STUDY HINTS
Regular segregation and the assortment of alleles in a heterozygote produce the familiar genotypic and phenotypic ratios that you investigated in Problem Set 4. In a real sense these ratios are also “hypotheses,” in that they are the expectations appropriate to a particular genetic situation. For example, a cross between two heterozygotes, Aa × Aa, yields a 3:1 phenotypic ratio among the offspring. Turning this around, if one finds a 3:1 phenotypic ratio in a family, it is reasonable to hypothesize that the parents were both heterozygotes. Ratios are therefore an important key to establishing the genetic basis of an unfamiliar trait.
Sex-linkage, multiple alleles, gene interactions, and maternal and cytoplasmic effects are natural complications that can modify these underlying patterns and ratios. The secret to solving these types of problems is to be familiar with the clues that are often embedded in modified ratios. A distinct difference between male and female phenotypic ratios leads one to consider the possibility of sex-linkage and/or a maternal or cytoplasmic involvement. Lethality would lead to truncated ratios (e.g., 1:2), whereas the gene interactions such as epistasis would merge certain genotypic classes into the same phenotypic class. Textbooks often describe a large number of these modified ratios, but for convenience we have summarized some of the most commonly encountered ones in Table 6.1.
TABLE 6.1 Summary of Modified Ratios
|
Ratio |
Possible Interpretations |
|
1:1 |
Monohybrid testcross: genotype or phenotype |
|
1:1:1:1 |
Dihybrid testcross: genotype or phenotype |
|
1:2:1 |
Monohybrid genotypic ratio |
|
Monohybrid phenotypic ratio for an incompletely dominant or codominant trait |
|
|
1:2 |
Monohybrid phenotypic or genotypic ratio in which the dominant or recessive homozygote dies (a recessive lethal) |
|
3:1 |
Monohybrid phenotypic ratio |
|
9:3:3:1 |
Dihybrid phenotypic ratio |
|
12:3:1 |
Dominant epistasis, in which one dominant locus masks the expression of the second locus |
|
9:3:4 |
Recessive epistatis, in which one recessive homozygous locus masks the expression of a second locus |
|
15:1 |
Either dominant trait sufficient to produce the phenotype |
|
9:7 |
Both dominant traits required to produce the phenotype |
In addition to traits that can be traced to nuclear genes and their interactions, phenotypes are often dependent upon cytoplasmic interactions. The behavior of nuclear genes is most familiar, because they follow the patterns summarized by Mendelian rules. Cytoplasmic genes, such as those in mitochondria or chloroplasts, on the other hand, are inherited in each generation through the maternal cytoplasm. Thus, although cytoplasmic genes persist within a family they do not segregate in a Mendelian fashion.
In contrast, maternal effects are often transient. Maternal effects involve genes that affect the cytoplasm, such as by coding for some material that is deposited in the egg and that functions until it is replaced by products of the offspring’s nucleus, early in development. For example, the pigment precursor kynurenine, produced by a dominant gene, A, causes brown-pigmented areas in larval and adult flour moths, Ephestia kuhniella. In a cross between Aafemales and aa males, all progeny are pigmented at first, but as they mature only the Aa larvae retain their pigmentation, because the kynurenine deposited in the egg cytoplasm by the mother cannot be replenished by the larval genome in aa individuals.
Another example, the direction of snail shell coiling in Limnaea, demonstrates that maternal effects can sometimes persist in an individual. Maternal effects and cytoplasmic inheritance can generally be distinguished by the fact that the extrachromosomal factors in cytoplasmic inheritance can persist for many generations.
IMPORTANT TERMS
Antibody
Antigen
Attached X
Autosome
Balanced lethal
Codominance
Complementary gene action
Cytoplasmic inheritance
Epistasis
Expressivity
Hemizygous
Heterogametic sex
Histocompatibility
Homogametic sex
Incomplete dominance
Isoallele
Lethal
Maternal effects
Nondisjunction
Overdominance
Parthenogenesis
Penetrance
Petite mutations
Phenocopy
Pleiotropism
Segregation distortion
Sex-influenced
Sex-limited
Sex-linked
PROBLEM SET 6
1. In a family, the father and mother have normal red–green color vision. Red–green color blindness is a sex-linked recessive trait in humans. If their son is color-blind and their daughter has normal color vision, give the genotypes of all members of the family.
