Which of the following is the genotype of the circled individual in generation ii?

In the smaller population --

In the larger population --

In the merged population --

Before the virus comes through, the frequency of the three genotypes is:

After the viral epidemic, the only cats left are homozygous dominant and heterozygotes:

Now the heterozygotes make up 2/3 of the surviving population, so the recessive allele makes up 1/3 of the total alleles in the population. Therefore, in the next generation the frequency of black cats will be (1/3)2 = 1/9.

Let the frequency of D = p, and the frequency of d = q, forward mutation rate = u, and back mutation rate = v

Then the change in p would include loss from forward mutation and gain from back mutation; likewise, change in q would include gain from forward mutation and loss from back mutation:

At equilibrium, change in p is exactly matched by change in q, so the change in p = 0 (as is the change in q)--

Therefore, at equilibrium, p = 0.00004/0.00016 = 0.25

Do a complementation test... the strain with the unknown mutation is crossed with the known torso mutant strain or the fs strain. If the unknown mutation (called mut in the diagram below) is in torso, the progeny of the cross will also have the same phenotype (tailless offspring) -- i.e., the unknown mutation fails to complement torso and therefore the unknown mutation is in torso. Alternatively, if the unknown mutation fails to complement fs, the mutation must be in fs.

If the female progeny from Cross #1 have tailless offspring, the unknown mutation must be in torso; if the female progeny from Cross #2 have tailless offspring, the unknown mutation must be in fs.

There's a catch--how do we deal with the problem that the progeny from the cross are going to be inviable? If conditional alleles (--see Answer 4 in Problem set 5) are available, there's an easy solution: do the cross and allow development of the resulting embryos at the permissive condition, to let the embryos develop, and then shift the young animals to the restrictive condition to look at the phenotype of their progeny.

If conditional alleles are not available, an alternative strategy is to cross heterozygotes and to ask if one quarter of the progeny show the phenotype:

The logic here is that if mut and torso are mutations in the same gene (for example), then Cross 1 is a monohybrid cross; one quarter of the progeny should be homozygous recessive, giving the mutant phenotype.

500 out of 20 million individuals are homozygous dd (where D = wildtype allele and d = disaccharide intolerance).

Let B = allele for beach-loving; b = bridge-loving

So in the next generation, the frequency of bridge-loving iguanas = q2 =0.08.

At this point, the alleles should be at Hardy-Weinberg frequencies, so the subsequent generation will not show a change.

This one can be solved only if we make the assumption that everyubody gets to mate, and that all crosses produce equal numbers of progeny.

While bridge-loving iguanas are homozygous and will give rise to bridge-loving iguanas only, beach-loving iguanas consist of homozygotes as well as heterozygotes. So we can set up a table as before, but this time only for frequencies of alleles B and b within the pool of beach-loving iguanas:

Beach-loving iguanas from these crosses = (0.48)+(0.192)+(0.096) = 0.768.

Bridge-loving iguanas = 1 - 0.768 = 0.232.

(If we didn't make the assumption stated at the beginning, we'd just be able to make the general conclusion that homozygosity is expected increase while heterozygosity would decrease.)

Among females, the distribution of genotype frequencies is the usual Hardy-Weinberg frequencies -- homozygous dominant = p2, heterozygotes = 2pq, homozygous recessive = q2 (where p = frequency of dominant allele, q = frequency of recessive allele).

But in males, there are no heterozygotes for X-linked traits -- males are hemizygous for such traits. Therefore, among males, p = frequency of the dominant phenotype; q = frequency of recessive phenotype.

Assuming that the allele frequencies are the same in men vs. women --

2(0.8)(0.2) x 0.22 = 0.0128

2(0.8)(0.2) x 2(0.8)(0.2) = 0.1024

Constitutively high beta-galactosidase.

Constitutively low.

Constitutively low (no transcription of lac operon)

Constitutively high

Constitutively low (same as iii)

A key point to note here is that various types of mutations can occur in any gene. A transcription activator can be mutated so that it is incapable of activation, or it be mutated so that it always activates, even when it's not supposed to. Likewise, a repressor could be mutated so that it never represses or so that it always represses.

The reg gene product must be a regulator of transcription of operon ABC. It could either be an activator or a repressor.

