BIOL 1406 Lecture Quiz Answers

1. A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15 chicks are gray, 6 are black and 8 are white. What is the simplest explanation for the inheritance of these colors in chickens? What offspring would you expect from the mating of a gray rooster and a black hen?

This appears to be an example of incomplete dominance for feather color. If we assume two alleles for feather color, black (B) and white (b), then the heterozygotes would be gray. If true, then the two original parents must have heterozyote geneotypes (Bb) since they were gray. A Punnett Square for a cross of Bb x Bb is shown below:

This means that the ratio of phenotypes should be:

1 - black (BB)
2 - gray (Bb)
1 - white (bb)

This is close to what was actually observed in the question (6 black, 15 gray, 8 white).

A cross between a gray (Bb) rooster and a black (BB) hen should produce the following:

2 - black
2 - gray
0 -white

2. In some plants, a true-breeding, red-flowered strain gives all pink flowers when crossed with a white-flowered strain: (RR (red) x rr (white) -------> Rr (pink). if flower position (axial or terminal) is inherited as it is in peas (meaning that axial is dominant), what will be the ratios of genotypes and phenotypes of the generation resulting from the following cross: axial-red (true breeding) x terminal-white. What will be the ratios of the F₂ generation?

Axial-red must have a genotype of AA RR (remember that it is true breeding so it must be homozygous); the terminal-white must have a genotype of aa rr. This is a dyhybrid cross. We must first determine what type of gametes these plants can form.

AA RR: Because of the law of independent assortment, these two genes will move independently in meiosis. The only type of gametes it can produce is A R (it only has these alleles).

aa rr: Similar reasoning concludes that this one can only produce a r gametes, thus all offspring must be heterozygous for both traits, Aa Rr with a pink, axial phenotype.

The F₂ generation will be the result of a cross between two heterozygotes (Aa Rr x Aa Rr). The gametes produced by each of these would be:

A R
A r
a R
a r

This would require a 4 x 4 Punnett Square

The genotypic ratios would be:

AA RR - 1
AA Rr - 2
AA rr - 1
Aa RR - 2
Aa Rr - 4
Aa rr - 2
aa RR - 1
aa Rr - 2
aa rr - 1

And the phenotypic ratios would be:

red-axial - 3
red-terminal - 1
pink-axial - 6
pink-terminal - 2
white-axial - 3
white-terminal - 1

3. A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino was crossed with a second black one, 7 blacks and 5 albinos were obtained. What is the best explanation for this genetic situation? Write genotypes for the parents, gametes and offspring.

Assume that there are two alleles for coat color, black (B) which is dominant and albino (b) which is recessive.

If the black guinea pig in the first cross is homozygous (BB thus can produce only B gametes) then crossing it with an albino (bb thus can produce only b gametes) would produce offspring that all were heterozygous (Bb) and all black.

If the black guinea pig in the second cross is heterozygous (Bb thus can produce equal numbers of B and b gametes) then crossing it with an albino (bb thus can produce only b gametes) would produce equal numbers of Bb and bb, thus a 1 to 1 ratio of black and white which is what was observed.

4. A man with group A blood marries a woman with group B blood. Their child has group O blood. What are the genotypes of these individuals? What other genotypes and in what frequencies, would you expect offspring from this marriage?

Since the first child is type O, he/she/it must have a genotype of OO. This means that the mother and the father must have a copy of the O allele, thus the father must be AO and the mother must be BO.

The father can make two different gametes, A and O; the mother can also make two different gametes, B and O. Below is a Punnett Square of this cross:

One-fourth of the offspring should have AB blood, One-fourth should have BO, One-fourth should have AO and One-fourth should have OO blood.

Note: these are not the official designations for these alleles. If you want the correct ones, substitute I^A for A, I^B for B, I^AB for AB and i for O.

5. The color pattern in a species of duck is determined by one gene with three alleles. Alleles H and I are codominant and allele i is recessive to both. How many phenotypes are possible in a flock of ducks that contains all the possible combinations of these three alleles?

First, let's determine all the possible combinations of alleles:

HI
Hi
HH
II
Ii
ii

We do not really know what the exact phenotypes are (what the colors are) but we do know whether the color that alleles represent are expressed.

HI would express the H and the I color (codominance) - 1 phenotype (imagine the Count from Sesame Street counting)

Hi and HH would express only the H color only- 2 phenotypes

II and Ii would express the I color only - 3 phenotypes

ii would express the i color only - 4 phenotypes.

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1. Suppose that two Dd Ee Ff Gg Hh individuals are mated. What is the predicted frequency of dd EE Ff gg Hh offspring from such a mating (do not use a Punnett square).

One should use the laws of probability to answer this question. The geneotype of the offspring is

dd EE Ff gg Hh

Therefore, both parents must donate a d to the offspring. Each parent can produce two types of gametes with respect to the d gene, D and d. The probability that a gamete will have a d gene is 1/2 for each parent. Therfore the probability that the offspring will have d is the product of 1/2 and 1/2 = 1/4. The probabilites that the offspring will have the indicated genotype for each gene is shown below"

dd 1/2 x 1/2 = 1/4
EE 1/2 x 1/2 = 1/4
Ff (1/2 x 1/2) + (1/2 x 1/2) = 2/4 = 1/2 (There are two ways in which the offspring can be Ff, either the mother donates the F or the father donates the F, thus the offspring can be either Ff or fF, so the probability of either is equal to the sum of the probabilities).
gg 1/2 x 1/2 = 1/4
Hh (1/2 x 1/2) + (1/2 x 1/2) = 2/4 = 1/2

The probability of all five of these occurring in a single offspring is equal the product of all these probabilites.