link to answer
2. If a normal man marries a woman who is heterozygous for a sex-linked recessive lethal mutation, what would be the expected sex ratio of their offspring?
link to answer
3. In a paternity suit, it is discovered that the mother and the child are both heterozygous for the MN blood group and are both blood type B. The suspected father has blood type A, group MN, and, in addition, is an albino and is heterozygous for the dominant nervous degenerative condition Huntington’s chorea. The accused man
(a) can,
(b) cannot be the father of the child.
link to answer
4. A woman with blood type O has a child with blood type O. She claims that a friend of hers is the child’s father.
(a) His blood type is A. Can he be excluded as the father on this evidence alone?
link to answer
(b) Does the fact that the accused man’s mother is type A and his father is type AB permit him to be excluded?
link to answer
(c) Does the additional information that his mother’s parents are both AB permit him to be excluded?
link to answer
link to answer
5. Assume that two different albino strains of mice give pigmented progeny when crossed with pigmented strains. In both cases, the F2 segregates three pigmented to one albino. When albino strain I is crossed to albino strain II, however, all progeny are pigmented. How could you explain these results, and what further cross could be made to test your hypothesis?
link to answer
6. Here are the results of two crosses of a short-bristle mutant in Drosophila.
|
(1) short-bristle × wild-type: |
||
|
36 short-bristle females |
32 short-bristle males |
|
|
42 wild-type females |
34 wild-type males |
n = 144 |
|
(2) short-bristle × short-bristle: |
||
|
39 short-bristle females |
33 short-bristle males |
|
|
21 wild-type females |
18 wild-type males |
n = 111 |
Crosses of type (2) invariably have fewer progeny than those of type (1).
How would you explain these results?
link to answer
7. Assume in humans a sex-linked locus with three alleles, A1, A2, and A3, and two independently assorting autosomal loci: B, with four alleles (B1, B2, B3, and B4), and C, with three alleles (C1, C2, and C3). Given this potential genetic variability, the number of genetically different males that are possible is
(a) 600,
(b) 11,
(c) 48,
(d) 180,
(e) 72,
(f) none of the above.
link to answer
8. Assume that a gene for baldness (B) is dominant to its allele for hair production (b) and that a gene for curly hair (Cy) is dominant to its allele for straight hair (cy). The two loci are not linked, and baldness shows dominant epistasis over hair form. In a cross of Bb Cycy × bb cycy, the expected progeny will be
(a) 12 bald: 3 curly: 1 straight,
(b) 9 bald: 4 curly: 3 straight,
(c) 4 bald: 3 curly: 1 straight,
(d) 2 bald: 1 curly: 1 straight,
(e) none of the above.
link to answer
9. There are at least 4 closely linked genes in the histocompatibility complex in humans. These genes are multiallelic, having as many as 35 codominant alleles. They are so closely linked that recombination among them can be disregarded. An individual is tested and found to have the genotype
A1A34B3B24C1C23D10D11
His wife has the genotype
A1A20B10B11C3C12D9D13
Their son is
A1A34B11B24C1C12D9D10
Diagram the cross that produced this son, and show the allele linkages (the haplo-type) on the chromosomes of the parents and their son.
link to answer
10. Assume that in rabbits there are three different independently assorting autosomal loci that affect coat color. A colorless-pigment precursor V is converted to a colorless precursor W by action of the Aallele. W is converted to tan pigment by action of the T allele, and the tan pigment is converted to black pigment by action of the B allele.
The homozygous recessive condition at each locus results in loss of enzyme activity for the reaction controlled by that gene. A cross of trihybrid bunnies, it AaTtBb × AaTtBb, would be expected to give rise to which of the following progeny?
