Genetics 371B, Autumn '99 -- PS1 answer key

Genetics 371B, Autumn 1999

Answer key to Problem set 1

1. (a) The parental brown eyed fly was homozygous. If the brown eye phenotype were dominant, all progeny of this fly should show the dominant phenotype (because they'd all inherit a brown-eye allele from him). However, that is clearly not the case -- when the F1 females were mated to him, the F2 progeny consisted of brown- and red-eyed flies in equal proportions. Therefore, brown eyes must be recessive to normal, red eyes.

(b) If R = normal, red eyes and r = brown eyes, the genotype of the parental male = rr . Since the F2 progeny showed the dominant and recessive phenotypes in 1:1 ratio, the F1 females must have been heterozygous Rr red-eyed flies. Therefore, the parental females must have been homozygous dominant RR flies.

2. (i) This plant must be homozygous WW , because it breeds true. WW x WW --> WW

(ii)

The cross White x Speckled gives white and speckled progeny in equal proportions. This results tells us that the speckled plant must be heterozygous SW, and that speckled is dominant to white (otherwise, the progeny would be all white, as the white plant is homozygous).
WW x SW --> WW & SW in 1:1 ratio.

(iii) Selfing Patchy plants gives Patchy and White in 2:1 ratio. Because the White phenotype segregates out of the Patchy plant, the plant must be heterozygous PW and therefore, Patchy must be dominant to White (else the plant wouldn't be Patchy). However, we'd expect a 3:1 Patchy:White ratio from selfing the plant, but we only see a 2:1 ratio, suggesting that Patchy is lethal when homozygous. PW x PW --> {PP, dead}, PW & WW in 2:1 ratio.

(iv) By exactly the same logic, Speckled must be lethal when homozygous. SW x SW --> {SS, dead}, SW & WW in 2:1 ratio.

(v) Here, the cross should be SW x PW --> SP, SW, PW, WW in 1:1:1:1 proportions. Of these, we have already encountered the last three; we don't know what to expect for SP, and in fact we don't see SP -- we only see the other three phenotypes in 1:1:1 ratio. Therefore, the SP genotype appears to be lethal also.

Putting these inferences together, we can say that Speckled and Patchy are each dominant to White, and that at least one copy of the White allele is required for viability.

3. Dmitri's mother and father must be AO and BO heterozygotes, respectively, because they have blood type A and B but have a daughter (Dmitri's sister) who is blood type O and therefore must be homoyzgous OO. Dmitri and his wife have a homozygous OO son, so they must both carry the O allele. Dmitri's wife must be AO, because she is blood type A and carries the O allele. Their second son is type B, so Dmitri's other allele must be B -- i.e., he is a BO heterozygote. The second son inherited a B allele from him; the maternal allele must be O (because his blood type is B).

4. The life cycle of the virus is: free virus --> infected cell --> free virus again. In free virus, the ratio of (A+G) to (C+T) is not 1.0 therefore, this DNA is not double-stranded, but must be single-stranded. After infection, the ratio becomes 1.0, indicating that the single-stranded genome must have become double-stranded. [In fact, the new base composition exactly matches what we should see if a complementary strand is added to the single-stranded viral DNA (i.e., if a T is added for every A in the single strand, and a G for every C, etc., to make a double-stranded DNA.) Thus, the A base content after synthesis of the complementary strand should be (20+12)/2 = 16%; the C content should be (28+40)/2 = 34%, etc., which exactly matches the observed base composition.] Free virus that is released from the infected cells must be single-stranded again; the sequence of that single strand must be identical to the original single strand, (and not the complement) otherwise the base composition would change.

1.	(a)	The parental brown eyed fly was homozygous. If the brown eye phenotype were dominant, all progeny of this fly should show the dominant phenotype (because they'd all inherit a brown-eye allele from him). However, that is clearly not the case -- when the F1 females were mated to him, the F2 progeny consisted of brown- and red-eyed flies in equal proportions. Therefore, brown eyes must be recessive to normal, red eyes.
	(b)	If R = normal, red eyes and r = brown eyes, the genotype of the parental male = rr . Since the F2 progeny showed the dominant and recessive phenotypes in 1:1 ratio, the F1 females must have been heterozygous Rr red-eyed flies. Therefore, the parental females must have been homozygous dominant RR flies.

2.	(i)	This plant must be homozygous WW , because it breeds true. WW x WW --> WW
	(ii)	The cross White x Speckled gives white and speckled progeny in equal proportions. This results tells us that the speckled plant must be heterozygous SW, and that speckled is dominant to white (otherwise, the progeny would be all white, as the white plant is homozygous). WW x SW --> WW & SW in 1:1 ratio.
	(iii)	Selfing Patchy plants gives Patchy and White in 2:1 ratio. Because the White phenotype segregates out of the Patchy plant, the plant must be heterozygous PW and therefore, Patchy must be dominant to White (else the plant wouldn't be Patchy). However, we'd expect a 3:1 Patchy:White ratio from selfing the plant, but we only see a 2:1 ratio, suggesting that Patchy is lethal when homozygous. PW x PW --> {PP, dead}, PW & WW in 2:1 ratio.
	(iv)	By exactly the same logic, Speckled must be lethal when homozygous. SW x SW --> {SS, dead}, SW & WW in 2:1 ratio.
	(v)	Here, the cross should be SW x PW --> SP, SW, PW, WW in 1:1:1:1 proportions. Of these, we have already encountered the last three; we don't know what to expect for SP, and in fact we don't see SP -- we only see the other three phenotypes in 1:1:1 ratio. Therefore, the SP genotype appears to be lethal also.
	Putting these inferences together, we can say that Speckled and Patchy are each dominant to White, and that at least one copy of the White allele is required for viability.