1/4 x 1/4 x 1/2 x 1/4 x 1/2 = 1/256

2. A man and a wife both have normal color vision but a daughter has green color blindness, a sex-linked recessive trait. The man sues his wife for divorce on grounds of infidelity. Can genetics provide evidence supporting his case?

If the parents have normal vision, then the genotype of the husband must be X⁺Y and the wife must be either X⁺X⁺ or X⁺X^c. The daughter is color blind therefore must be X^cX^c. This means the daughter must receive a colorblind gene from each parent. The wife must be X⁺X^c (the only way she can be the mother as to donate an X^c and still have normal vision). However, the husband does not have a X^c allele to donate (if he did, then he would be colorblind) and so cannot father a colorblind daughter. Now, who is the father?

3. In cats, short hair is dominant over long hair; the gene involved is autosomal (not on the sex chromosome). An allele, B¹, of another gene, which is sex-linked, produces yellow coat color; the allele B² produces black coat color; and the heterozygous combination of B¹/B² produces tortoiseshell (calico) coat color. If a long-haired black male is mated with a tortoiseshell female homozygous for short hair, what kind of kittens will be produced in the F₁ generation. If the F₁ cats are allowed to interbreed freely, what are the chances of obtaining a long-haired yellow male?

A long haired black male must have a genotype of B²Y ss and the female must be B¹B² SS (do not forget that the B gene is on the X-chromosome - it is sex-linked). The father can produce the following gametes: B²s or Ys; the mother can produce B¹S or B²S. This cross is shown in the Punnett Square below:

Thus there would be four different types of offsping:

Females would be either tortoiseshell/shorthair or black/shorthair (probability of 1/2 for each)

Males would be either yellow/shorthair or black shorthair (probability of 1/2 for each).

If these cats then mated, is there a way to get a long-haired yellow male? The genotype of such a cat would be B¹Y ss. Either of the F₁ males could father this offspring with 1/4 of all sperm would have both the Y and the s. Only 1/2 of the females could mother such a monstrosity, the one with a genotype of B²B¹Ss. To produce the long-haired yellow male, the gametes from the mother must contain B¹s, there is a probability of 1/4 of such aa gamete. The total probability would be:

1 x 1/4 x 1/2 x 1/4 = 1/32 (remember we must include the probability that the correct female mates, a 1/2 chance; either male would do, so there is probability of 1 that the correct male mates).

4. In Drosophila melanogaster there is a dominant allele for gray body color and a dominant allele of another gene for normal wings. The recessive alleles of these two genes result in black body color and vestigial wings. Flies homozygous for gray body and normal wings were crossed with flies that had black bodies and vestigial wings. The F₁ progeny were then test crossed, with the following results:

Gray, normal 236
Black, vestigial 253
Gray, vestigial 50
Black, normal 61

Are these linked? Defend your answer. If so, how far apart are they (based on these results only).

A cross between a homozygous gray body and normal wings (b⁺b⁺ vg⁺vg⁺) with a fly with black body, vestigial wings (bb vgvg) would produce only gray, normal flies that were all heterozygous for each trait (b⁺b vg⁺vg). This is because both parental flies were homozygous for both traits and therefore they each could produce only one type of gamete each (what are the genotypes of the gametes produced by each of the parental flies?).

As you need to remember, a test cross is a specific cross where a fly (in this case the F₁ progeny which are b⁺b vg⁺vg) is crossed with a homozygous recessive fly (bb vgvg). The question is asking if these genes are linked so we must prepare apunnett square for the case of them being linked as well as being not-linked. These are shown below (Note: if the genes are linked, then the wild type genes must be on the same chromosome and the mutant genes must be on the same chromosome in this case as the parental flies were homozygous. This is not always true.)

If linked, then the cross should produce the following:

b⁺b vg⁺vg x bb vgvg

The possible gametes of the fly (the one that is heterozyous for each trait) is either b⁺ vg⁺ or b vg (remember, this is if linked), while the possible gametes for the homozygous recessive fly (bb vgvg) is only b vg.

Therefore if the genes are linked, we would expect to see equal numbers of gray/normal and black/vestigial flies.

If the genes were not linked, then there would be 4 different types of gametes from the F₁flies: b⁺ vg^+, b vg, b⁺ vg and b vg⁺. The only possible gametes for the homozygous recessive fly (bb vgvg) is still only b vg.

Thus, equal numbers of 4 different phenotypes would be expected if the genes were no linked.

When the flies were actually mated, they produced gray normal and black vestigial flies in equal numbers (as if they were linked), but a small number of flies were gray vetigial and black normal - not enough to say that the genes were not linked. The data suggests that the genes are linked and that the most likely explanation for the gray vetigial and black normal is that crossing over occurred.

The percentage of cross overs is used as the map distance. This experiment counted a total of 600 flies of which a total of 50 + 61 showed crassing over. The percentage of crossing over is equal to 111/600 x 100 = 18.5% so these two genes are 18.5 map units apart.

5.The recombination frequency between linked genes A and B is 40%; between B and C, 20%; between C and D, 10%; between C and A, 20%; between D and B, 10%. What is the sequence of the genes on the chromosome?

This chromosome contains these 4 genes. We know that A and B are 40 units apart and C is 20 units from B. There are two possible ways this can be done:

or

If the first possiblility is true then the C gene would have to be 60 units from the A gene while they would only be 20 units away in the second case. The question states that A and C were 20 units apart, therefore the second diagram must be correct.

D is 10 units from C, the two possible arrangements consistant with this data are shown below:

or

If the first one is correct, then B and D would be 10 units apart; if the second one is right, then B and D wold be 20 units apart. The question tells us that they are 10 units apart, therefore, the first one must be the correct order of these genes on the chromosome.

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