(a) 12 black: 3 tan: 1 white,
(b) 9 black: 3 tan: 4 white,
(c) 4 black: 3 tan: 9 white,
(d) 27 black: 21 tan: 16 white,
(e) 27 black: 9 tan: 28 white,
(f) 21 black: 16 tan: 27 white,
(g) none of the above
link to answer
11. In a certain plant species, an F2 generation produced 113 round seeds, 75 oval seeds, and 12 curled seeds. One of the original homozygous parent strains had round seeds, and the other had curled seeds. What is the minimum number of genes that would adequately explain the data, and what genotypic ratio and explanation best fit these results?
link to answer
12. The presence of dominant alleles at two different loci (A– B–) in poultry results in walnut comb. Pea comb is produced by the genotype aaB–, rose comb by A – bb, and single comb by aabb. If you were given a rooster with a walnut comb, known to be heterozygous for both genes, and a rose-comb hen from a true-breeding strain, what cross or crosses, and how many generations, would it take to obtain true-breeding pea-comb poultry?
link to answer
13. When two genes interact, phenotypic ratios are modified from the 9:3:3:1 independent segregation ratio. What will the phenotypic ratios be when
(a) one recessive homozygote suppresses the expression of an independently assorting gene?
link to answer
(b) either recessive homozygote suppresses the expression of the other locus?
link to answer
(c) the presence of one dominant allele suppresses the expression of the other locus?
link to answer
link to answer
14. The autosomal genes Curly (Cy) and Plum (Pm) in Drosophila are independently segregating, phenotypically dominant mutations that are recessive lethals. In other words, they have a distinct phenotype as heterozygotes but are lethal as homozygotes. When Curly and Plum are separately crossed to wild flies and the F1 mutant progeny are intercrossed, what will the resulting phenotypic ratio be?
link to answer
15. Assume that in mice there are a recessive lethal gene (d) and a recessive gene (f) that, when homozygous, results in female sterility. The two loci are both on chromosome 3. Assume that you have available one strain carrying the lethal allele and another strain with the f allele together with a crossover inhibitor that prevents exchange between the d and f loci. Describe the balanced lethal stock that can be made using these two chromosomes. Consider this a “brain-teaser” problem.
link to answer
Maternal Effects and Cytoplasmic Inheritance
16. In some of the first experiments carried out using Drosophila, Calvin Bridges crossed white-eyed Drosophila females to red-eyed males. The progeny are red-eyed females and white-eyed males. Which of the following one or more statements is true?
(a) The male is the heterogametic sex.
(b) The white-eyed allele is located on the X chromosome.
(c) Red eyes (the wild-type) is dominant to white eyes.
link to answer
17. The genetic makeup of a population of mice has four different alleles for a certain sex-linked gene and five different alleles for an autosomal locus. Basing your analysis upon these two loci, what is the number of genetically different males in the population?
link to answer
18. Some years ago, the well-known movie star Charlie Chaplin was named in a paternity suit by an actress. Her blood type was A, the child’s was B, and Chaplin’s was O. The first jury was hung; the second found him guilty. What do you think of the verdict, and why?
link to answer
19. The ABO locus in humans has four common alleles: A1, A2, B, and O. These are sometimes called the “public” antigenic variants. There are also several less common alleles at this locus, some so rare that they may very well be found in only one family or a very few families. These are called “private” antigenic variants. A man who is A9B (where A9 is a hypothetical “private” allele) has been wrongfully accused of fathering a child with blood type A. Should he look forward to serological tests with trepidation or with confidence?
link to answer
20. A man with type O blood marries a woman with type A. One of their children has type B blood, and the other is blood type AB. Assuming that the man is indeed the father of these children, what hypothesis or hypotheses might you propose to explain his phenotype and those of his children?
link to answer
21. In cockatiels (medium-sized parrots), gray birds often produce a clutch that has white chicks as well as gray ones. The white chicks from such a cross are always females. White parents produce only white chicks. What is the probable mode of inheritance of color in these birds?
link to answer
22. In a hypothetical rodent species, a gene that produces floppy ears was found segregating in the population. When two floppy-eared rodents are crossed, a ratio of 2 floppy ears to 1 straight ears is found in the offspring. If a floppy-eared rodent is crossed to a straight-eared individual, a ratio of approximately 1:1 is produced. What is the probable mode of inheritance of ear phenotype in this species?
link to answer
23. Assume there is a strain of moths in which the D gene makes the wings solidly colored, while those homozygous for the recessive dd have spotted wings. The A gene blocks synthesis of pigment, with aahaving pigmented wings. What would be the expected phenotypic ratio from a dihybrid cross?
