). Both parents are “bent,”
yet some progeny are “normal.” From cross 1, it is Xlinked. The trait is inherited in a sex—specific manner
— all sons have the mother’s phenotype. In the following table, the Y chromosome is stated; the X is implied. Parents Progeny
Cross Female Male Female Male
1 b/b B/Y B/b b/Y
2 B/b b/Y B/b, b/b B/Y, b/Y
3 B/B b/Y 8/19 3/ Y
4 b/b b/ Y b/ b b/Y
5 8/3 B/Y B/B B/Y
6 B/b B/Y B/B, B/b B/Y, b/Y 20. Unpacking the Problem 1. 2. Normal is used to mean wild type, or red eye color and long wings. Both line and strain are used to denote purebreeding ﬂy stocks, and the
words are interchangeable. Your choice. Three characters are being followed: eye color, wing length, and sex. For eye color, there are two phenotypes: red and brown. For wing length,
there are two phenotypes: long and short. For sex, there are two phenotypes: male and female. The F1 females designated normal have red eyes and long wings.
The F1 males that are called shortwinged have red eyes and short wings. The F2 ratio is
3/8 red eyes, long wings 44 Chapter Three 3/8 red eyes, short wings
1/8 brown eyes, long wings
1/8 brown eyes, short wings 9. Because there is not the expected 9:3:3:1 ratio, one of the factors that
distorts the expected dihybrid ratio must be present. Such factors can be sex
linkage, epistasis, genes on the same Chromosome, environmental effect,
reduced penetrance, or a lack of complete dominance in one or both genes. 10. With sex linkage, traits are inherited in a sexspecific way. With autosomal
inheritance, males and females have the same probabilities of inheriting the
trait. 11. The F2 does not indicate sexspecific inheritance. 12. The F1 data does show sex—specific inheritance—all males are short— winged, like their mothers, while all females are normal—winged, like their
fathers. 13. The F1 suggests that long is dominant to short and red is dominant to
brown. The F2 data show a 3 red : 1 brown ratio indicating the dominance
of red but a 1 : 1 long : short ratio indicative of a test cross. Without the F1 data, it is not possible to determine which form of the wing character is
dominant. 14. If Mendelian notation is used, then the red and long alleles need to be
designated with uppercase letters, for example R and L, while the brown (r)
and short (I) alleles need to be designated with lowercase letters. If
Drosophila notation is used, then the brown allele may be designated with a
lowercase b and the wildtype (red) allele with a b+; the short winglength gene with a s and the wildtype (long) allele with an s+. (Genes are often
named after their mutant phenotype.) 15. To deduce the inheritance of these phenotypes means to provide all genotypes for all animals in the three generations discussed and account for
the ratios observed. Solution to the Problem Start this problem by writing the crosses and results so that all the details
are clear. P brown, short female x red, long male
F1 red, long females red, short males 21. Chapter Three 45 These results tell you that redeyed is dominant to browneyed and since
both females and males are redeyed, that this gene is autosomal. Since
males differ from females in their genotype with regard to wing length, this
trait is sexlinked. Knowing that Drosophila females are XX and males are
XY, the longwinged females tell us that long is dominant to short and that
the gene is Xlinked. Let B = red, b = brown, S = long, and S = short. The
cross can be rewritten as follows: P b/b ; 5/5 x 3/3 ; S/Y F1 1/2 B/b ; S/S females
1/2 B/b ; s/Y males F2 1/1() 3/8 ; S/s
1/16 3/3 ; s/s
1/8 B/b ; 5/5
1/8 B/b ; 3/8
1/16 b/b ; S/S
1/16 17/19 ; s/S
1/16 3/3 ; S/Y
1/16 8/3 ; S/Y
1/8 3/19 ;S/Y
1/8 B/b ; S/Y
1/16 Mi) ;S/Y
1/16 b/b ; S/Y red, long, female
red, short, female
red, long, female
red, short, female
brown, long, female
brown, short, female
red, long, male red, short, male red, long, male red, short, male