Possibility 1 -- reg is an activator of transcription. An "always on" mutant phenotype must be the result of a mutant activator that constitutively (and inappropriately) activates transcription. The mutant phenotype is expected to be dominant, because even if normal protein is being made and only activates transcription when appropriate, the mutant protein will always activate transcription. The "never on" mutant phenotype in this scenario must the result of mutant activator protein that fails to activate -- and this phenotype will be recessive.

Possibility 2 -- reg is a repressor of transcription. The "always on" phenotype must be a recessive mutation that allows transcription when appropriate; the "never on" phenotype must be from a dominant mutation that always represses transcription.

Looking at the actual results, we see that the the data support possibility 1 and not possibility 2: the "always on" phenotype is dominant and the "never on" phenotype is recessive. Therefore, reg must be an activator of ABC gene transcription.

The 9:7 F2 ratio indicates that we are dealing with a dihybrid ratio (the fractions go in sixteenths). The 9:7 ratio can be derived from the standard 9:3:3:1 ratio if we postulate the following --

• Progeny that have at least one dominant allele for each gene show the dominant phenotype (normal vision) -- giving 9/16 progeny with normal vision
• Progeny that lack a dominant allele for any gene show the recessive phenotype (blindness) -- giving 7/16 blind progeny

If we call the two genes D and E, the parents were ddEE and DDee; the F1 progeny are DdEe (and can therefore see); the F2 progeny are:

D_E_	D_ee	ddE_	ddee
9/16 normal vision	3/16 blind	3/16 blind	1/16 blind

If the F1 crickets were crossed to homozygous recessive crickets (i.e., DdEe x ddee) the progeny would be DdEe, Ddee, ddEe, and ddee in equal proportions -- i.e., 1/4 of the progeny will be able to see, and 3/4 will be blind.

As with any independently assorting pair of genes,we can look at the the ratios for the two genes independently. With respect to presence or absence of color (gene E), the progeny are 1/2 colored (black or brown) and half yellow. Therefore, with respect to gene E, the parents were Ee and ee. With respect to black vs. brown (gene B), the brown parent has to be bb and the other parent must be Bb (there must be at least one B allele to give black progeny; it cannot be BB, or there would be no brown progeny). [Another way to think about this is that of the non-yellow progeny, half are brown and half black; therefore, the parents must be bb and Bb giving 1:1 B_ and bb progeny.]

Thus, the parents must be bbEe (brown) and Bbee (yellow).

Remember that a mutation in the last gene in the pathway can only be rescued by the final product; a mutation in the next-to-last gene can be rescued by the last two compounds in the pathway, etc. Thus, the pathway is:

(Thiazole rescues thi-1, so the problem with thi-1 must be thiazole synthesis; likewise, the problem with thi-2 must be pyrimidine synthesis. thi-3 is rescued only be thiamine, so it must be the last step, the point of convergence of the two branches.)

• B is required for any color, so it must be required for conversion of white to color -- it is epistatic to A and to C.
• A appears to be required for conversion of a red intermediate to orange -- absence of A gives a red color instead of orange.
• C is not required for pigment production, but rather, appears to be needed to prevent pigment production in a portion of the flower, keeping that portion white.
• B- and C- have opposite phenotypes (no color vs. too much color) so their interaction must be negative.

Putting all this together, we can come up with at least two pathways that can each explain the data -- one in which C regulates B directly, and one in which C acts to convert some pigmented areas back to white.

Clearly, in either model, there must be some other gene that controls which portion of the flower are gong to express C (to make white) and which repress C (to allow color).

SIC and CLB mutations have opposite effects, and CLB is epistatic to SIC, so SIC must be an inhibitor of CLB. By the same logic, CLN is an inhibitor of SIC. An alternative pathway--

-- is also possible, but the cln-sic- double mutant phenotype argues against it. This double mutant shows too much DNA synthesis. If CLN is required for CLB function, as this second pathway implies, then the double mutant should show no DNA synthesis. So the data are most consistent with the pathway at the top.

