6. Father's genotype = AaMother's genotype = Aa%3D%3DAAA Aa2 NormalGenotype RatioPhenotype Ratio eA lbino Coin GeneticsSuppose a father and mother are both heterozygous (Aa), as in question #6above. When you completed that Punnett Square, you were predicting thepercentage or fraction of possibility that this couple's children would bealbino (aa), or produce melanin (Aa)(AA).From this pairing there are three possible genotype outcomes AA, Aa, andaa. Your Punnett square from #6 tells you the percentage chance that theseparents will have children with those genotypes, write the percentages here5).7. AA =% Aa =% aa =%This mating scenario can easily be simulated using two-sided coins, wheretails represent the recessive allele that controls pigment production (a), andheads represent the dominant allele (A). Each coin then represents aheterozygous parent (Aa).When tossing the coin, there is a 50% chance for either tails up or heads up.This represents the 50-50 chance that an egg or sperm produced by theparent will include an a allele or an A allele. To simulate a mating betweentwo heterozygous (Aa) parents, the students will toss two coins and theresult of this pair of coin tosses will indicate the pair of alleles contributed byan egg and a sperm to the baby that results from that mating.For this exercise, you will need two coins. Each one will represent a parent inthis mating pair. Every time you toss the two coins, a baby is made with thegenotype represented by the donation of each parent (side of coin-heads ortails).The first coin toss will indicate the pair of alleles in the first child producedby a mating of two heterozygous (Aa) parents. Make a check in the box inthe table below under the appropriate genotype. Then, make three morepairs of coin tosses to determine the genetic makeup for the second, thirdand fourth children in this family. Record how many of these 4 children hadeach of the 3 possible combinations (AA, Aa, or aa) in the first row of thetable labeled "first family of 4 children" in the table below.Now make 4 more pairs of coin tosses to indicate the alleles in a secondfamily of 4 children. Record the resulting baby genotypes in the second rowin the table below. Repeat this and record the results in the third and fourthrows of the table below. When you are finished, you should have made atotal of 16 babies (16 separate coin tosses representing 4 families of 4). Fillout the table below with your results.

Name: 6. Father's genotype = AaMother's genotype = Aa%3D%3DAAA Aa2 NormalGe
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Description: 6. Father's genotype = AaMother's genotype = Aa%3D%3DAAA Aa2 NormalGenotype RatioPhenotype Ratio eA lbino Coin GeneticsSuppose a father and mother are both heterozygous (Aa), as in question #6above. When you completed that Punnett Square, you were p

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From this pairing there are three possible genotype outcomes AA, Aa, and aa. Your Punnett square from #6 tells you the percentage chance that these parents will have children with those genotypes, write the percentages here 5). 7. AA = % Aa = % aa = %3D Thic mating cconario

From this pairing there are three possible genotype outcomes AA, Aa, and aa. Your Punnett square from #6 tells you the percentage chance that these parents will have children with those genotypes, write the percentages here 5). 7. AA = % Aa = % aa = %3D Thic mating cconario

Biology: The Unity and Diversity of Life (MindTap Course List)

15th Edition

ISBN:9781337408332

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter13: Observing Patterns In Inherited Traits

Section: Chapter Questions

Problem 5SQ: The offspring of the cross AA aa are ________. a. all AA b. all aa c. all Aa d. half are AA and...

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