brown, long, male
brown, short, male The final phenotypic ratio is 3/8 red, long
3/8 red, short
1/8 brown, long
'1/8 brown, short with equal numbers of males and females in all classes. Because photosynthesis is affected and the plants are yellow, rather than
green, it is likely that the chloroplasts are defective. If the defect maps to
the DNA of the chloroplast, the trait will be maternally inherited. This fits
the data that all progeny have the phenotype of the female parent and not
the phenotype of the male (pollen—donor) parent. If the defect maps to the DNA of the chloroplast, the trait will be maternally
inherited. Use pollen from sickly, yellow plants and cross to emasculated 46 22. 23. 24. 25. 26. Chapter Three ﬂowers of a normal dark greenleaved plant. All progeny should have the
normal dark green phenotype. c. The chloroplasts contain the green pigment chlorophyll and are the site of
photosynthesis. A defect in the production of chlorophyll would give rise to
all the stated defects. Maternal inheritance of chloroplasts results in the greenwhite color variegation
observed in Mirabilis. Cross 1: variegated female x green male —> variegated, green, or white
progeny
Cross 2: green female x variegated male —> green progeny In both crosses, the pollen (male contribution) contains no chloroplasts and thus
does not contribute to the inheritance of this phenotype. Eggs from a variegated
female plant can be of three types : contain only “green” chloroplasts, contain
only “white” chloroplasts, or contain both (variegated). The offspring will have
the phenotype associated with the egg’s chloroplasts. The crosses are
Cross 1:
Cross 2: stopstart female x wildtype male —> all stopstart progeny wildtype female x stopstart male —> all wildtype progeny
mtDNA is inherited only from the “female” in Neurospora. The genetic determinants of R and S are showing maternal inheritance and are therefore cytoplasmic. It is possible that the gene that confers resistance maps
either to the mtDNA or chNA. The hypothesis is that the organism being tested is a dihybrid with
independently assorting genes and that all progeny are of equal Viability. The
expected values would be that phenotypes occur with equal frequency. There are
four phenotypes so there are 3 degrees of freedom. x2 = Z (observedexpected)2/expected x2 = [(230—233)2 + (210—233)2 + (240—233)2 + (250—233)2]/233
= 3.75; p value between 0.1 and 0.5, nonsignificant; hypothesis cannot be
rejected The hypothesis is that the organism being tested is a dihybrid with
independently assorting genes and that all progeny are of equal viability. The
expected values would be that phenotypes occur in a 9 : 3 : 3 : 1 ratio. There are
four phenotypes so there are 3 degrees of freedom. x2 = Z (observedexpected)2/expected 27. 28. 29. 30. 31. Chapter Three 47 x2 = (178—180)2/180 + (62—60)2/6o + (56—60)2/60 + (24—20)2/20
= 1.156; p > 0.50, nonsignificant; hypothesis cannot be rejected When results of a cross are sexspecific, sex linkage should be considered. In
moths, the heterogametic sex is actually the female while the male is the
homogametic sex. Assuming that dark (D) is dominant to light (d), then the data
can be explained by the dark male being heterozygous (D/d) and the dark female
being hemizygous (D). All male progeny will inherit the D allele from their
mother and therefore be dark while half the females will inherit D from their
father (and be dark) and half will inherit d (and be light). a. This inheritance pattern is diagnostic for organelle inheritance. These
crosses indicate the mutant gene resides in the mitochondria. (Neurospora
does not have chloroplasts.) b. All progeny from this cross will have the “maternal” trait stopper but the
nic3 allele should segregate 1:1 in the octad—four spores will be stp ; nic3 and four spores will be stp ; nic3+.