The patches probably arose by mitotic recombination. Recombination between the two loci would give lone spots of the recessive phenotype of the more centromere-distal locus, while recombination between the centromere and the two loci would give twin spots. From this logic, we can conclude that the rd locus must be closer to the centromere than the b locus. (An alternative explanation for the lone spots is mitotic nondisjunction, but that wouldn't explain the twin spots.)

Because the twin spots and lone spots occurred in 6:5 ratio, the centromere-rd distance and the rd-b distance must be in the approximate ratio of 6:5:

Lone spots of rd phenotype could arise either from mitotic nondisjunction or by mitotic recombination with double crossover, one crossover between the centromere and rd and one crossover between rd and b.

The strain described in lecture had the dominant alleles for yellow and singed in trans. If the dominant alleles are in cis, a crossover between the centromere and the teo genes can give a single spot that has both recessive phenotypes:

The map is:

If a recombinant sector has phenotype a alone, then the crossover must have occurred between a and all the other genes; if a sector has phenotypes a and b, then b must be between a and all the other genes, etc.

*Corrected*

There must have been two crossovers -- one between B and D and one between F and G, as shown:

The fact that chromosomal material gets stretched between the spindle poles and breaks suggests that a dicentric chromosome must be formed -- indicating that the inversion must be a paracentric inversion:

Note that the gene order shown is arbitrary; the question does not give information on the correct gene order or even which parent had the inversion. All we can say is that there must have been an odd number of crossovers in the inversion loop.

What about the second cross? If the translocation had been to an autosome, the exceptional son would have had some wildtype daughters. The fact that the wildtype phenotype segregates exclusively with male progeny indicates that the translocation must have been to the Y chromosome.

NOTE: Only the gametes producing the aberrant (unexpected) offspring are shown for cross 1. Most of the gametes were normal, producing the expected offspring.

In the first litter, the mom was XrXr and the dad was XRY. The mom could transmit only Xr; the XR allele must have come from the dad along with Y, so nondisjunction must have occurred in the dad.

In the second litter, the mom was XRXr and the dad was XRY. The male calico kitten could have received XRXr from the mom (ND in mom) or XRY from the dad (ND in dad)-- it's not possible to distinguish between these possibilities.

If the two T-allele bearing homologs are called T1 and T2, and the two t-allele bearing homologs are t1 and t2, there are three possible sets of pairings, giving the gametes shown:

The gamete genotypes are TT, Tt and tt in 1:4:1 ratio, or 5:1 ratio of T_:tt. If mated to tttt plants (whose gametes will all be tt), the progeny are expected to be T___:tttt (i.e., tall and short) in 5:1 ratio.

The absence of the non-parental types and the semi-sterility suggest that the explanation may be a translocation. One possible configuration is shown:

The "adjacent" pattern of segregation would give Tt and Dd gametes, while the "alternate" pattern would give TD and td. The Tt and Dd gametes would be inviable, so the only viable progeny would have TD and td phenotypes -- the parental types.

Note that there is more than one configuration that would fit the results. For example, the t and d alleles need not be on the translocated segment.

The math

Case 1: probability of a chance match = (0.01)(0.02)(0.003)(0.01)(0.07)(0.04)(0.13)(0.08)(0.04)(0.05)32 = 1.1 x 10-14 -- i.e., the probability of getting this combination of alleles just by chance is about 1 in 100 trillion.

Case 2: probability of a chance match = (0.2)(0.4)(0.15)(0.35)(0.4)(0.3)(0.3)(0.6)(0.2)(0.3)32 = 1.7 x 10-4 -- i.e., a little over 1 in 6000.

(The factor of 32 comes in because there are two ways of getting each heterozygous combination, and 5 loci we're looking at = 25 = 32.) We'll see more of this factoring-in-2 business when we get to population genetics.)

Note: Your answer does not need to be this long-winded!

The strategy here is to look at the dominant trait and ask: does any one allele of the polymorphic trait preferentially segregate with the dominant trait? (And conversely, does an allele preferentially segregate with the recessive allele? This question is harder to address, because in this pedigree there are seven sources of the recessive allele -- two copies from I-1, one copy from I-2, and two each from II-1 and II-5 -- so it's harder to track the recessive allele. In contrast, there's only one source of the dominant trait -- I-2 -- so it's a lot easier to track.)