(a) 1:1:1:1;
(b) 12:3:1;
(c) 9:3:4;
(d) 3:1;
(e) 15:1.
link to answer
24. Two gerbils are heterozygous for a dominant “long tail” mutation that has 50 percent penetrance. If these two gerbils are mated together, what phenotypic ratio of “long tail” to “normal tail” would you expect in the next generation?
link to answer
25. If the mother’s haplotype for four linked genes that show multiple allelism is A4A7B8B2C4C4D9D1 and her child is A7A9B8B5C4C3D9D9, which of the following could not be the biological father?
(a) A7A9B7B5C4C3D9D1
(b) A9A10B5B8C9C3D9D1
(c) A4A9B1B5C9C6D9D3
(d) A2A9B5B8C4C3D2D9
(e) Any of the above could theoretically be the biological father.
link to answer
26. In the snail Limnaea, dextral (right) coiling is dominant to sinistral (left) coiling and shows a maternal effect. What are the possible genotypes for direction of coiling in a phenotypically dextral snail?
link to answer
27. When Mendel crossed plants heterozygous for genes determining height (Tt × Tt), he got 3/4 tall (TT and Tt) and 1/4 short (tt) in the next generation. Does a comparable genetic ratio result in Limnaeafrom a cross of Dd × Dd? Please explain.
link to answer
28. The variegation that is seen in plants such as Mirabilis, where leaves or parts of leaves are full green, lighter green, variegated green and white, or completely white, provides evidence that chloroplasts (which are capable of replication)
(a) are distributed more or less at random to daughter cells during mitosis,
(b) behave much like chromosomes during mitosis, in that they divide and separate to opposite poles of the cell.
link to answer
29. In a plant species capable of either self-fertilization or outcrossing, two pure-breeding, white flower strains occur. One strain has white flowers because of a dominant cytoplasmic gene. The other has white flowers because of a recessive chromosomal gene. If reciprocal crosses are made between these two strains, what phenotypes and what proportions would be expected in the progeny? If the F1 were self-fertilized, what ratios and phenotypes would you expect?
link to answer
30. In a certain bird species, two color morphs are known: gray and red. Red chicks are often produced by gray parents, but gray chicks can also be produced by red parents. There are also pure-breeding birds that only produce offspring the color of the parents. If a female from a pure-breeding red line is mated to a male from a pure-breeding gray line, the F1 are all red. If the reciprocal cross is made, the F1 are all gray. If the F1 female progeny from either cross are mated to males from either the pure-breeding gray strain or the pure-breeding red strain, the offspring are always red. However, if the F1 males are mated to the pure-breeding lines, the offspring are phenotypically the color of the pure-breeding line. What is the probable mode of inheritance of color in these birds?
link to answer
ANSWERS TO PROBLEM SET 6
1. Sons get their X chromosome from their mothers. The combination of a normal mother with a color-blind son tells us that the mother was heterozygous (Cc) and that the son is hemizygous (c). The father is phenotypically normal for color vision and is therefore C. The daughter is either CC or Cc, each with a 1/2 probability.
2. The genotypes may be written as Aa female × A male (where a is lethal). The possible offspring will consist of 1/4 AA females: 1/4 Aa females: 1/4 A males: 1/4 a males (a ratio of 2 normal females: 1 normal male: 1 lethal-carrying male [dead]). The sex ratio of the children will be 2 females: 1 male.
3. The MN blood-group system shows codominance, where MM = M blood group, NN = N blood group, and MN = the MN blood group. In the ABO system, IA and IB are both dominant to i and are codominant to each other. ABO genotypes are summarized as follows:
|
IAIA and IAi |
A |
blood type |
|
IBIB and IBi |
B |
blood type |
|
IAIB |
AB |
blood type |
|
ii |
O |
blood type |
Albinism is an autosomal recessive condition, and Huntington’s chorea is an autosomal dominant condition. The Huntington’s and albino conditions are of little or no value here, since albino carriers are relatively infrequent, and, in this case, even if the mother is a carrier (Aa), only half the children from the alleged parentage will be albino. The average age for the onset of Huntington’s chorea is 35 to 40, so that condition is useless as a marker in this instance. Even if the young child carries the Huntington’s allele from the alleged father, the child will not be expected to manifest it at such an early age. So we have to follow the blood types:
The man cannot be excluded, since half of this couple’s progeny would be expected to be MN, and if he is IAi (one of the possible genotypes that will be expressed as an A blood type), then the child could have inherited the i allele from him.