a. There should be nine classes—0, 1, 2, 3, 4, 5, 6, 7, 8 “doses” b. There should be 13 classes—0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, '11, 12 “doses” There are three ways for a self cross of this genotype to give rise to just one dose
— one dose ofR1 and none ofR2 or R3; one dose ong and none of R1 or R3; or one dose of R3 and none of R1 or R2. The chance of inheriting one dose of R]
(from RI/rl x RI/rl) is 1/2. The chance of no doses of R2 (from RZ/r2 >< Rz/I’Z) is 1/4 as is the chance of no doses of R3 (from R3/r3 x R3/r3). Therefore, the
desired outcome ofjust one dose is 3 x 1/2 x 1/4 x 1/4 = 3/32. a. Both the gametophyte and the sporophyte are closer in shape to the mother
than the father. Note that a size increase occurs in each type of cross. b. Gametophyte and sporophyte morphology are affected by extranuclear
factors. Leaf size may be a function of the interplay between nuclear
genome contributions. c. If extranuclear factors are affecting morphology while nuclear factors are
affecting leaf size, then repeated backcrosses could be conducted, using the
hybrid as the female. This would result in the cytoplasmic information
remaining constant while the nuclear information becomes increasingly like
that of the backcross parent. Leaf morphology should therefore remain
constant while leaf size would decrease toward the size of the backcross
parent. 48 32. 33. 34. Chapter Three The goal here is to generate a plant with the cytoplasm of plant A and the
nuclear genome predominantly of plant B. Remember that the cytoplasm is
contributed by the egg only. So using plant A as the maternal parent, cross to B
(as the paternal parent) and then backcross the progeny of this cross using plant
B again as the paternal parent. Repeat for several generations until Virtually the
entire nuclear genome is from the B parent. If the variegation is due to a chloroplast mutation, then the phenotype of the
offspring will be controlled solely by the phenotype of the maternal parent. Look
for ﬂowers on white branches and test to see if they produce seeds that grow
into all white plants regardless of the source of the pollen. To test whether a
dominant nuclear mutation is responsible for the variegation, cross pollen from
ﬂowers on white branches to green plants (or ﬂowers on green branches of same
plant) and see if half the progeny are white and half are green (assuming the
dominant mutation is heterozygous) or all white, if the dominant mutation is
homozygous. Progeny plants inherited only normal chNA (lane 1); only mutant chNA
(lane 2); or both (lane 3). In order to get homoplasmic chNA (all chloroplasts
containing the same DNA), seen in lanes 1 and 2, segregation of chloroplasts
had to occur. CHALLENGING PROBLEMS 35. a. Before beginning the specific problems, calculate the probabilities
associated with each jar.
jar l p(R) 600/(600 + 400) = 0.6
p(W) = 400/(600 + 400) = 0.4
jar 2 p(B) = 900/(900 + 1.00) = 0.9
p(W) 100/(900 + 100) = 0.1
jar 3 p(G) 10/(10 + 990) = 0.01
p(W) 990/(10 + 990) = 0.99 (1) p(R, B, G) = (0.6)(0.9)(0.01) = 0.0054 (2) p(W, W, W) = (0.4)(0.l)(0.99) = 0.0396 (3) Before plugging into the formula, you should realize that, while
white can come from any jar, red and green must come from specific jars (jar 1 and jar 3). Therefore, white must come from jar 2:
p(R, W, G) = (0.6)(0.1)(0.01) = 0.0006 (4) p(R, W, W) = (0.6)(0.1)(0.99) = 0.0594 Chapter Three 49 (5) There are three ways to satisfy this:
R, W,W or W, B,W or W,W,G
= (0.6)(0.1)(0.99) + (0.4)(0.9)(0.99) + (0.4)(0.1)(0.01)
= 0.0594 + 0.3564 + 0.0004 = 0.4162 (6) At least one white is the same as 1 minus no whites:
p(at least 1 W) = 1 — p(no W) = 1 —p(R, B, G)
= 1 — (0.6)(0.9)(0.01) = 1 — 0.0054 = 0.9946 b. The cross is R/r >< R/r. The probability of red (R/—) is 3/4, and the probability of white (r/r) is 1/4. Because only one white is needed, the only
unacceptable result is all red. In )1 trials, the probability of all red is (3/4)”. Because the probability of
failure must be no greater than 5 percent: (3/4)" < 0.05 n > 10.41, or 11 seeds c. The p(failure) = 0.8 for each egg. Since all eggs are implanted
simultaneously, the p(5 failures) = (0.8)5. The p(at least one success) = l —
(0.8)5 = '1 — 0.328 = 0.672 36. Use the following symbols: Gene function Dominant allele Recessive allele
color R = red r = yellow
loculed L = two 1 = many
height H = tall h = dwarf The starting plants are purebreeding, so their genotypes are
red, twoloculed, dwarf R/R ; L/L ; h/h and
yellow, manyloculed, tall r/r ; l/l ; H/H. The farmer wants to produce a pure—breeding line that is yellow, twoloculed,
and tall, which would have the genotype r/r ; L/L ;H/H. The two purebreeding starting lines will produce an F1 that will be R/r ; L/l ; H/h. By doing an F1 x F1 cross, 1/64 of the F2 progeny should have the correct
genotype (% r/r x 1/4 UL x ]/4 H/H). The probability of NOT getting that is (l —
I/64)", where n is number of progeny scored. For that to be less than 5 percent, It 50 37. Chapter Three = 191. So we need at least 191 progeny to start with, and by selecting yellow,
twoloculed, and tall plants from these progeny, the known genotype will be r/r ;
L/— ; H/—. To identify how many of these are required for further testing (by test
cross): the probability of being homozygous dominant for both (given that we
are selecting only from those plants with dominant phenotypes) is 1/3 X 1/3 = 1/9.
