Name, please BSC2011-03 Assignment 4 (Genetics Problem Set I!) Spring 2018 P as,
ID: 202083 • Letter: N
Question
Name, please BSC2011-03 Assignment 4 (Genetics Problem Set I!) Spring 2018 P as, show all work "parrniai genoppn, Punneer squers, maliplication, alliion, "thought pnenr", and 1. What is the total umher of diffierent kinds of genotypes that are possible among offispring if the following two parents are crossed, assuming that the genes for each trait are found en different chromosomes? FIRrHhGG FFRrHhgs De a Punmett square for each ipene on SCRAP PAPER, but "NO" branch diagram is necessary because 1 am NOT asking for what all the possible genotypesare- just the pumker of different genotypes [see sample elicker Q SHOW MATH ANSWER 2. A brother and sister have Type O and Type AB blood, respectively (a) What are the parents gemotypes (show siblings genotypes and SHOW your lagic to determine what the parents genotypes must be: USE NO PUNNETT SQUARESI One parent, Oher parent:-- Brother: Sister (b) What are the chances that the NEXT child of the same parents will be a Type A or Type B daughter? (NOW show a Punnett square AND your MATH) ANSWER: 3.li addition tothe gene coding for blood type A. BorO·there is another gene that codes forthe blood factor that determines whether a person hasa"orbloed phenotype for this trait. Two alleles are possible, Rhe and Rh-, and the Rh+ allele is dominant to the Rh-allele tinfo given here, you will need tow ifor your etan GIVEN: A type AB+ woman and a type B-man have a type A- son (a) Using LOGIC (no Punnett squaresn what is each family member's blood genotype both s? Mom: 1st Son: (b) What is the probability that their next child will be a type B-son? (show 2 Punnett squares & all math ANSWER:-Explanation / Answer
1.
Parental cross: FfRrHhGG X FFRrHhgg
Number of possible gametes from parent 1 = 2 X 2 X 2 X 1= 8
Number of possible gametes from parent 2 = 1 X 2 X 2 X 1= 4
Total number of genotypes possible = 8 X 4 = 32
2.
Brother = OO
Sister = AB
Parental cross: AO X BO
Progeny: AB AO BO OO
The probability for type A or B blood group = 1/2
The probability for their child to be female = 1/2
Total probability = 1/2 X 1/2 = 1/4
3.
Parental cross: AB +/- X BO -/-
Gemetes: (A+)(A-)(B+)(B-) X (B-)(O-)
Progeny: (AB +/-) (AO +/-) (AB -/-)(AO -/-)(BB +/-)(BO +/-)(BB -/-)(BO -/-)
Their son = (AO -/-)
The probability for a B-ve group = 1/8
The probability for the enxt child to be a son = 1/2
Total probability = 1/8 X 1/2 = 1/16 = 0.0625 = 6.25%
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