Question

1. a man who is homozygous for tongue rolling is crossed with a woman who is heterozygous for tongue rolling. (remember tongue rolling is dominant to not tongue rolling.) although both of their tongues, what is the percent chance that they have a child who cant?
2. in guinea pigs, short hair is dominant to long hair. cross two heterozygous guinea pigs. determine the phenotypic and genotypic ratios. then determine how many guinea pigs, if 16 are born, will have long hair.
3. a woman who has no finger hair wants to marry and have children with a man who does have mid - finger hair. they do not know his genotype though. (remember finger hair is dominant to no finger hair.) determine the likelihood that they have a child with no finger hair like their mom. include all possible options.
4. in chimpanzees, straight fingers are dominant to bent fingers. if two chimps are crossed and have offspring that are 50% bent fingers and 50% straight fingers, what must have been the genotypes of the parents?
5. in humans, dimples are dominant to no dimples and long eyelashes are dominant to short. a man with no dimples and short eyelashes mates with a woman who is heterozygous for both traits. determine the phenotypic ratio of their offspring. also determine what percentage of their offspring will have the same genotype as their mother.

Explanation:

Step1: Define alleles

Let \(T\) represent the allele for tongue - rolling (dominant) and \(t\) represent the allele for non - tongue - rolling (recessive). The man is homozygous for tongue - rolling, so his genotype is \(TT\). The woman is heterozygous, so her genotype is \(Tt\).

Step2: Set up Punnett square

The possible gametes from the man are all \(T\), and the possible gametes from the woman are \(T\) and \(t\). The Punnett square has one row with \(T\) (from man) and two columns with \(T\) and \(t\) (from woman).

Step3: Analyze genotypes of offspring

The genotypes of the offspring are \(TT\) and \(Tt\). There are no \(tt\) genotypes. So the probability of having a non - tongue - rolling child (\(tt\)) is \(0\%\).

for question 15:

Step1: Define alleles

Let \(S\) represent the allele for short hair (dominant) and \(s\) represent the allele for long hair (recessive). The two heterozygous guinea pigs have the genotype \(Ss\).

Step2: Set up Punnett square

The possible gametes from each parent are \(S\) and \(s\). The Punnett square has 4 boxes: \(SS\), \(Ss\), \(Ss\), \(ss\).

Step3: Calculate ratios

The genotypic ratio is \(SS:Ss:ss = 1:2:1\). The phenotypic ratio is short - hair (\(SS + Ss\)): long - hair (\(ss\)) \(= 3:1\).

Step4: Calculate number of long - haired guinea pigs

If 16 guinea pigs are born, and the proportion of long - haired (\(ss\)) is \(\frac{1}{4}\), then the number of long - haired guinea pigs is \(\frac{1}{4}\times16=4\).

for question 16:

Step1: Define alleles

Let \(F\) represent the allele for finger hair (dominant) and \(f\) represent the allele for no finger hair (recessive). The woman has no finger hair, so her genotype is \(ff\). The man has finger hair, his genotype could be \(FF\) or \(Ff\).

Step2: Case 1: Man is \(FF\)

If the man is \(FF\), all of the offspring will have the genotype \(Ff\) and will have finger hair. The probability of having a child with no finger hair is \(0\).

Step3: Case 2: Man is \(Ff\)

Set up a Punnett square with the woman's gametes (\(f\)) and the man's gametes (\(F\) and \(f\)). The genotypes of the offspring are \(Ff\) and \(ff\) with a ratio of \(1:1\). The probability of having a child with no finger hair is \(50\%\).

for question 17:

Step1: Define alleles

Let \(B\) represent the allele for straight fingers (dominant) and \(b\) represent the allele for bent fingers (recessive).

Step2: Analyze offspring phenotypes

Since the offspring are \(50\%\) bent fingers (\(bb\)) and \(50\%\) straight fingers, one parent must be heterozygous (\(Bb\)) and the other must be homozygous recessive (\(bb\)). A cross between \(Bb\times bb\) gives \(Bb:bb = 1:1\) which is consistent with the offspring phenotypes.

Answer:

0%