The source of the dominant trait in this pedigree is I-1. We know that he is heterozygous for the dominant trait (because he has an unaffected daughter, II-3). So if D = dominant and d = recessive, he is Dd. He has alleles 13 and 20 at PS1, and 21 and 27 at PS2. So we can ask if one of these four alleles segregates with the dominant trait (i.e., do people who show the dominant phenotype -- and therefore, inherited allele D -- also get one of those four alleles preferentially?)

Let's look at PS1 first. There are 11 affected (Dd) individuals not counting I-2. Of these, three have inherited allele 13 (II-2, III-3, III-5) and four have inherited allele 20 (II-4, III-10, III-12, III-14). The remainder have inherited neither 13 nor 20 (these individuals have inherited alleles from people marrying into the family). So allele D of I-2 has co-segregated about half the time with allele 13 of PS1 and about half the time with allele 20. Therefore, PS1 does not appear to be linked to D/d -- the two loci appear to be segregating independently of each other.

Now let's look at PS2. Here, the alleles in individual I-2 are 21 and 27. Of the eleven Dd progeny, ten also have allele 21; only one has allele 27. Additionally, of the six dd progeny, only one has allele 21; the remainder have other alleles. Therefore, it appears that in this pedigree, allele 21 is preferentially found with allele D; the two loci (D/d and PS2)are probably linked. In this particular case, I-1 must have allele D and allele 21 of PS2 on the same homologue (in cis, or in "attraction phase").

The fraction of control (sugar-water-treated) crosses that lack wildtype male progeny is the background rate of spontaneous mutagenesis. This rate is = 13/(6255 + 13) = 0.002/generation

We can now compare the rate of mutation in the other groups to see if any treatment causes an increase above this background rate. Thus --

Food color #1: 76/(4821 + 76) = 0.016/generation -- this rate is higher than the background rate, so Food color #1 is mutagenic.

Food color #2: 18/(9361 + 18) = 0.002/generation -- this rate is no higher than the background rate, so Food color #2 is not mutagenic.

Food color #3: 91/(5382 + 91) = 0.017/generation -- this rate is higher than the background rate, so Food color #3 is mutagenic.

Remember that normal diploid cells have two copies of the gene for Enzyme E (Gene E) and two copies of the gene for Enzyme Z (Gene Z). So, assuming that the amount of enzyme in the cell scales linearly with the gene copy number, each copy of Gene E contributes 30 units of enzyme E activity (so a diploid produces 60 units of Enzyme E), and each copy of Gene Z contributes 50 units of Enzyme Z activity. So a cell line that has a duplication of a gene should produce three copies' worth of enzyme. For Enzyme E, the duplication should result in cells that produce ~90 units, and for Enzyme Z, a duplication of the gene should result in ~150 units.

To find the location of Gene E, we look for cell lines that produce ~90 units of Enzyme E, and ask, what's common between these duplications. We see from the table that cell lines 1, 5, and 6 all produce ~90 units of Enzyme E (while the other cell lines produce the normal ~60 units). So these three cell lines must have duplications of Gene E. The band that is common to these three duplicaitons is band 2 -- that must be the location of Gene E.

For Enzyme Z, cell lines 2 through 6 all produce ~150 units instead of the standard 100 units. The band common to the duplications in these lines is band 5; Gene Z must be located there.

Although we could postulate multiple deletions in each progeny class, the most parsimonious explanation is that each progeny class has a single deletion that uncovers recessive alleles at multiple adjacent loci. So if two recessive traits are uncovered, the genes for those two traits must be next to each other on the chromosome. Using that logic --

• Strain #1 uncovers a and c, so gene a and gene c must be neighbors
• Likewise, a and b must be neighbors; a must be between b and c (the order so far is b-a-c)
• f is next to c, so the order is b-a-c-f
• de is next to f (but the order of d and e is not known yet)
• and from strain #6, e is next to f, so the completed gene order is b-a-c-f-e-d

With respect to the disease, the boy must be homozygous recessive (because achondroplasia is dominant). If A = achondroplasia and a = unaffected, the boy is aa.

With respect to the polymorphic locus, one allele has 12 repeats of CA, while the other has 7 repeats--so his genotype for the polymorphic site is 7,12. (Or 12,7.)

