371B Autumn 2000 -- PS1 answer key

Genetics 371B, Autumn 2000

Answer key, Problem set 1

(a)

Non-vaulting is dominant, vaulting is recessive -- cross (iii) is a heterozygote x heterozygote cross giving a 3:1 ratio of dominant: recessive phenotypes. Vaulting progeny were obtained even though neither parent was a vaulter, so the vaulting allele must have been present in the two parents.


(b)


To give the vaulting response, the rabbit has to have vaulting ability and the ability to hear the whistle. Ability to vault is recessive; ability to hear the whistle is dominant. Therefore, Anabelle's genotype must have been ppHH (homozygous recessive for vaulting ability -- because vaulting is recessive to non-vaulting -- and homozygous dominant for hearing; remember that you have been told that this animal is a homozygote).

Mesquite, being a homozygous non-vaulter, could potentially have one of three genotypes:

PPHH

PPhh

pphh <-- but this possibility can be ruled out because this genotype would give vaulter progeny when crossed with Anabelle (i.e., ppHH x pphh --> ppHh, which would be vaulter) but we are told that the F₁ progeny are non-vaulters.

Therefore, Mesquite is either PPHH or PPhh.


(c)


Mesquite must be PPhh. If his genotype were PPHH, then the original cross with Anabelle would be:

ppHH (Anabelle) x PPHH (Mesquite) --> PpHH (F1 progeny)

and the F₁ x F₁ cross would be:

PpHH x PpHH --> 3:1 ratio of non-vaulters : vaulters in F₂

(Since both sides of the cross are homozygous dominant for H, this cross behaves just like a monohybrid cross.) However, we are told that vaulters make up 3/16 of the F₂, so this hypothesis does not fit the observation -- Mesquite must be PPhh. We can confirm this conclusion by writing out the whole experiment with Mesquite's genotype as PPhh:

ppHH (Anabelle) x PPhh (Mesquite) --> PpHh (F₁ progeny)

PpHh x PpHh --> 9/16 P_H_ : 3/16 P_hh : 3/16 ppH_ : 1/16 pphh (F₂ progeny)

This hypothesis predicts that 3/16 of F₂ (i.e., the ppH_ class) should be vaulters, and that is what is seen, confirming that Mesquite's genotype must be PPhh.

2.	(a)	David = II-2 Samantha = II-3 Matt = II-6

	(b)	If the trait is autosomal recessive: III-1 could be either AA or Aa, with the probability of each being 1/2. For her to have an affected child, she has to be heterozygous (probability = 1/2) and she has to transmit the recessive allele a (probability = 1/2). Therefore the probability of her child being affected = `(1/2)(1/2) = 1/4`.

	(c)	If the trait is autosomal dominant: David's grandchild would have to inherit the dominant allele A from III-2 (the only source of the dominant allele in that couple); the probability of that = `1/2`.
	(d)	The trait could not be X-linked recessive, because an affected woman (II-3) has an unaffected son (III-3) -- if it were an X-linked recessive trait, both copies of her X chromosome would carry the recessive allele, and her son would inherit one of them and therefore show the trait also.

(a)

The son's genotype is X^hX^hY. We don't know whether the extra X chromosome came from his mother or his father**. Assuming that one X came from the mother, she must be carrying at least one copy of the recessive hemophilia allele. So we can list the possible combinations of parental genotypes and then see which one(s) give a probability = 1/2 for an affected child, male or female.

**We are assuming that at least one X came from his mother, although it is possible to imagine complex scenarios where the mother contributed no X and the father contributed both X chromosomes and the Y -- that would require aberrant meioses in the mother and in the father.