Therefore, the probability of a plant not being homozygous for both is 8/9. We
want the probability of all plants tested not being homozygous for both to be less than 5 percent, or (8/9)" < 0.05. If n = 26, p = .047. So at least 26 of the
yellow, twoloculed, and tall progeny should be testcrossed to a l/l; h/h parent
to determine which are homozygous for the two dominant traits. (Note: several
[/1 ; h/h testers are likely to be recovered among the 191 F2 progeny generated. So you will only need one test cross for each candidate.) For each testcross, the plant will obviously be discarded if the testcross reveals a
heterozygous state for the gene in question. If no recessive allele is detected,
then the minimum number of progeny that must be examined to be 95 percent
confident that the plant is homozygous, is based on the frequency of the dominant phenotype if heterozygous, which is 1/2. In n progeny, the probability
of obtaining all dominant progeny in a testcross, given that the plant is
heterozygous, is (1/2)". To be 95 percent confident of homozygosity, the following formula is used, where 5 percent is the probability that it is not
homozygous: (1/2)n = 0.05 n = 4.3, or 5 phenotypically dominant progeny must be obtained from each
testcross to be 95 percent confident that the plant is homozygous. 3. Because each gene assorts independently, each probability should be
considered separately and then multiplied together for the answer. For (1) 1/2 will be A, 3/4 will be B, 1/2 will be C, 3/4 will be D, and 1/2 will
be E.
1/2 X 3/4 X 1/2 X 3/4 X 1/2 = 9/128 For (2) 1/2 will be a, 3/4 will be B, 1/2 will be c, 3/4 will be D, and 1/2 will
be e. 1/2 X 3/4 X 1/2 X 3/4 X 1/2 = 9/128
For (3) it is the sum of (1) and (2) = 9/128 + 9/128 = 9/64 For (4) it is 1 — (part 3) = 1 — 9/64 = 55/64 b. For (1) 1/2 will be A/a, 1/2 will be B/b, 1/2 will be C/c, 1/2 will be D/d, and
1/2 will be E/e. 38. Chapter Three 51 1/2 x 1/2 X 1/2 X 1/2 X 1/2 = 1/32 For (2) 1/2 will be a/a, 1/2 Will be B/b, 1/2 Will be 6/6, 1/2 Will be D/d, and
1/2 will be 6/6.
1/2 X 1/2 X 1/2 X 1/2 X 1/2 = 1/32 For (3) it is the sum of (1) and (2) = 1/16
For (4) it is 1 —— (part 3) = 1_ 1/16 :15/16 Cataracts appear to be caused by a dominant allele because affected people
have affected parents. Dwarfism appears to be caused by a recessive allele
because affected people have unaffected parents. Both traits appear to be
autosomal. Using A for cataracts, a for no cataracts, B for normal height, and b for
dwarfism. The genotypes are: 111 : a/a ;B/b, a/a ;B/b, A/a ;B/—, a/a ; B/—, A/a ;B/b, A/a ;B/b, a/a ; B/—,
a/a ;B/—, 61/61 ; b/b The mating is a/a ; b/b (IV1) x A/— ; B/— (IV—5). Recall that the probability
of a child’s being affected by any disease is a function of the probability of
each parent carrying the allele in question and the probability that one
parent (for a dominant disorder) or both parents (for a recessive disorder)
donate it to the child. Individual IV1 is homozygous for these two genes, therefore, the only task is to determine the probabilities associated with
individual IV5. The probability that individual [V5 is heterozygous for dwarfism is 2/3,.
