Answer:
a: 1/4
b: 1/2
c: 1/4
d: 0
Explanation:
Since the wife had a color-blind father (X^cY), she should be a carrier for the disease. Her genotype would be "X^cX". Fathers do not transmit the X-linked disorders to their sons, so the genotype of the husband is "XY".
a. The probability that the couple's first child will be a normal son: 1/4
b. The probability that the couple's first child will be a normal daughter: 1/2
c. The probability that the couple's first child will be a color-blind son: 1/4
d. The probability that the couple's first child will be a color-blind daughter: Zero
The answer ==is insertion and error occurs during the process motosed
Because it can cause serious infections in the lungs, blood and our brains...It may also cause urinary tract and infections in wounds.
Answer:
c. 1:2:1
The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.
Explanation:
If flower color were determined by a gene showing incomplete dominance, the possible genotypes and phenotypes are as follows:
- RR- red
- ww - white
- Rw - pink
If pink sweet peas are self-pollinated, then a cross between two heterozygous individuals is done (Rw x Rw).
<u>From this cross the expected ratios are:</u>
- 1/4 RR (red)
- 2/4 Rw (pink)
- 1/4 ww (white)
So the null hypothesis is that the observed results exhibit a 1:2:1 ratio.
<h3><u>Chi square test</u></h3>
<u>The observed frequencies were:</u>
Total 150
<u>The expected frequencies for our null hypothesis are:</u>
- 1/4 x 150 = 37.5 Red
- 2/4 x 150 = 75 Pink
- 1/4 x 150 = 37.5 white
The degrees of freedom (DF) are calculated as number of phenotypes - 1; in this case DF = 3-1 = 2.
If we look at the Chi square table, for 2 DF and a probability of p0.05, the critical value is 5.991
Our X^2 value of 0.5067 is less than the critical value, so we do not reject the null hypothesis. The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.