4.
(a) The mother is ii, the child is ii, and the man can be IAIA or IAi. In the latter case, he could be the father, so he cannot be excluded.
(b) His mother is either IAIA or IAi, and his father is IAIB. He could be IAi if his mother had transmitted the i allele. So this information will not exclude him from being the child’s father.
(c) If his mother’s parents are both AB, then his mother must be IAIA, and he, in turn, must be IAIA. This additional information will permit him to be excluded.
5. The segregation pattern described in the first part of this problem shows us that the albino trait is inherited as an autosomal recessive trait in both strains. The progeny of albino I crossed with albino strain II are all normal. The only way to account for this is for the albino phenotype to be produced by a different genetic locus in each strain. The cross is therefore
A test of this hypothesis would involve crossing the F1 progeny to produce an F2. The F2 would include
![9/16W−C−9/16pigmentd3/16W−cc3/16wwC−1/16wwcc]−7/16albino](genetic-analysis.files/image084.jpg)
6. In cross (1), the short-bristle trait is clearly not recessive to wild-type, but the mutant flies used here are not homozygous for the short-bristle trait, since the F1 segregates 1 mutant (68 flies) to 1 wild-type (76 flies). In cross (2), the short-bristle parents are again heterozygous for the mutant and wild-type alleles. Instead of the expected ratio of 3 short: 1 wild-type, we get 72 short: 39 wild-type, something close to a 2:1 ratio. With the clue of fewer progeny in cross (2), it appears that the dominant short-bristle allele is lethal when homozygous. The 2:1 ratio is therefore a modification of the 1:2:1 ratio, with one homozygous class dying. Cross (1) = Ss × ss, which gives 1/2 Ss (short) and 1/2 ss (wild). Cross (2) = Ss × Ss, which gives 1/4 Ss (die): 1/2 Ss (short): 1/4 ss (wild).
7.
(d) Since males have a single X chromosome, there are three A genotypes available to them: A1A2, and A3. The genotypes available at the B locus include all combinations of the four alleles taken two at a time, since it is an autosomal locus. There are 10 such combinations (B1B1, B1B2, B1B3, etc.). The formula to determine this is
where n is the number of alleles. The combinations can, of course, be written out and tallied. For the C locus there are six combinations:
Since these are all assorting independently of each other, we multiply the number of possible genotypes at each locus, and the answer is 3 · 10 · 6 = 180.
8.
(d) Dominant epistasis of the B (baldness) allele means that when B is present, one cannot tell whether the hair is curly (Cy –) or straight (cycy), since there is no hair at all. The gametes and progeny expected from this cross are
|
1/4 B Cy |
1/4 B cy |
1/4 b Cy |
1/4 b cy |
|
|
All b cy |
Bb Cycy |
Bb cycy |
bb Cycy |
bb cycy |
|
Trait |
Bald |
Bald |
Curly |
Straight |
9. Since we can rule out recombination among the four histocompatibility loci, we can consider each ABCD combination as a single segregating unit. The puzzle is, therefore, to arrange the alleles so that one haploid set from the mother and one from the father will account for the genotype of the offspring.
10.
(e) This cross consists of three separate monohybrid crosses:
Each gives a ratio of 3 dominant to 1 recessive, with all assorting independently. In the reaction
each step requires at least one dominant allele at the appropriate locus. Working back from the end point, black coat color requires the genotype A – T – B – (that is, at least one dominant allele at each locus), which has the probability 3/4 · 3/4 · 3/4 = 27/64. Tan coat color results from the genotype A – T – bb (that is, the last step is blocked), and occurs with the probability 3/4 · 3/4 · 1/4 = 9/64. An individual homozygous for the recessive allele at the A and/or Tloci (aaT –, A – tt, or aatt) will be white. These make up the remainder of the progeny:
11. This ratio is approximately 9:6:1, the ratio one would expect of an interaction in which both dominants are needed to produce round seeds (A – B –). Either dominant alone will produce oval seeds (A – bbor aaB –), and the double homozygous recessive will have curled seeds (aabb). The loci are assorting independently.