Thus the probability that she has the 1) allele and will pass it to her child is
2/3 X 1/2 = 1/3 The probability that individual IV—S is homozygous for cataracts is 1/3; the probability that she is heterozygous is 2/3. If she is homozygous for the
allele that causes cataracts, she must pass it to her child or if she is heterozygous for cataracts, she has a probability of 1/2 of passing it to her
child. The probability that the first child is a dwarf with cataracts is the
probability that the child inherits the A and b alleles from its mother which is (1/3 X 1)(2/3 X 1/2) + (2/3 x 1/2)(2/3 X 1/2) = 2/9. Alternatively, you can
calculate the chance of inheriting the b allele (2/3 x 1/2) and not inheriting
the a allele (1— 1/3,) or 1/3 x 2/3 = 2/9_ 52 Chapter Three The probability of having a phenotypically normal child is the probability
that the mother donates the a and B (or not b) alleles, which is (2/3 x 1/2)(1— 1/3) = 2/9. 39. a. and b. Begin with any two of the three lines and cross them. If, for example,
you began with a/a ;B/B ; C/C x A/A ; b/b ; C/C, the progeny would all be
A/a ;B/b ; C/C. Crossing two of these would yield: 9 A/— ; B/— ; C/C
3 61/61 ;B/— ; C/C
3 A/— ; b/b ; C/C
1 a/a ; b/b ; C/C The a/a ; b/b ; C/C genotype has two of the genes in a homozygous
recessive state and occurs in 1/16 of the offspring. If that were crossed with
A/A ; B/B ; c/c, the progeny would all be A/a ; B/b ; C/c. Crossing two of
them (or “selfing”) would lead to a 27:9:9:9:3:3:3:1 ratio, and the plant
occurring in 1/64 of the progeny would be the desired a/a ; b/b ; c/c. There are several different routes to obtaining a/a ; b/b ; c/c, but the one
outlined above requires only four crosses. 40. First, draw the pedigree. Let the genes be designated by the pigment produced by the normal allele : red
pigment, R; green pigment, G; and blue pigment, B. Recall that the sole X in males comes from the mother, while females obtain an
X from each parent. Also recall that a difference in phenotype between sons and
daughters is usually due to an X—linked gene. Because all the sons are colorblind
and neither the mother nor the daughters are, the mother must carry a different
allele for colorblindness on each X Chromosome. In other words, she is
heterozygous for both Xlinked genes, and they are in repulsion: R g/r G. With
regard to the autosomal gene, she must be B/—. Because all the daughters are normal, the father, who is colorblind, must be able
to complement the defects in the mother with regard to his X chromosome.
Because he has only one X with which to do so, his genotype must be R G/Y ; 41. 42. Chapter Three 53 b/b. Likewise, the mother must be able to complement the father’s defect, so she
must be 3/3. The original cross is therefore P Rg/rG;B/B><RG/Y;b/b F1 Females Males
R g/R G ;B/b R g/Y ;B/b
r G/R G ;B/b r G/Y ;B/b a. The pedigree clearly shows maternal inheritance. b. Most likely, the mutant DNA is mitochondrial. In the following schematic drawings, chromosomes (or chromatids) that are
radioactive are indicated be the grains that would be observed after
radioautography. After the second mitotic division, a number of outcomes are possible due to the random alignment and separation of the radioactive and non—
radioactive chromatids. 54 Chapter Three Prophase of
ﬁrst mitosis Telophase of
ﬁrst mitosis Prophase of i]
second mitosis /\ Chapter Three 55 Telophase of second mitosis PIObablllty and OR and OR
and L \ 1/4 and OR and 5 6 Chapter Three 43. Prophase of
first meiotic
division /\
.3
\ / /\ 44. In a diploid cell, expect two chromosomes (a pair of homologs) to each
have a single locus of radioactivity. Expect many regions of radioactivity scattered throughout the
chromosomes. The exact number and pattern would be dependent on the
specific sequence in question, and where and how often it is present within
the genome. The multiple copies of the genes for ribosomal RNA are organized into
large tandem arrays called nucleolar organizers (NO). Therefore, expect
broader areas of radioactivity compared to (a). The number of these regions
would equal the number of NO present in the organism. Each chromosome end would be labeled by telomeric DNA. The multiple repeats of this heterochromatic DNA are organized into large
tandem arrays. Therefore, expect broader areas of radioactivity compared to
(a). Also, there may be more than one area in the genome of the same
simple repeat. 45. 46. Chapter Three 57 The following is meant to be examples of what is possible. It is also possible,
for instance, that more than one band would be present in (a), depending on the
position of the restriction sites within the sequence complementary to the probe
used. H :1









 For ((1) and (e), the specifics or where the DNA is cut relative to the telomeric
DNA or heterochromatic DNA will effect what is observed. Assuming the
restriction sites are not within the telomeric or heterochromatic DNA, then (d)
will be similar to (b), and (c) will have one or several very large bands. (1) Impossible: the alleles of the same genes are on nonhomologous
chromosomes (2) Meiosis II (3) Meiosis II (4) Meiosis II (5) Mitosis (6) Impossible: appears to be mitotic anaphase but alleles of sister
chromatids are not identical (7) Impossible: too many chromosomes (8) Impossible: too many chromosomes (9) Impossible: too many chromosomes (10) Meiosis I (11) Impossible: appears to be meiosis of homozygous a/a ;B/B (12) Impossible: the alleles of the same genes are on nonhomologous chromosomes 58 47. 48. Chapter Three Recall that each cell has many mitochondria, each with numerous genomes.