Therefore, his overall genotype for these two loci is aa 7,12.

Each enzyme by itself gives two fragments. Therefore, each enzyme must have a single cut site in the bacteriophage genome, such that each enzyme cuts the DNA into two.

Ava I must be cutting 12 kb from one end (i.e., 48 kb from the other end); Bam HI cuts 10 kb from one end, and Cla I cuts 18 kb from one end. The question is, which end are we measuring from -- we know that Ava I cuts 12 kb from one end, but Bam HI might be cutting 10 kb from the other end. To get that information we look at the double digests.

Let's look at Ava I + Bam HI. We know that Ava I by itself is going to generate a 12 kb fragment and a 48 kb fragment. We now see in this double digest that Bam HI leaves the 48 kb fragment untouched -- we're still seeing a 48 kb fragment. In contrast, the 12 kb fragment released by Ava I has been cut by Bam HI to a 10 kb fragment and a 2 kb fragment. Therefore, Bam HI site must be within the 12 kb Ava I fragment. The map we have so far is:

We can do a similar analysis for Cla I. In the Ava I + Cla I double digest, we see that Cla I does not cut within the 12 kb AvaI fragment (because if it had cut within the 12 kb fragment we'd be seeing smaller-than-12 kb fragments, which we don't). However, Cla I does cut within the 48 kb Ava I fragment to release a 30 kb fragment and an 18 kb fragment. We already know that Cla I cuts 18 kb from one end of the genomic DNA molecule -- therefore there is only one way to place the Cla I site on the map, as shown:

The map predicts that a Bam HI + Cla I double digest should give 10 kb, 32 kb, and 18 kb fragments -- which according to information we are given is true.

The primers are :

5'-TGCTCTGGAT-3' and 5'-TCCGAGAGCC-3', which correspond to the yellow, boxed segments (immediately flankng the greyed segment) below:

Note added 10/26/99: The way the question is worded, it is actually possible to amplify an even smaller fragment, by choosing primers within the grayed segment as shown below:

In this case, only the grayed segment would be amplified, giving a product length of 26 bp.

Someone who is homozygous normal will have two identical copies of the allele that has all four Xba I sites -- i.e., digestion of their DNA with Xba I and hybridization with the indicated probe should detect three fragments, of sizes 3 kb, 5 kb, and 7 kb.

In contrast, a carrier (a heterozygote with one normal and one disease allele) will have one allele that has 4 Xba I sites and one allele that lacks one or two of the middle Xba I sites (see table below). Their DNA, when cut and probed similarly, will also pick up the same three fragments (3 kb, 5 kb, 7 kb) because of the one normal allele. However, the other allele will give different products,which will be seen in addition to the normal digestion products (asterisks indicate absence of Xba I sites):

If the two loci are unlinked, gametes of the different possible genotypes are equally probable; the eight possible progeny genotypes are equally possible, as shown below:

We are not given phase information here -- i.e., we don't know whether allele configuration in the father is {D 8 & d 18}, or {D 18 & d 8}. (Does the mother's allele configuration matter in this question?) Different outcomes will be seen depending on the phase, as shown below. The gametes produced by the mother will be d, 7 and d, 15 in equal proportions, as in (b).

Phase (allele configuration) in father:
{D 8 & d 18}	{D 18 & d 8}
Gamete genotypes (frequencies): D, 8 (0.4)d, 18 (0.4)D, 18 (0.1)d, 8 (0.1)	Gamete genotypes (frequencies): D, 18 (0.4)d, 8 (0.4)D, 8 (0.1)d, 18 (0.1)

(from 1998)

The Lod score graph tells us that the pedigree data favor a map distance of 5 cM between Gene 1 and PS1; a map distance of 15 cM between Gene 1 and PS2; a map distance of 10 cM between Gene 1 and PS3, etc. A map that is consistent with these interpretations is:

Putting this information together -- A/a, D/d and F/f are in the same linkage group; B/b and E/e are in a separate linkage group. The linkage relationships can be depicted as:

The parental genotypes are TTFF x ttff to give TtFf. Therefore, the parental genotypes for gametes made by the F1 plants are TF and tf.

If the two loci are unlinked, we expect the four progeny phenotypes (TF, Tf, tF, and tf) in equal proportions. Because there are 1000 proteny total, we expect 250 of each phenotype if the loci are unlinked.