Possible parental genotypes

The cross

Predicted outcome

X^hX^h and X^HY

X^H

X^h

X^HX^h	X^hY
X^HX^h	X^hY

X^h

Probability of daughter being affected = 0

Probability of son being affected = 1

... this is not the correct set of parental genotypes

X^hX^h and X^hY

X^h

X^hX^h	X^hY
X^hX^h	X^hY

X^h

Probability of affected child = 1

... this is not the correct set of parental genotypes

X^HX^h and X^HY

X^H

X^HX^H	X^HY
X^HX^h	X^hY

X^h

Probability of daughter being affected = 0

Probability of son being affected = 1/2

... this is not the correct set of parental genotypes

X^HX^h and X^hY

X^h

X^H

X^HX^h	X^HY
X^hX^h	X^hY

X^h

Probability of daughter being affected = 1/2

Probability of son being affected = 1/2

... this set of parental genotypes fits the data -- the correct parental genotypes are X^HX^h and X^hY

(There's also a much shorter way of saying the same thing --

For a son to have a 50% chance of being affected, the mom has to be heterozygous. If she were homozygous dominant, the son would have 0% chance of being affected, as he would inherit the dominant allele from her; if the mom were homozygous recessive, the son would have a 100% chance of being affected. As for the father's genotype, if the father had the dominant allele, his daughter would inherit it and she would have 0% chance of being affected. Therefore, for the daugher to have a chance of being affected, the father must be X^hY.

(b)

Knowing the parental genotypes (from part (a) above) and still assuming that there was just one aberrant meiosis, the possible combinations of egg and sperm that could have given rise to the affected X^gX^gY child are:

X^hX^h egg and Y sperm

or

X^h egg and X^hY sperm

(c)

Since the son has normal color vision, he must have received the allele for normal color vision (X^G) from the dad. Therefore, the combination of egg and sperm that gave rise to the XXY child was: X^gh egg and X^GhY sperm.

(a)

In females, long index fingers is dominant, so in a heterozygote x heterozygote "cross" the probability of a daughter showing the dominant phenotype is expected to be 3/4.

The question can also be interpreted to mean: What is the probability that a child will be female and with long index fingers. In that case, we also have to factor in the probability of a female child, so the overall probability would be (3/4)*(1/2) = 3/8. (Either answer is okay as long as an explanation is provided.)

(b)

The probability of a child being a daughter with long index fingers = 3/8 (see above)

The probability of a child being a son with long index fingers = (probability of a child being male)*(probability of a male having long index fingers) = (1/2)*(1/4) = 1/8.

Therefore, the overall probability of a child having long index fingers = 3/8 + 1/8 = 1/2.

This answer can be confirmed by writing out the cross as Punnett squares -- shown below with F^L and F^S as the alleles for long and short index fingers, respectively, and X and Y as the sex chromosomes to show sex determination.

The parents are F^LF^SXX and F^LF^SXY. The gametes are shown in the top row (from male) and the left column (from female). Children with long index fingers are shown in shaded boxes.

F^LX

F^SX

F^LY

F^SY

F^LX

F^LF^LXX	F^LF^SXX	F^LF^LXY	F^LF^SXY
F^LF^SXX	F^SF^SXX	F^LF^SXY	F^SF^SXY

F^SX

(a)

(i) 0.03 pg

(ii) 0.03 pg

(iii) 0.06 pg

(iv) 0.015 pg

(b)

Flask #1: Both the parental and the newly synthesized DNA contain radioactive adenine. Since adenine constitutes 30% of the bases in the DNA, the fraction of the bases consisting of radioactive adenine = 0.3.

Flask #2: Here, the parental DNA is non-radioactive; only the newly synthesized DNA contains radioactive adenine. Therefore, only half the adenines will be radioactive; the fraction radioactive will be 0.15. (The reasoning is outlined below with a snippet of DNA ten bp long.)

`TTTTTTCCCC` `AAAAAAGGGG`	S phase `---------->`	`TTTTTTCCCC` `AAAAAAGGGG`
		`TTTTTTCCCC` `AAAAAAGGGG`

...the radioactive adenines in this illustration are in red, underlined. You can count up the adenines to see that there are 6 radioactive adenines out of a total of 40 bases.