When crossing two heterozygous plants What is the probability of producing a homozygous recessive plant?

(16 pts total)

In plant species #1, the three genes are linked and located on chromosome 1 the gene order is B-T-F, and each interval is 1 cM.

Blue flower color is dominant, so in a cross between two heterozygotes, 3/4 of the progeny will produce blue flowers. It doesn't matter that B is linked to T and F; when we look at B/b alone, it's still a standard monohybrid cross.

Half the progeny in a heterozygote x heterozygote monohybrid cross are expected to be heterozygous -- so 1/2 the progeny will be heterozygous for fragrance.

Because BtF and bTf are the most abundant gamete genotypes, we know that that must be the allele combinations present on the homologs in the parent:

The most likely phenotype is blue flowers, tall, fragrant (those being the dominant phenotypes).

Since these genes are assorting independently, we can multiply the probabilities of the individual outcomes to get the overall probability. For each trait, we expect 3/4 of the progeny to show the dominant phenotype. Therefore, the probability of the progeny showing all three dominant phenotypes = (3/4)(3/4)(3/4) = 27/64.

(i) Can you tell from these data whether loci B and T are linked? If they are linked, what is the map distance between them? BRIEFLY explain your reasoning.

Since only one allele type (B) of the flower color gene is present, we cannot tell if there is linkage between B and T -- the same gamete types will be produced whether the genes are linked or not:

If B and T are not linked: genotype BB Tt will produce BT and Bt gametes

If B and T are tightly linked, with no crossovers: BT/Bt genotype will produce BT and Bt gametes

If B and T are linked, and there is crossing over: BT/Bt genotype will product BT and Bt gametes

(ii) Can you tell from these data whether loci T and F are linked? If they are linked, what is the map distance between them? BRIEFLY explain your reasoning.

(Ignoring the flower color gene here...)

Tf and tF gametes greatly outnumber TF and tf. Therefore, the genes must be linked; the parental genotype must be Tf/tF, while the recombinant gametes are TF and tf.

The recombinants account for 10% of the gametes, so the map distance between T/t and F/f is 10 cM.

(24 pts total)

In humans, glucose-6-phosphate dehydrogenase (G6PD) deficiency and red-green colorblindness are both X-linked recessive traits; the genes are linked at a distance of 6 cM. The following questions are all based on the pedigree shown below.

Allele designations: g = G6PD-deficient, G = normal G6PD

d = red-green colorblind, D = unaffected (normal vision)

The genotypes should be accurate with respect to the alleles present and should also show which alleles are present on which homolog (the way the genotypes are shown for I-1 and I-2).

III-1 and III-4 are both recombinants. They both get their X chromosomes from their mom (II-1). Neither of her X chromosomes carries both dominant or both recessive alleles, so the only way III-1 and III-4 could get both dominant or both recessive alleles is if their X chromosome is recombinant.

III-2 inherited one X chromosome (Gd) from her dad. Her other X chromosome, which she got from II-1, could either be Gd (if there was no recombination) or gd (if it is recombinant). Since the G-D map distance is 6 cM, 6% of the gametes are expected to be recombinant. So the probability that she is heterozygous = probability that she received a recombinant X chromosome = 6%.

[Note: Of the gametes produced by II-1, the parental types combined (Gd and gD) add up to 94%, and recombinant ones combined (GD and gd) add up to 6%. However, we know that she did not transmit gD or GD, because her daughter is colorblind. So of the remaining possibilities, Gd accounts for 94% and gd accounts for 6%.]

You are given two strains of the worm C. elegans. One strain is homozygous for a recessive allele that makes the worm short, called dumpy (d). The other strain is homozygous for a recessive allele that makes the worms uncoordinated (unc). The two genes are unlinked.

What crosses can you perform to obtain a strain that is homozygous recessive for both genes? Show the genotypes of all parents/offspring and the ratios of offspring phenotypes from each cross you propose. (12 pts)

Again, we'd cross the two starter strains to get F1 progeny that are heterozygous for both loci. The F1 progeny genotype would be:

Only the recombinant d unc gametes made by this strain would be capable of generating the desired F2 phenotype (fully homozygous recessive). The total fraction of recombinant gamtes = 1/10 (or 10%), so the fraction that will be this particular genotype of recombinant gamete = 1/20.

Therefore the fraction homozygous recessive offspring will be (1/20)(1/20) = 1/400.

The Australian echidna comes in two varieties, one variety with soft spines and one with hard spines. Spine hardness/softness is controlled by a single gene.

Since hard x hard always gives only hard progeny (i.e., hard breeds true), hard must be homozygous (e.g., hh).

If that's the case, then soft must be heterozygous (Hh) -- otherwise they wouldn't give hard and soft progeny when crossed to hard.

We know from part (a) that soft is heterozygous. Therefore, a soft x soft cross must be:

giving a 1:2:1 ration of HH:Hh:hh. We already know that Hh and hh must be soft and hard, respectively. Since those are all the progeny we see -- se never see a different phenotype associated with HH, and we know that HH cannot be soft (otherwise we'd see true-breeding soft echidnas, which we never see) -- we must presume that the HH genotype is lethal

(Note that for this series of questions, we are not making any statement about dominant or recessive with respect to hard/soft spines.)

Since both parents are homozygous hh, we can ignore that gene and just look at the snout length gene. Here, we are seeing that a cross between two short-snout echidnas gives a 1:2:1 ratio of progeny with respect to snout length. Therefore, short snout length cannot be homozygous (since it does not breed true)--it must be heterozygous. If L and l are the snout length alleles:

LL = long, Ll = short, ll = no snout... a case of incomplete dominance.