Question

(d) Bob is homozygous for the lof mutation in ADH and he gets red-face. What are his possible geno- types at the both genes? aa LL aa ll 3 op fions pn (e) Following up on this, he mates with Susan who is wild-type (does not get red face). What could be Susan's genotypes? AaLL AALL (4 aphins mat AALL If Susan is true-breading for the Bob's and Susan's genotypes gane and 50% of their babies get red face, what must be So, she is eihur AALR Must be Aus aaLL x AA LL 08 wid hye aal l AALl F frpe Bob has h ke aa ll.. or I CAnt be ts dE hen all oa ics woald be wild tyre

Hi can someone please explain how you’d get 100%, 75% and 50% wild type for part F? A detailed explanation would be much aappreciated. thank you!

Answer 1

Given, Allele "A" codes for ADH and "L" codes for ALDH. Mutant forms are "a" and "l" respectively.

Homozygous mutation in either gene leads to "red-face" phenotype. Thus, aa__ and __ll individuals will have the red-face phenotype.

This means, we can have following inference:

Wild type genotype	Red face genotypes possible
AALL, AALl, AaLL, AaLl.	aaLL, aaLl, AAll, Aall, aall

Bob has red face phenotype due to a mutation in ADH gene, thus his genotype can be aaLL , aaLl or aall.

Susan is true breeding for ADH , thus she has AA alleles at first locus. At second locus she can be either LL or Ll. (She cannot be AAll as then she would also have red-face phenotype).

It is given that upon mating of Bob and Susan, 50% offspring got red face. This is possible only if the offsprings received recessive alleles from both parents.

Since Susan is AA for first locus according to question, so she should be Ll at second locus. Thus, genotype of Susan is AALl.

Genotype of Bob can be either aaLL or aaLl or aall.

We have three possibilities to check. Let us use punett square to check the offsping outcomes in each case. as follows:

CASE 1:

Susan, AALl x aaLL, Bob

Gametes are AL, Al from Susan and aL from BOB. Cross can be shown as: AL Al aL AaLL (wild type) AaLl (wild type)

OUTCOME: All offspring will have wild type face. 100% wild type.

Thus, Bob cannot be aaLL.

CASE 2:

Susan, AALl x aaLl (Bob)

Gametes are AL and Al from Susan and aL, al from Bob. The cross can be shown as : AL Al aL AaLL (wild type) AaLl wild type) al AaLl (wild type) Aall (red face) In this case, 3/4 are wild types and 1/4 are red face. 3/4 x 100 = 75% wild type 1/4 x 100 = 25 % red face

OUTCOME: 75% are wild type and 25% are red faced.

Thus, Bob cannot be aaLl.

CASE 3:

Susan AALl x aall , Bob

Gametes are AL and Al from Susan and "al" from Bob.

AL Al al AaLl (wild type) Aall (red face) Thus, half (50%) will be wild type and 50% will have red face phenotype.

Thus, Bob is "aall" and Susan is "AALl", so that their offsprings are 50% wild type and 50% red face.