If the two loci are linked at at map distance of 44 cM, we expect 44% of the gametes to be recombinant -- i.e., 44% of the progeny should show the recombinant (non-parental) phenotype. As shown above, the parental types are TF and tf, so we expect 44% of the progeny to add up to Tf and tF, or 22% each. The parental types then should be 56% of the progeny = 28% each. So for 1000 progeny, we expect 280 each of TF and tf, and 220 each of Tf and tF.

Clearly, the observed progeny numbers don't match either scenario. So let's do a chi-square analysis on the two data sets, for the two sets of expectations, and see if we can find statistical evidence against either model.

Scenario 1 -- the loci are unlinked

The flaw is that the F1 progeny, although heterozygous for sneezy and jumpy, are homozygous for itchy. So while recombination between sneezy and jumpy can be detected, there's no way to detect recombination involving itchy.

... no change in genotype (ijs and i++ giving i++ and ijs)

Note: i = itchy, j = jumpy, s = scratchy; only one chromatid of each homolog is shown

He should be using a fully heterozygous (ijs/+++ in any cis/trans configuration) and a homozygous recessive (ijs/ijs) for his mapping cross.

Assuming that we are starting with the dominant alleles in cis in the heterozygote (i.e., +++/ijs), then parental, double-crossover, and single-crossover products can be predicted as follows:

Finding the correct gene order

H = Hairy, P = purple, T = Thorny; and lower case denotes the recessive phenotypes.

The parental non-crossover (NCO) allele combinations are Hpt and hPT (these being the most abundant progeny phenotypes), while the double-crossover (DCO) classes are HPT and hpt. To find the correct gene order, we start with the known NCO types, and see if a double crossover yields the known DCO types. If it doesn't, the order must be wrong; we try a different gene order (the critical information is the gene in the middle). Trial and error (trying each of the three genes in the middle) establishes that H must be the middle gene:

Products of single crossover (SCO) in P-H interval are phT and PHt. Products of SCO in H-T interval are pHT and Pht. Note the correction! (SCO classes were reversed. --10/19/99)

Now we can start calculating map distances:

P-H map distance = percent recombinants in this interval = (SCO in P-H) + DCO) as percent of total progeny = (150 + 132 + 18)/2500 = 300/2500 = 0.12, or 12 cM.   H-T map distance = percent recombinants in this interval = (101 + 81 + 18)/2500 = 200/2500 = 0.08, or 8 cM.   The completed map is:   P/p         H/h      T/t|------------|--------|     12 cM      8 cM  Coefficient of coincidence = (observed DCO/(Expected DCO)Observed DCO = 18Expected DCO = (0.12)(0.08)(2500) = 24Coefficient of coincidence = 18/24 = 0.75.

The important thing to remember is that in order to map the genes, we need to be able to detect recombination, and that in order to detect recombination, one of the parents has to be fully heterozygous. Here, the genes are on the X chromosome -- so that parent by default has to be the female (the male only has one X -- no recombination there).

There are a couple of ways of setting this up. One option is to make the female heterozygous, and to have recessive alleles on the male's X chromosome. Then the males and females would consist. To generate heterozygous females, we could cross homozygous dominant females (+++/+++) with recessive males (abc/Y); the females in the resulting progeny would be heterozygotes. When these females are crossed with abc/Y males, the progeny (males and females) would show the non-recombinant phenotypes (abc and +++) as well as the 6 recombinant types: a++ and +bc, ab+ and ++c, a+c and +b+.

A different option is to cross the heterozygous females with males showing the dominant phenotypes (+++/Y). Then the female progeny would all show the dominant phenotypes, and would be ignored; the male progeny would get the single X from the female, and show the same parental and recombinant phenotypes listed above.

For an example -- see Question 1998-2 in Questions from yesteryear.

You just have to realize that because the ratio of phenotypes is very different in females vs. males, the mode of inheritance must be sex-linked--specifically, these are X-linked genes. Other than that, the procedure is the same as above--you use just the male progeny to follow the recombination that occurred in the female parent. (If you are confused--DRAW THE CROSS! You know that the genes are on the X chromosome; you know the parental genotypes.)