12. The walnut-comb rooster is heterozygous for both loci (AaBb). He is crossed to a true-breeding rose-comb strain, which must have the genotype AAbb. The genotype one is trying to obtain is aaBB, true-breeding pea comb. The F1 progeny include four different genotypes:
One step would then be to mate the original AaBb rooster to walnut-comb F1 hens. Half of such matings would be to AaBb hens. In their offspring, 3/16 will have pea combs, but you still need to take an additional step before you will have a pure-breeding line, since only 1/3 of the pea-comb chickens will be homozygotes (aaBB). The other 2/3 are aaBb. Here a testcross would be useful. If one mated each of the pea-comb chickens to a single-comb (aabb) sibling, one could eliminate from further consideration any chickens that produce single-comb chicks, because these would have had to be heterozygous at the Blocus. Those that only produce pea-comb chicks could be mated to each other, since they would be homozygous at both loci.
13.
(a) 9:3:4
(b) 9:7
(c) 12:3:1
14. Since these are dominant, independently assorting genes, the cross is
A common error is to forget that both parents carry both genes, so the full genotype must be included for each, even though the problem refers to just one gene per parent. In the offspring, half will be Cy and half will be cy+, where the plus sign designates the wild-type (normal) allele. The normal allele is recessive since the mutation is dominant. Independently of these phenotypes, half will be Pm and half will be pm+. There will therefore be four progeny phenotypes when both loci are considered: 1/4 Curly: 1/4 Curly Plum: 1/4 Plum: 1/4 Wild-type.
15. When a female mouse that is d + / + f is crossed to the same type of male, the progeny will be 1/4 d + / d +, which die in utero; 1/2 d + / + f, which are viable and fertile as females; and 1/4 + f / + f, which are sterile females but fertile males. The only viable and fertile females are the d + / + f individuals, which make up 2/3 of the females. There are two types of viable and fertile males: 2/3 are d + / + f / and 1/3 are + f / + f. When the d + / + f females are mated to the same type of brother, the progeny will again be 2/3 d + / + f and 1/3 + f / + f. When the d + / + f females are mated with their + f / + f brothers, the progeny consist of 1/2 d + / + f and 1/2 + f / + f. So the only viable and fertile mice will be d + / + f females and d + / + f and + f / + f males, generation after generation. The recessive lethal gene (as well as the f gene) is maintained in a balanced stock.
16. All three statements are correct.
17. For the sex-linked locus, the mice will be hemizygous and will have any of four different X chromosome genotypes. For the autosomes, however, the five alleles of the autosomal locus can occur in all combinations of pairs of alleles (A1A1, A1A2, A1A3, and so forth). There are 15 such combinations (remember that A2A3, for example, is the same as A3A2). The total number of genotypes is the product of these independent combinations, giving a total of 60 different male genotypes. Note, too, that the number of female genotypes would be much larger (150), since one would have various combinations of sex-linked alleles to take into consideration.
18. He should have been exonerated. The child could be IBIB or IBi. Since the mother had no IB allele, the child must be IBi, and the mother must be IAi. The father must have either the B or AB blood type to contribute the IB allele. Chaplin, with the O blood type (ii), could not have fathered the child.
19. He should look forward to the tests with confidence, since the child will almost certainly not have the rare A9 allele. If this man were the father, the child would have received his A9 allele, since the child’s blood type is not B or AB. Note that you could use the symbols IA, IB, IA9, and so forth, if you prefer.
20. The best explanation would be that the father carries a gene that masks the expression of his blood-type locus. This occurs in individuals homozygous for the rare Bombay locus (hh). These have an O blood type, because the Bombay locus is epistatic to the ABO blood-type antigen alleles. Alternatively, a germ-line mutation could have occurred in the man, changing i to IB.
21. This factor is sex-linked. In birds, the female is the heterogametic sex. The sex chromosomes in the male are denoted WW, while those in the female are WZ. Thus, in a cross between a pair of gray birds in which white female chicks were found, the male must have been heterozygous for the recessive white allele.
22. The floppy-ear phenotype is a recessive lethal, but morphologically dominant condition. Thus a cross between two floppy-eared individuals is always a cross between two heterozygotes. If we define F to represent the floppy-ear allele and f the straight-ear allele, Ff × Ff gives an F1 composed of 1/4 FF, which die; 1/2 Ff, which have floppy ears; and 1/4 ff, which have straight ears. Of the surviving rodents, 2/3 have floppy ears. The cross of floppy to straight ears is a testcross yielding a 1:1 ratio.