Also recall that cytoplasmic segregation is routinely found in mitochondrial
mixtures within the same cell. The best explanation for this pedigree is that the mother in generation I
experienced a mutation in a single cell that was a progenitor of her egg cells
(primordial germ cell). By chance alone, the two males with the disorder in the
second generation were from egg cells that had experienced a great deal of
cytoplasmic segregation prior to fertilization, while the two females in that
generation received a mixture. The spontaneous abortions that occurred for the first woman in generation 11
were the result of extensive cytoplasmic segregation in her primordial germ
cells: aberrant mitochondria were retained. The spontaneous abortions of the
second woman in generation [I also came from such cells. The normal children
of this woman were the result of extensive segregation in the opposite direction:
normal mitochondria were retained. The affected children of this woman were
from egg cells that had undergone less cytoplasmic segregation by the time of
fertilization, so that they developed to term but still suffered from the disease. a. Let B = brachydactylous, b = normal, T = taster and t = nontaster. The
genotypes of the couple are B/b ; T/t for the male and b/b ; T/t for the
female. b. For all four children to be brachydactylous, the chance is (1/2)4 = 1/1(,. c. For none of the four children to be brachydactylous, the chance is (1/2)4 =
1/16 d. For all to be tasters, the chance is (3/4)4 = 81/256.
e. For all to be nontasters, the chance is (1/4)4 = 1/256. f. For all to be brachydactylous tasters, the chance is (1/2 x 3/4)4 = 81/4096. g. Not being a brachydactylous taster is the same 1— (the chance of being a
brachydactylous taster) or 1 — (1/2 x3/4) = 5/8. The chance that all four children are not brachydactylous tasters is (5/8)4 = 625/4096. h. The chance that at least one is a brachydactylous taster is 1 — (the chance of
none being a brachydactylous taster) or 1 — (5/8)4. 49. Chapter Three 59 For the following, S will signify cytoplasm of a malesterile line and N will
signify cytoplasm of a nonmale—sterile line. Rf will signify the dominant nuclear
restorer allele and rf, the recessive nonrestorer allele. a. If S rf/rf (malesterile plants) are crossed with pollen from N Rf/Rf plants,
the offspring will all be S Rf/rf and male fertile. If these offspring are then
crossed with pollen from N rf/rf plants, half the offspring will be S Rf/rf
(malefertile) and half will be S rf/rf (malesterile). The S cytoplasm will not be altered or affected even though the maternal offspring parent plant
was Rf/rf. b. The cross is S rf/rf x N Rf/Rf so all the progeny will be S Rf/rf and male
fertile. c. The cross is S Rf/rf x N rf/rf so half the progeny will be S Rﬂrf (male—
fertile) and half will be S rﬂrf(ma1e—sterile). d. i. The cross is S rfI/rfI ;rf—2/rf2 x NRfI/rfI ;Rf2/rf2 The progeny will be: 1/4 S Rf—I/rfI ; RfZ/rfZ (male—fertile)
1/4 S RfI/rfI ; rf2/rf2 (malefertile)
1/4 S rfI /rf1 ;Rf2/rf2 (male—fertile) 1/4 S rf—I/rfI ; rf—Z/rf2 (malesterile) ii. The cross is S rfI/rf—I ;rf2/rf2 x NRfI/RfI ;rf2/rf2
The progeny will all be: S RfI/rfI ; rf2/rf2 (malefertile) iii. The cross is S rfI /rf1 ;rf2/rf2 x NRfI/rfI ;rf2/rf2
The progeny will be: 1/2 S RfI /rf1 ; rf2/rf2 (malefertile)
1/2 S rfI /rf1 ; rfZ/rfZ (malesterile) vi. The cross is S rfI/rfI ; rfZ/rfZ x N RfI/rfI ;Rf2/Rf—2
The progeny will be: 1/2 S RfI /rf1 ;Rf2/rf2 (male—fertile)
1/2 S rfI /rf1 ;Rf2/rf2 (malefertile) ...
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This note was uploaded on 03/19/2008 for the course BIOL 202 taught by Professor Kieberhogan during the Spring '08 term at UNC.
 Spring '08
 KieberHogan
 molecular biology, Genetics

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