The parental types (most abundant in the male progeny) are s+ sn+ fu and s sn fu+. The DCO products are s+ sn fu+ and s sn+ fu. Therefore, the s locus must be in the middle; the parental types can be re-written as sn+ s+ fu and sn s fu+.

A single crossover between sn and s would give sn+ s fu+ and sn s+ fu; a single crossover between s and fu would give sn+ s+ fu+ and sn s fu.

Now we can calculate the percent recombinant types for each interval:

Genotype of female parent = sn+ s+ fu / sn s fu+ (This notation-- a set of alleles, then a slash "/" then another set of alleles-- is standard notation to show that the first set of alleles is on one homolog and the second set of alleles following the slash is on the second homolog.)

Genotype of male parent = sn+ s+ fu+/Y

Map of the region:

We know that the final genotype has to be gaY. Therefore, at anaphase I, the Y chromosome has to segregate with the homologs carrying the recessive alleles, as diagrammed:

Note: In the interests of simplicity, crossing over has been ignored here. Also, the relative sizes of chromosomes and locations of genes is fictitious.

Again, affected children have unaffected parents, so the trait cannot be dominant.

It could be autosomal recessive. If so, however, we would have to assume that the disease is fairly common, because heterozygotes would have to enter the pedigree on five separate occasions (I-1/I-2, II-2, III-8, IV-2, and IV-7). Furthermore, only men have been affected in theis pedigree, arguing against a simple autosomal recessive pattern.

The fact that only men are affected in this pedigree suggests sex-linkage. But affected men have unaffected sons, so it is not Y-linked. It could be X-linked recessive -- but the trait appears to be passed on from father to son in one instance (IV-8 to V-6). So it could be X-linked recessive only if IV-7 is a carrier.

It could also be sex-limited (phenotype expressed in men), but as with autosomal recessive, we'd have to assume that the disease is common.

And as with #6, it could be sex-influenced, dominant in males but recessive in females.

Because it is possible to explain this pedigree either as autosomal recessive/sex-limited (if the disease is common) or as X-linked recessive (if the disease is relatively rare), we cannot deduce anything about how common or rare the disease is from just the pedigree. We could as a matter of parsimony say that the most probable mode of inheritance is X-linked recessive or sex-influenced, but leave open the possibility that it is autosomal recessive or sex-limited if the disease proves to be common.

The F1 x F1 cross would give squiggly eye females, squiggly eye males, normal females, and normal males in 1:1:1:1 ratio, as shown below:

where S = squiggly-eye and + is normal

The cross here is AABbDdee x AaBbddEe. We are asked to calculate what fraction of the progeny will have the phenotype ABde. Because these are independently assorting traits, we can calculate the fraction of progeny that will have phenotype A, then the fraction that will have phenotype B, etc., then multiply these fractions to get the fraction that has all the desired phenotypes. Thus --

Therefore, the fraction of progeny expected to be phenotype ABde is (1)(3/4)(1/2)(1/2) = 3/16.

Here, we have to find the fraction of progeny that will have the genotype AabbddEe. Using the same logic as above--

Therefore, the fraction of progeny expected to be genotype AabbddEe is (1/2)(1/4)(1/2)(1/2) = 1/32.

For a dihybrid cross, we expect to see a 9:3:3:1 ratio of phenotypes in the offspring--clearly not the case here. Because nothing is mentioned about males vs. females, we have to assume that this is not a sex-linked gene. To sort out the puzzle, therefore, we could begin by looking at each phenotype separately and seeing if that helps.

The observed progeny are creeper white, creeper yellow, normal white, and normal yellow chickens in 6:2:3:1 ratio. Let's look at creeper vs. normal separately from yellow vs. white. When we do that, we find that the ratio is 8 creeper : 4 normal , i.e., 2:1 creeper: normal. Hmmm. Where have we seen a heterozygote x heterozygote cross giving a 2:1 ratio before? That's right, if creeper is dominant over normal and creeper is lethal when homozygous, we'd get a 2:1 ratio of creeper : normal in the progeny.

How about white vs. yellow? Here, the ratio is 9 white : 3 yellow, a simple 3:1 ratio. Therefore, white must be dominant and yellow is recessive.