23. Since presence of the dominant A gene blocks pigment synthesis, segregation of the color gene is masked in all genotypes homozygous or heterozygous for A. This is an example of dominant epistasis. The dihybrid genotypes segregate in the normal 9:3:3:1 ratio, and the resulting phenotypic ratio comes from combining those with the same appearance. The answer is (b) 12:3:1.
24. 3 long tail to 5 short tail. The probability of inheriting at least one dominant allele is 3/4. If half of these show the dominant trait, 3/8 will have long tails. The rest, 5/8, will have short tails. Thus, the ratio of 3/8 to 5/8.
25. By discounting the alleles that must have come from the mother (A7B8C4D9), the father must have contributed A9B5C3D9. The only prospective father who could not have provided these alleles is choice (c). He does not carry allele C3.
26. The ovum is one of the mother’s cells, and prior to fertilization it already has its future mitotic spindle orientation (and ultimately the coiling of the adult snail) determined in response to her genotype. It is independent of the genotype of the fertilized egg. The only thing that one can say with certainty about a dextral snail is that its mother had at least one D allele and that the genotype of the male parent was immaterial. So any of the following crosses will give rise only to dextral snails: DD or Dd female × DD, Dd, or dd males. It therefore follows that the genotype of a phenotypically dextral snail can be DD, Dd, or dd.
27. Yes, but not in the next generation, as it does in peas and other organisms. The progeny will all be coiled dextrally, since the mother had a D allele. However, the progeny will be 1/4 DD: 1/2 Dd: 1/4 dd. When these, as females, produce progeny, 3/4 of them (DD and Dd) will produce only dextral snails, and 1/4 will be dd and produce only sinistral snails, for an overall 3:1 ratio.
28.
(a) If they behaved as chromosomes, daughter cells would get the same proportions of normal and defective chloroplasts, with little or no variegation. The variegation supports the idea of random separation of normal and defective chloroplasts, with some cells (and larger regions of the leaves) getting more (darker green) and some getting fewer (lighter green to white) of the normal chloroplasts.
29. Since one of the genes is cytoplasmic, it will have an effect on the phenotype only when transmitted by the female. Let us represent the dominant chromosomal color allele as A, with aa being white, and represent the cytoplasmic factor as C, with c being color. In the reciprocal crosses, different F1 phenotypes would be expected:
When the F1 individuals from the left-hand cross are self-fertilized, all would be white because of the cytoplasmic factor. When the F1 on the right are self-fertilized, 1/4 would be white because of the segregation of the chromosomal locus.
30. There is a maternal effect such that the offspring phenotypically express the genotype of the mother, with red dominant to gray.
CROSSWORD PUZZLE 6
Sex-Linkage and Interactions
Across
1. One gene with more than one phenotypic effect
6. Any chromosome other than the sex chromosomes
7. Proportion of phenotypic variation in a given trait that results from genetic segregation
8. Interference of one gene with the expression of another gene
9. Separation of homologous chromosomes during first meiotic division
14. Term for the sex that produces only one kind of gamete relative to the sex chromosomes
17. One gene with more than one phenotypic effect
18. Environmental mimic of a phenotype typically associated with a specific genotype
19. Form of linkage in which alleles of two genes occur together more often than predicted by their frequencies
Down
2. Mutation that causes death before reproductive age
3. Frequency with which a gene manifests itself in the phenotype of the heterozygote
4. Effect through which the genotype of mother is expressed in the offspring rather than the offspring expressing the offspring’s own genotype
5. Form of dominance in which the phenotype of the heterozygote falls outside the range of either homozyogote
10. Type of dominance in which the phenotype of the heterozygote is intermediate between those of the homozygotes
11. Irregular distribution of chromosome or chromatids due to mishaps in cell reproduction
12. Genes present only once in the genotype, such as the genes of the X chromosome in the human male
13. Genes that are found on the X chromosome and not on the Y chromosome
14. Term for the sex that produces two kinds of gametes
15. Degree of phenotypic expression in those showing a given phenotype
16. Type of dominance in which two alleles express products in the heterozygote independently