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# heterozygous punnett square

Question:If two heterozygous brown-eyed individuals with dimpled chins were have to have children, punnett square will be 4 x 4 squares. Assume both independent assortment and segregation are occuring?

Answers:If you are wanting to know how to set the Punnett square up: I'm making a lot of assumptions here.... #1 - Brown eyes is dominant to whatever color.. #2 - Dimples are dominant to non dimples. Brown = B Not brown = b Dimples = D No dimples = d Across the 4 boxes on top use: BD, Bd, bD, bd Use the same for the boxes going down. I'm not doing the Punnett Square but I think your final ratio should be 9:3:3:1 Fill it out and count them. Good luck

Question:1. Show the Punnett Square cross results between two Dalmatian dogs that are heterozygous for spots? (For Dalmatian dogs, the spotted condition is dominant to non-spotted) 2. In humans, brown eyes are dominant over blue eyes. A woman with blue eyes marries a man with brown eyes. This man's father had blue eyes. (Please answer all 3 questions) a) What would be the genotype of the man that married the blue-eyed woman? b) What would be the genotype of the blue-eyed woman? c) What will be the ratio of children with blue eyes from this marriage? (please show the Punnett Square cross)

Answers:1. S=dominant allele for spots s=recessive allele for no spots Ss x Ss = SS, Ss, Ss, ss 25% offspring would be homozygous dominant with spots (SS) 50% would be heterozygous with spots (Ss) 25% would be homozygous recessive with no spots (ss) 2. B=dominant allele for brown eyes b= recessive allele for blue eyes. Woman is bb since she has blue eyes. If the mans father had blue eyes, the father would be bb too. So since the many has brown eyes, he has to be Bb. a. Bb b. bb. c. Bb x bb= Bb, bb, Bb, bb. 50% are Bb, brown eyes 50% are bb, blue eyes.

Question:QUESTION: how do I know the genotype for the male and female if the problem doesn't state whether or not they are homozygous or heterozygous. Q: In humans, there is a type of blindness due to a dominant gene. Normal vision is the result of a recessive gene. Migraine headaches are due to a dominant gene, and normal (no headaches) is recessive. A male who is blind and does not suffer from headaches marries a woman who has normal vision and suffers from migraines. Could they produce a child with normal vision who does not suffer from heacahes? If yes, can the probability of such child be determined? If B=blindness, b=normal vision, N=migraines and n=no headaches What is the male and females "codes"???? The male would either be: BBnn or Bbnn and the female could be bbNn or bbNN Do I do Punnett Squares for each possibility? Then how do I figure out the probability... or do you assume they are either homozygous or heterozygous............. please help!

Answers:Yes, they could. Whether the genotypes indicate male or female (which they don't) is irrelevant, since the alleles are not sex-linked. You don't need to do Punnett squares for each; they are both heterozygous for the genes in question. If B=blind and M=migraines (I use "M" because migraines is a dominant allele), the male would be Bbmm and the female would be bbMm. Look at the genes separately: Could BB x bb produce normal vision, bb? No, but Bb x bb could (with 50% probability). Could MM x mm produce no migraines, mm? No, but Mm x mm could (with 50% probability) Proof: ______Bm_______bm___ bM__BBMm____BbMm__ bm__Bbmm____bbmm___ child with normal vision, no migraines; 25% which makes sense, since 50% x 50% = 25%