Putting it all together, the cross is CcWw x CcWw (where C= creeper, dominant, c= normal, recessive; W = white, dominant, w = yellow, recessive) and CC offspring die:

This one is a little tricky. A common mistake is to misinterpret the question to think that the first two children are boys -- when in fact, all we know is that at least two children (in any order) are boys. If the sex of the children is written in order of birth as B (for boy) or G (girl), the possible 3-children families with at least 2 boys are:

Only one of the four possible sets has all three children being boys -- so if you know that two of the children are boys, the probability that all three are boys is 1/4.

We use the binomial distribution to solve this one. Because this is a recessive disorder, and both parents are heterozygous, the probability of an affected child is 1/4. Therefore, let a = probability of unaffected child = 3/4, and b = probability of affected child = 1/4.

The equation then is

For a family with exactly 2 affected and 4 unaffected children, we use the term 15a4b2 (the exponents indicating the number of a=unaffected and b=affected children). Substituting the probabilities of unaffected and affected children, we get:

p(2 affected children) = 15a4b2 = 15(3/4)4(1/4)2 = 1215/4096 = 0.297.

But an easier way is to find the probability of less than two affected children, then subtract that value from 1 --

(Try it. The longer expression 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 will give the same result.)

This being a dihybrid cross, we expect a 9:3:3:1 ratio of tall purple : tall white : short purple: short white. For 3200 progeny, the expected numbers are:

What are the possibilities here?

Possibility # 1: the cross was homozygous purple x homozygous purple; there should be no white-flower progeny

Possibility #2: the cross was heterozygote x heterozygote; 1/4 of the progeny should make white flowers.

If the seed merchant picks just one seed at random and grows it up, and it makes white flowers -- she knows it must have been a heterzygote x heterozygote cross. However, if she picks one seed, and it makes a purple-flower plant -- can she then say that it must have been a homozygote x homozygote cross? No, because even in a heterozygote x heterozygote cross, 3/4 of the progeny will be purple, so she has a 3/4 chance of picking a purple progeny even if white progeny are present--i.e., she has a 1/4 (=0.25) probability of missing a white progeny.

Suppose she picks two seeds? Then the probability that both will be purple (if it was indeed a dihybrid cross) = (3/4)(3/4) = 9/16; the probability that she has missed a white progeny plant has dropped to 7/16 = 0.4375. So that's the question -- how many seeds should she sample if she wants the probability of accidentally missing a white-flower seed to drop below 2%. In other words, she needs to sample n seeds such that

(3/4)n = 0.02

or, n(log(0.75)) = log(0.02)

n = 13.6

So if she samples 14 seeds and they all grow up to make purple flowers, there is < 2% probability that white flower seeds are present but missed just due to chance.

To know the probability that IV-1 will be affected, we need to know the genotypes of the parents, III-4 and III-5. In turn, we have to know the genotypes of their parents, and so on.

Because I-1 and I-2 are unaffected but have an affected daughter (II-1), they must both be carriers -- genotype Dd (where D = dominant, unaffected; d = recessive, affected). II-3 is D_, with a 1/3 chance of being DD and 2/3 chance of being Dd. II-5 and II-6 are both Dd (because they are unaffected but have an affected son, III-9).

III-4 is unaffected; the only way she can have an affected child is she is heterozygous Dd. What is the probability of that? She (III-4) has a father who is DD and a mother who has a 2/3 chance of being Dd. Therefore, the probability that III-4 is Dd is (1/2)(2/3) = 1/3. Likewise, III-5 has to be heterozygous Dd for their child to be affected. The probability that III-5 is heterozygous Dd is 2/3 (he could be DD or Dd, with a 2/3 chance of being Dd -- just as with II-3).

Therefore, the chance that they will have an affected child = (1/4)(1/3)(2/3) = 1/18.

The only way a tall plant can yield short progeny after selfing (i.e., mating with itself) is if the tall plant is heterozygous. Therefore, what the question is asking is: what fraction of the tall plants are heterozygous? (Refer to the crosses shown in answer 2 for these questions.)

Note: You are not looking for tall plants that give only short progeny upon selfing (is that even possible?)--you are looking for tall plants that will give any short progeny on selfing.